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Math: Number Theory
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Updated on: 04 Feb 2019, 02:52
NUMBER THEORYThis post is a part of [ GMAT MATH BOOK] created by: Bunueledited by: bb, walker Get The Official GMAT Club's App  GMAT TOOLKIT 2. The only app you need to get 700+ score! [ iOS App] [ Android App]  Definition
Number Theory is concerned with the properties of numbers in general, and in particular integers. As this is a huge issue we decided to divide it into smaller topics. Below is the list of Number Theory topics.
GMAT Number Types
GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers. INTEGERS
Definition
Integers are defined as: all negative natural numbers \(\{...,4, 3, 2, 1\}\), zero \(\{0\}\), and positive natural numbers \(\{1, 2, 3, 4, ...\}\).
Note that integers do not include decimals or fractions  just whole numbers.
Even and Odd Numbers
An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. An even number is an integer of the form \(n=2k\), where \(k\) is an integer.
An odd number is an integer that is not evenly divisible by 2. An odd number is an integer of the form \(n=2k+1\), where \(k\) is an integer.
Zero is an even number.
Addition / Subtraction: even +/ even = even; even +/ odd = odd; odd +/ odd = even.
Multiplication: even * even = even; even * odd = even; odd * odd = odd.
Division of two integers can result into an even/odd integer or a fraction.
ZERO: 1. 0 is an integer. 2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer except 0 itself. IRRATIONAL NUMBERS
Fractions (also known as rational numbers) can be written as terminating (ending) or repeating decimals (such as 0.5, 0.76, or 0.333333....). On the other hand, all those numbers that can be written as nonterminating, nonrepeating decimals are nonrational, so they are called the "irrationals". Examples would be \(\sqrt{2}\) ("the square root of two") or the number pi (\(\pi=\)~3.14159..., from geometry). The rationals and the irrationals are two totally separate number types: there is no overlap.
Putting these two major classifications, the rationals and the irrationals, together in one set gives you the "real" numbers. POSITIVE AND NEGATIVE NUMBERS
A positive number is a real number that is greater than zero. A negative number is a real number that is smaller than zero.
Zero is not positive, nor negative.
Multiplication: positive * positive = positive positive * negative = negative negative * negative = positive
Division: positive / positive = positive positive / negative = negative negative / negative = positive Prime Numbers
A Prime number is a natural number with exactly two distinct natural number divisors: 1 and itself. Otherwise a number is called a composite number. Therefore, 1 is not a prime, since it only has one divisor, namely 1. A number \(n > 1\) is prime if it cannot be written as a product of two factors \(a\) and \(b\), both of which are greater than 1: n = ab.
• The first twentysix prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101
• Note: only positive numbers can be primes.
• There are infinitely many prime numbers.
• The only even prime number is 2, since any larger even number is divisible by 2. Also 2 is the smallest prime.
• All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form \(6n1\) or \(6n+1\), because all other numbers are divisible by 2 or 3.
• Any nonzero natural number \(n\) can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors.
• Prime factorization: every positive integer greater than 1 can be written as a product of one or more prime integers in a way which is unique. For instance integer \(n\) with three unique prime factors \(a\), \(b\), and \(c\) can be expressed as \(n=a^p*b^q*c^r\), where \(p\), \(q\), and \(r\) are powers of \(a\), \(b\), and \(c\), respectively and are \(\geq1\). Example: \(4200=2^3*3*5^2*7\).
• Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq{\sqrt{n}}\). Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime. Note that, it is only necessary to try dividing by prime numbers up to \(\sqrt{n}\), since if n has any divisors at all (besides 1 and n), then it must have a prime divisor.
• If \(n\) is a positive integer greater than 1, then there is always a prime number \(p\) with\(n < p < 2n\). Factors
A divisor of an integer \(n\), also called a factor of \(n\), is an integer which evenly divides \(n\) without leaving a remainder. In general, it is said \(m\) is a factor of \(n\), for nonzero integers \(m\) and \(n\), if there exists an integer \(k\) such that \(n = km\).
• 1 (and 1) are divisors of every integer.
• Every integer is a divisor of itself.
• Every integer is a divisor of 0, except, by convention, 0 itself.
• Numbers divisible by 2 are called even and numbers not divisible by 2 are called odd.
• A positive divisor of n which is different from n is called a proper divisor.
• An integer n > 1 whose only proper divisor is 1 is called a prime number. Equivalently, one would say that a prime number is one which has exactly two factors: 1 and itself.
• Any positive divisor of n is a product of prime divisors of n raised to some power.
• If a number equals the sum of its proper divisors, it is said to be a perfect number. Example: The proper divisors of 6 are 1, 2, and 3: 1+2+3=6, hence 6 is a perfect number.
There are some elementary rules: • If \(a\) is a factor of \(b\) and \(a\) is a factor of \(c\), then \(a\) is a factor of \((b + c)\). In fact, \(a\) is a factor of \((mb + nc)\) for all integers \(m\) and \(n\).
• If \(a\) is a factor of \(b\) and \(b\) is a factor of \(c\), then \(a\) is a factor of \(c\).
• If \(a\) is a factor of \(b\) and \(b\) is a factor of \(a\), then \(a = b\) or \(a=b\).
• If \(a\) is a factor of \(bc\), and \(gcd(a,b)=1\), then a is a factor of \(c\).
• If \(p\) is a prime number and \(p\) is a factor of \(ab\) then \(p\) is a factor of \(a\) or \(p\) is a factor of \(b\).
Finding the Number of Factors of an Integer
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
Greatest Common Factor (Divisior)  GCF (GCD)
The greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more nonzero integers, is the largest positive integer that divides the numbers without a remainder.
To find the GCF, you will need to do primefactorization. Then, multiply the common factors (pick the lowest power of the common factors).
• Every common divisor of a and b is a divisor of gcd(a, b). • a*b=gcd(a, b)*lcm(a, b)
Lowest Common Multiple  LCM
The lowest common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple both of a and of b. Since it is a multiple, it can be divided by a and b without a remainder. If either a or b is 0, so that there is no such positive integer, then lcm(a, b) is defined to be zero.
To find the LCM, you will need to do primefactorization. Then multiply all the factors (pick the highest power of the common factors).
Perfect Square
A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is an perfect square. There are some tips about the perfect square: • The number of distinct factors of a perfect square is ALWAYS ODD. • The sum of distinct factors of a perfect square is ALWAYS ODD. • A perfect square ALWAYS has an ODD number of Oddfactors, and EVEN number of Evenfactors. • Perfect square always has even number of powers of prime factors.
Divisibility Rules
2  If the last digit is even, the number is divisible by 2.
3  If the sum of the digits is divisible by 3, the number is also.
4  If the last two digits form a number divisible by 4, the number is also.
5  If the last digit is a 5 or a 0, the number is divisible by 5.
6  If the number is divisible by both 3 and 2, it is also divisible by 6.
7  Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.
8  If the last three digits of a number are divisible by 8, then so is the whole number.
9  If the sum of the digits is divisible by 9, so is the number.
10  If the number ends in 0, it is divisible by 10.
11  If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21(9+8+6+9)=11, 11 is divisible by 11, hence 9,488,699 is divisible by 11.
12  If the number is divisible by both 3 and 4, it is also divisible by 12.
25  Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25. Factorials
Factorial of a positive integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to n. For instance \(5!=1*2*3*4*5\).
• Note: 0!=1. • Note: factorial of negative numbers is undefined.
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a nonnegative integer \(n\), can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that \(5^k\leq{n}\).
It's easier if you look at an example:
How many zeros are in the end (after which no other digits follow) of \(32!\)? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
Hence, there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
Finding the number of powers of a prime number \(p\), in the \(n!\).
The formula is: \(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}\) ... till \(p^x\leq{n}\)
What is the power of 2 in 25!? \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\)
Finding the power of nonprime in n!:
How many powers of 900 are in 50!
Make the prime factorization of the number: \(900=2^2*3^2*5^2\), then find the powers of these prime numbers in the n!.
Find the power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}\)\(=25+12+6+3+1=47\)
= \(2^{47}\)
Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)
=\(3^{22}\)
Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\)
=\(5^{12}\)
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!. Consecutive Integers
Consecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and 2, 1, 0, 1, are consecutive integers.
• Sum of \(n\) consecutive integers equals the mean multiplied by the number of terms, \(n\). Given consecutive integers \(\{3, 2, 1, 0, 1,2\}\), \(mean=\frac{3+2}{2}=\frac{1}{2}\), (mean equals to the average of the first and last terms), so the sum equals to \(\frac{1}{2}*6=3\).
• If n is odd, the sum of consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.
• If n is even, the sum of consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.
• The product of \(n\) consecutive integers is always divisible by \(n!\). Given \(n=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.
Evenly Spaced Set
Evenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. The set of integers \(\{9,13,17,21\}\) is an example of evenly spaced set. Set of consecutive integers is also an example of evenly spaced set.
• If the first term is \(a_1\) and the common difference of successive members is \(d\), then the \(n_{th}\) term of the sequence is given by: \(a_ n=a_1+d(n1)\)
• In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula \(mean=median=\frac{a_1+a_n}{2}\), where \(a_1\) is the first term and \(a_n\) is the last term. Given the set \(\{7,11,15,19\}\), \(mean=median=\frac{7+19}{2}=13\).
• The sum of the elements in any evenly spaced set is given by: \(Sum=\frac{a_1+a_n}{2}*n\), the mean multiplied by the number of terms. OR, \(Sum=\frac{2a_1+d(n1)}{2}*n\)
• Special cases: Sum of n first positive integers: \(1+2+...+n=\frac{1+n}{2}*n\)
Sum of n first positive odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n1\). Given \(n=5\) first odd positive integers, then their sum equals to \(1+3+5+7+9=5^2=25\).
Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n\)\(=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\).
• If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is 13, so the sum is 13*5 =65. FRACTIONS
Definition
Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.
Fraction can be expressed in two forms fractional representation \((\frac{m}{n})\) and decimal representation \((a.bcd)\).
Fractional representation
Fractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a parttowhole relationship in terms of a numerator (the part) and a denominator (the whole).
• The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, \(\frac{9}{7}\), 9 is the numerator and 7 is denominator.
• Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. \(\frac{1}{3}\) is a proper fraction.
• Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. \(\frac{5}{2}\) is improper fraction.
• An integer combined with a proper fraction is called mixed number. \(4\frac{3}{5}\) is a mixed number. This can also be written as an improper fraction: \(\frac{23}{5}\)
Converting Improper Fractions
• Converting Improper Fractions to Mixed Fractions: 1. Divide the numerator by the denominator 2. Write down the whole number answer 3. Then write down any remainder above the denominator Example #1: Convert \(\frac{11}{4}\) to a mixed fraction. Solution: Divide \(\frac{11}{4} = 2\) with a remainder of \(3\). Write down the \(2\) and then write down the remainder \(3\) above the denominator \(4\), like this: \(2\frac{3}{4}\)
• Converting Mixed Fractions to Improper Fractions: 1. Multiply the whole number part by the fraction's denominator 2. Add that to the numerator 3. Then write the result on top of the denominator Example #2: Convert \(3\frac{2}{5}\) to an improper fraction. Solution: Multiply the whole number by the denominator: \(3*5=15\). Add the numerator to that: \(15 + 2 = 17\). Then write that down above the denominator, like this: \(\frac{17}{5}\)
Reciprocal
Reciprocal for a number \(x\), denoted by \(\frac{1}{x}\) or \(x^{1}\), is a number which when multiplied by \(x\) yields \(1\). The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\). To get the reciprocal of a number, divide 1 by the number. For example reciprocal of \(3\) is \(\frac{1}{3}\), reciprocal of \(\frac{5}{6}\) is \(\frac{6}{5}\).
Operation on Fractions
• Adding/Subtracting fractions:
To add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator.
To add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions
• Multiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators.
• Dividing fractions: Change the divisor into its reciprocal and then multiply.
Example #1: \(\frac{3}{7}+\frac{2}{3}=\frac{9}{21}+\frac{14}{21}=\frac{23}{21}\)
Example #2: Given \(\frac{\frac{3}{5}}{2}\), take the reciprocal of \(2\). The reciprocal is \(\frac{1}{2}\). Now multiply: \(\frac{3}{5}*\frac{1}{2}=\frac{3}{10}\).
Decimal Representation
The decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....).
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are nonnegative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).
Converting Decimals to Fractions
• To convert a terminating decimal to fraction: 1. Calculate the total numbers after decimal point 2. Remove the decimal point from the number 3. Put 1 under the denominator and annex it with "0" as many as the total in step 1 4. Reduce the fraction to its lowest terms
Example: Convert \(0.56\) to a fraction. 1: Total number after decimal point is 2. 2 and 3: \(\frac{56}{100}\). 4: Reducing it to lowest terms: \(\frac{56}{100}=\frac{14}{25}\)
• To convert a recurring decimal to fraction: 1. Separate the recurring number from the decimal fraction 2. Annex denominator with "9" as many times as the length of the recurring number 3. Reduce the fraction to its lowest terms
Example #1: Convert \(0.393939...\) to a fraction. 1: The recurring number is \(39\). 2: \(\frac{39}{99}\), the number \(39\) is of length \(2\) so we have added two nines. 3: Reducing it to lowest terms: \(\frac{39}{99}=\frac{13}{33}\).
• To convert a mixedrecurring decimal to fraction: 1. Write down the number consisting with nonrepeating digits and repeating digits. 2. Subtract nonrepeating number from above. 3. Divide 12 by the number with 9's and 0's: for every repeating digit write down a 9, and for every nonrepeating digit write down a zero after 9's.
Example #2: Convert \(0.2512(12)\) to a fraction. 1. The number consisting with nonrepeating digits and repeating digits is 2512; 2. Subtract 25 (nonrepeating number) from above: 251225=2487; 3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.
Rounding
Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.
Example: 5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5. 5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5. 5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.
Ratios and Proportions
Given that \(\frac{a}{b}=\frac{c}{d}\), where a, b, c and d are nonzero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained.
\(\frac{b}{a}=\frac{d}{c}\)  invertendo
\(\frac{a}{c}=\frac{b}{d}\)  alternendo
\(\frac{a+b}{b}=\frac{c+d}{d}\)  componendo
\(\frac{ab}{b}=\frac{cd}{d}\)  dividendo
\(\frac{a+b}{ab}=\frac{c+d}{cd}\)  componendo & dividendo EXPONENTS
Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number \(a\) multiplied \(n\) times can be written as \(a^n\), where \(a\) represents the base, the number that is multiplied by itself \(n\) times and \(n\) represents the exponent. The exponent indicates how many times to multiple the base, \(a\), by itself.
Exponents one and zero: \(a^0=1\) Any nonzero number to the power of 0 is 1. For example: \(5^0=1\) and \((3)^0=1\) • Note: the case of 0^0 is not tested on the GMAT.
\(a^1=a\) Any number to the power 1 is itself.
Powers of zero: If the exponent is positive, the power of zero is zero: \(0^n = 0\), where \(n > 0\).
If the exponent is negative, the power of zero (\(0^n\), where \(n < 0\)) is undefined, because division by zero is implied.
Powers of one: \(1^n=1\) The integer powers of one are one.
Negative powers: \(a^{n}=\frac{1}{a^n}\)
Powers of minus one: If n is an even integer, then \((1)^n=1\).
If n is an odd integer, then \((1)^n =1\).
Operations involving the same exponents: Keep the exponent, multiply or divide the bases \(a^n*b^n=(ab)^n\)
\(\frac{a^n}{b^n}=(\frac{a}{b})^n\)
\((a^m)^n=a^{mn}\)
\(a^{m^n}=a^{(m^n)}\) and not \((a^m)^n\)
Operations involving the same bases: Keep the base, add or subtract the exponent (add for multiplication, subtract for division) \(a^n*a^m=a^{n+m}\)
\(\frac{a^n}{a^m}=a^{nm}\)
Fraction as power: \(a^{\frac{1}{n}}=\sqrt[n]{a}\)
\(a^{\frac{m}{n}}=\sqrt[n]{a^m}\)
Exponential Equations: When solving equations with even exponents, we must consider both positive and negative possibilities for the solutions.
For instance \(a^2=25\), the two possible solutions are \(5\) and \(5\).
When solving equations with odd exponents, we'll have only one solution.
For instance for \(a^3=8\), solution is \(a=2\) and for \(a^3=8\), solution is \(a=2\).
Exponents and divisibility: \(a^nb^n\) is ALWAYS divisible by \(ab\). \(a^nb^n\) is divisible by \(a+b\) if \(n\) is even.
\(a^n + b^n\) is divisible by \(a+b\) if \(n\) is odd, and not divisible by a+b if n is even.
LAST DIGIT OF A PRODUCT
Last \(n\) digits of a product of integers are last \(n\) digits of the product of last \(n\) digits of these integers.
For instance last 2 digits of 845*9512*408*613 would be the last 2 digits of 45*12*8*13=540*104=40*4=160=60
Example: The last digit of 85945*89*58307=5*9*7=45*7=35=5?
LAST DIGIT OF A POWER
Determining the last digit of \((xyz)^n\):
1. Last digit of \((xyz)^n\) is the same as that of \(z^n\); 2. Determine the cyclicity number \(c\) of \(z\); 3. Find the remainder \(r\) when \(n\) divided by the cyclisity; 4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.
• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base. • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4. • Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6. • Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.
Example: What is the last digit of \(127^{39}\)? Solution: Last digit of \(127^{39}\) is the same as that of \(7^{39}\). Now we should determine the cyclisity of \(7\):
1. 7^1=7 (last digit is 7) 2. 7^2=9 (last digit is 9) 3. 7^3=3 (last digit is 3) 4. 7^4=1 (last digit is 1) 5. 7^5=7 (last digit is 7 again!) ...
So, the cyclisity of 7 is 4.
Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of \(127^{39}\) is the same as that of the last digit of \(7^{39}\), is the same as that of the last digit of \(7^3\), which is \(3\).
ROOTS
Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4.
General rules: • \(\sqrt{x}\sqrt{y}=\sqrt{xy}\) and \(\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}\).
• \((\sqrt{x})^n=\sqrt{x^n}\)
• \(x^{\frac{1}{n}}=\sqrt[n]{x}\)
• \(x^{\frac{n}{m}}=\sqrt[m]{x^n}\)
• \({\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}\)
• \(\sqrt{x^2}=x\), when \(x\leq{0}\), then \(\sqrt{x^2}=x\) and when \(x\geq{0}\), then \(\sqrt{x^2}=x\)
• When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.
That is, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5. Even roots have only a positive value on the GMAT.
• Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{64} =4\).
• For GMAT it's good to memorize following values: \(\sqrt{2}\approx{1.41}\) \(\sqrt{3}\approx{1.73}\) \(\sqrt{5}\approx{2.24}\) \(\sqrt{6}\approx{2.45}\) \(\sqrt{7}\approx{2.65}\) \(\sqrt{8}\approx{2.83}\) \(\sqrt{10}\approx{3.16}\) PERCENTS
A percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred"). It is often denoted using the percent sign, "%", or the abbreviation "pct". Since a percent is an amount per 100, percents can be represented as fractions with a denominator of 100. For example, 25% means 25 per 100, 25/100 and 350% means 350 per 100, 350/100.
• A percent can be represented as a decimal. The following relationship characterizes how percents and decimals interact. Percent Form / 100 = Decimal Form
For example: What is 2% represented as a decimal? Percent Form / 100 = Decimal Form: 2%/100=0.02
• Percent change
General formula for percent increase or decrease, (percent change):
\(Percent=\frac{Change}{Original}*100\)
Example: A company received $2 million in royalties on the first $10 million in sales and then $8 million in royalties on the next $100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next $100 million in sales?
Solution: Percent decrease can be calculated by the formula above: \(Percent=\frac{Change}{Original}*100=\) \(=\frac{\frac{2}{10}\frac{8}{100}}{\frac{2}{10}}*100=60%\), so the royalties decreased by 60%.
• Simple Interest Simple interest = principal * interest rate * time Example: If $15,000 is invested at 10% simple annual interest, how much interest is earned after 9 months? Solution: $15,000*0.1*9/12 = $1125
• Compound Interest \(Balance(final)=principal*(1+\frac{interest}{C})^{time*C}\), where C = the number of times compounded annually. If C=1, meaning that interest is compounded once a year, then the formula will be: \(Balance(final)=principal*(1+interest)^{time}\), where time is number of years. Example: If $20,000 is invested at 12% annual interest, compounded quarterly, what is the balance after 2 year? Solution: \(Balance=20,000*(1+\frac{0.12}{4})^{2*4}=20,000*(1.03)^8=$25,335.4\) ORDER OF OPERATIONS  PEMDAS
Perform the operations inside a Parenthesis first (absolute value signs also fall into this category), then Exponents, then Multiplication and Division, from left to right, then Addition and Subtraction, from left to right  PEMDAS.
Special cases: • An exclamation mark indicates that one should compute the factorial of the term immediately to its left, before computing any of the lowerprecedence operations, unless grouping symbols dictate otherwise. But \(3^2!\) means \((3^2)! = 9!\) while \(2^{5!} = 2^{120}\); a factorial in an exponent applies to the exponent, while a factorial not in the exponent applies to the entire power.
• If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^{m^n}=a^{(m^n)}\) and not \((a^m)^n\)  Get The Official GMAT Club's App  GMAT TOOLKIT 2. The only app you need to get 700+ score! [ iOS App] [ Android App]  Finding the Sum of the Factors of an Integer
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers. The sum of factors of \(n\) will be expressed by the formula: \(\frac{(a^{p+1}1)*(b^{q+1}1)*(c^{r+1}1)}{(a1)*(b1)*(c1)}\)
Example: Finding the sum of all factors of 450: \(450=2^1*3^2*5^2\)
The sum of all factors of 450 is \(\frac{(2^{1+1}1)*(3^{2+1}1)*(5^{2+1}1)}{(21)*(31)*(51)}\)\(=\frac{3*26*124}{1*2*4}=1209\)
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Originally posted by Bunuel on 23 Dec 2009, 18:36.
Last edited by Bunuel on 04 Feb 2019, 02:52, edited 26 times in total.
Edited.




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Re: Math: Number Theory
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24 Jan 2010, 00:44
defoue wrote: Hi buddies, going through this awesome contribution, I got some issues to understand the following : would you please help to understand by giving some extra examples?
1. Finding the power of nonprime in n!:
How many powers of 900 are in 50!
Make the prime factorization of the number: 900=2^2*3^2*5^2, then find the powers of these prime numbers in the n!.
Find the power of 2: \frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47
= 2^{47}
Find the power of 3: \frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22
=3^{22}
Find the power of 5: \frac{50}{5}+\frac{50}{25}=10+2=12
=5^{12}
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
2. I did not get the tips 3 or 4 regarding the pefect square whare it says a perfect square has an odd nbr of odd powers and an even nbr of even power what aboute 36 which is 2^{2} x 3^{2} I am sure I am missing something here..
Thx guys 1. It's highly unlikely that this concept will be tested in GMAT. But still: Suppose we have the number \(18!\) and we are asked to to determine the power of \(12\) in this number. Which means to determine the highest value of \(x\) in \(18!=12^x*a\), where \(a\) is the product of other multiples of \(18!\). \(12=2^2*3\), so we should calculate how many 2s and 3s are in \(18!\). Calculating 2s: \(\frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16\). So the power of \(2\) (the highest power) in prime factorization of \(18!\) is \(16\). Calculating 3s: \(\frac{18}{3}+\frac{18}{3^2}=6+2=8\). So the power of \(3\) (the highest power) in prime factorization of \(18!\) is \(8\). Now as \(12=2^2*3\) we need twice as many 2s as 3s. \(18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a\). So \(18!=12^8*a\) > \(x=8\). 2. A perfect square ALWAYS has an ODD number of Oddfactors, and EVEN number of Evenfactors.Let's take your example \(36\). \(36=2^2*3^2\), the number of factors of \(36\) is \((2+1)(2+1)=9\): 1, 2, 3, 4, 6, 9, 12, 18, 36. 1, 3, 9  THREE ODD factors  "ODD number of Oddfactors"; 2, 4, 6, 12, 18, 36  SIX EVEN factors  "EVEN number of Evenfactors". Perfect square always has even number of powers of prime factorsLets take again \(36\). \(36=6^2=2^2*3^2\), the prime factors of \(36\) are 2 and 3, their powers are 2 and 2, which are even. OR \(144=12^2=2^4*3^2\), here again the powers (4 and 2) are even. Hope it's clear.
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03 Jan 2010, 09:30
Bunuel wrote: Prime Numbers
• Verifying the primality of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m<\sqrt{n}\). Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.
Pardon my ignorance, but I am not very much aware of this term Primality. Could you please explain this in detail? Haven't heard of this before! Bunuel wrote: Converting Decimals to Fractions
• To convert a mixedrecurring decimal to fraction: 1. Write down the number consisting with nonrepeating digits and repeating digits. 2. Subtract nonrepeating number from above. 3. Divide 12 by the number with 9's and 0's: for every repeating digit write down a 9, and for every nonrepeating digit write down a zero after 9's.
Example #2: Convert \(0.2512(12)\) to a fraction. 1. The number consisting with nonrepeating digits and repeating digits is 2512; 2. Subtract 25 (nonrepeating number) from above: 251225=2487; 3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.
What exactly do you mean by mixed recurring decimal? In the example taken, you have the number written as \(0.2512(12)\). Does this mean the number is \(0.2512121212.....\) ? Please clarify...! Bunuel wrote: EXPONENTS
Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number \(a\) multiplied \(n\) times can be written as \(a^n\), where \(a\) represents the base, the number that is multiplied by itself \(n\) times and \(n\) represents the exponent. The exponent indicates how many times to multiple the base, \(a\), by itself.
Exponents one and zero: \(a^0=1\) Any nonzero number to the power of 0 is 1. For example: \(5^0=1\) and \(0^0=1\)
You say any non zero number to the power of 0 is 1.... but in the example below you show \(0^0 = 1\).. Is it a typo or is there something more to it which I do not understand! Please clarify! Overall it's an awesome summary of all important points needed in Maths.... Thanks Bunuel and the others! I see that you are still working on this as of now! Any plan by when would you complete this Bible!




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04 Jan 2010, 00:29
jeeteshsingh wrote: Bunuel wrote: Prime Numbers
• Verifying the primality of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m<\sqrt{n}\). Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.
Pardon my ignorance, but I am not very much aware of this term Primality. Could you please explain this in detail? Haven't heard of this before! Bunuel wrote: Converting Decimals to Fractions
• To convert a mixedrecurring decimal to fraction: 1. Write down the number consisting with nonrepeating digits and repeating digits. 2. Subtract nonrepeating number from above. 3. Divide 12 by the number with 9's and 0's: for every repeating digit write down a 9, and for every nonrepeating digit write down a zero after 9's.
Example #2: Convert \(0.2512(12)\) to a fraction. 1. The number consisting with nonrepeating digits and repeating digits is 2512; 2. Subtract 25 (nonrepeating number) from above: 251225=2487; 3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.
What exactly do you mean by mixed recurring decimal? In the example taken, you have the number written as \(0.2512(12)\). Does this mean the number is \(0.2512121212.....\) ? Please clarify...! Bunuel wrote: EXPONENTS
Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number \(a\) multiplied \(n\) times can be written as \(a^n\), where \(a\) represents the base, the number that is multiplied by itself \(n\) times and \(n\) represents the exponent. The exponent indicates how many times to multiple the base, \(a\), by itself.
Exponents one and zero: \(a^0=1\) Any nonzero number to the power of 0 is 1. For example: \(5^0=1\) and \(0^0=1\)
You say any non zero number to the power of 0 is 1.... but in the example below you show \(0^0 = 1\).. Is it a typo or is there something more to it which I do not understand! Please clarify! Overall it's an awesome summary of all important points needed in Maths.... Thanks Bunuel and the others! I see that you are still working on this as of now! Any plan by when would you complete this Bible! 1. Verifying the primality; checking whether the integer is a prime number or not. 2. Mixed recurring decimal: 0.1366666... or 7,7485656565656... In decimal 0.1366666... 13 is nonrepeating part and 6 is repeating part. Repeating (recurring) decimal 0.111.. can also be written as 0.(1). Another example 1.123333333... can be written as 1.12(3) 3. Yes there is a typo. Edited. As for 0^0, in some sources it's equals to 1, some mathematicians say it's undefined. You won't need this for GMAT. 4. I'll try to finish this topic soon.
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07 Apr 2010, 02:12
fruit wrote: Thanks! It was very very helpful! Kudos! But I have a question:
How many powers of 900 are in 50!
Make the prime factorization of the number: \(900=2^2*3^2*5^2\), then find the powers of these prime numbers in the n!.
Find the power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\)
= \(2^{47}\)
Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)
=\(3^{22}\)
Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\)
=\(5^{12}\)
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
Why do we take just 5 from {2,3,5} and why do we need divide 12 by 2 to get the result?
Thanks in advance! \(50!=900^xa=(2^2*3^2*5^2)^x*a\), where \(x\) is the highest possible value of 900 and \(a\) is the product of other multiples of \(50!\). \(50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b\), where \(b\) is the product of other multiples of \(50!\). So \(x=6\). Below is another example: Suppose we have the number \(18!\) and we are asked to to determine the power of \(12\) in this number. Which means to determine the highest value of \(x\) in \(18!=12^x*a\), where \(a\) is the product of other multiples of \(18!\). \(12=2^2*3\), so we should calculate how many 2s and 3s are in \(18!\). Calculating 2s: \(\frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16\). So the power of \(2\) (the highest power) in prime factorization of \(18!\) is \(16\). Calculating 3s: \(\frac{18}{3}+\frac{18}{3^2}=6+2=8\). So the power of \(3\) (the highest power) in prime factorization of \(18!\) is \(8\). Now as \(12=2^2*3\) we need twice as many 2s as 3s. \(18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a\). So \(18!=12^8*a\) > \(x=8\).
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18 Jan 2010, 09:52
Hi buddies, going through this awesome contribution, I got some issues to understand the following : would you please help to understand by giving some extra examples?
1. Finding the power of nonprime in n!:
How many powers of 900 are in 50!
Make the prime factorization of the number: 900=2^2*3^2*5^2, then find the powers of these prime numbers in the n!.
Find the power of 2: \frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47
= 2^{47}
Find the power of 3: \frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22
=3^{22}
Find the power of 5: \frac{50}{5}+\frac{50}{25}=10+2=12
=5^{12}
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
2. I did not get the tips 3 or 4 regarding the pefect square whare it says a perfect square has an odd nbr of odd powers and an even nbr of even power what aboute 36 which is 2^{2} x 3^{2} I am sure I am missing something here..
Thx guys



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11 Aug 2010, 18:07
utfan2424 wrote: Does this relationship breakdown at some point? I thought this was great and was just experimenting and looked at 21! (calculated in excel) and it ends with 5 zeros. Using the methodology you described above it should have 4 zeros. Am I missing something or did I make a mistake somewhere? You made everything right 21! ends with 21/5=4 zeros. It's excel: it makes rounding with such a huge numbers thus giving incorrect result.
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14 Sep 2010, 06:38
cheetarah1980 wrote: There seems to be a discrepancy in what some study guides consider consecutive integers. In Kaplan Premier 2011 consecutive integers are defined as numbers that occur at a fixed interval or exhibit a fixed pattern. However, on the Kaplan Free Practice Test I got a DS question wrong because it didn't consider evenly spaced numbers to necessarily be consecutive. Your definition also separates the two. Could anyone clarify which is correct so I know for the actual GMAT. Thanks! When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1: ... x3, x2, x1, x, x+1, x+2, ...7, 6, 5 are consecutive integers. 2, 4, 6 ARE NOT consecutive integers, they are consecutive even integers. 3, 5, 7 ARE NOT consecutive integers, they are consecutive odd integers. 2, 5, 8, 11 ARE NOT consecutive integers, they are terms of arithmetic progression with common difference of 3. All sets of consecutive integers represent arithmetic progression but not viseversa. Hope it's clear.
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11 Aug 2010, 17:12
Bunuel wrote: fruit wrote: Thanks! It was very very helpful! Kudos! But I have a question:
How many powers of 900 are in 50!
Make the prime factorization of the number: \(900=2^2*3^2*5^2\), then find the powers of these prime numbers in the n!.
Find the power of 2: \(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\)
= \(2^{47}\)
Find the power of 3: \(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)
=\(3^{22}\)
Find the power of 5: \(\frac{50}{5}+\frac{50}{25}=10+2=12\)
=\(5^{12}\)
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
Why do we take just 5 from {2,3,5} and why do we need divide 12 by 2 to get the result?
Thanks in advance! \(50!=900^xa=(2^2*3^2*5^2)^x*a\), where \(x\) is the highest possible value of 900 and \(a\) is the product of other multiples of \(50!\). \(50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b\), where \(b\) is the product of other multiples of \(50!\). So \(x=6\). Below is another example: Suppose we have the number \(18!\) and we are asked to to determine the power of \(12\) in this number. Which means to determine the highest value of \(x\) in \(18!=12^x*a\), where \(a\) is the product of other multiples of \(18!\). \(12=2^2*3\), so we should calculate how many 2s and 3s are in \(18!\). Calculating 2s: \(\frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16\). So the power of \(2\) (the highest power) in prime factorization of \(18!\) is \(16\). Calculating 3s: \(\frac{18}{3}+\frac{18}{3^2}=6+2=8\). So the power of \(3\) (the highest power) in prime factorization of \(18!\) is \(8\). Now as \(12=2^2*3\) we need twice as many 2s as 3s. \(18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a\). So \(18!=12^8*a\) > \(x=8\). Does this relationship breakdown at some point? I thought this was great and was just experimenting and looked at 21! (calculated in excel) and it ends with 5 zeros. Using the methodology you described above it should have 4 zeros. Am I missing something or did I make a mistake somewhere?



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28 Oct 2010, 22:28
\(36=6^2=2^2*3^2\) Powers of 2 & 3 are even (2) Posted from my mobile device
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01 Nov 2010, 16:16
shrive555 wrote: If n is a positive integer greater than 1, then there is always a prime number P with n<P<2n
n<p<2n can someone please explain this with example .
Thanks The result you are referring to is a weak form of what is known as Bertrand's Postulate. The proof of this result is beyond the scope of the GMAT, but it is easy to show some examples. Choose any n>1, you will always find a prime number between n & 2n. Eg. n=5, 2n=10 ... p=7 lies in between n=14, 2n=28 ... p=19 lies in between n=20, 2n=40 ... p=23 lies in between
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25 Sep 2012, 10:43
conty911 wrote: Bunuel wrote: NUMBER THEORY
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a nonnegative integer \(n\), can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that \(5^k<n\).
It's easier if you look at an example:
How many zeros are in the end (after which no other digits follow) of \(32!\)? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
Hence, there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
I noticed in case the number (n) is multiple of \(5^k\) and we have to find number of trailing zero zeroes, then it will be \(5^k<=n\) rather \(5^k<n\) no of trailing zeros in 25! =6 \(\frac{25}{5}+\frac{25}{5^2}= 5+1\); Please correct me, clarify if i'm wrong. Thanks The highest power of a prime number "k" that divides any number "n!" is given by the formula n/K + n/k^2+n/k^3.. (until numerator becomes lesser than the denominator). Remember to truncate the remainders of each expressionE.g : The highest number of 2's in 10! is 10/2 + 10/4 + 10/8 = 5 + 2 + 1 = 8 (Truncate the reminder of each expression) As a consequence of this, the number of zeros in n! is controlled by the presence of 5s. Why ? 2 reasons a) 10 = 5 x 2, b) Also in any n!, the number of 5's are far lesser than the number of 2's. Think about this example. The number of cars that you make depends on the number of engines. You can have 100 engines and 1000 cars, but you can only make 100 cars (each car needs an engine !) 10 ! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 Lets factorize each term ... 10! = (5 x 2) x(3x3)x(2x2x2)x7x(2x3)x(5)X(2x2)x1 the number of 5s = 2 The number of 2s = 7 The number of zeros in 10! = the total number of 5s = 2 (You may use a calc to check this 10! = 3628800) hence in any n! , the number of 5's control the number of zeros.As a consequence of this, the number of 5's in any n! is n/5 + n/25 + n/125 ..until numerator becomes lesser than denominator. Again, i want to emphasize that this formuala only works for prime numbers !! So to find the number of 10's in any n!, DO NOT DIVIDE by 10 ! (10 is not prime !) i.e DONT do n/10 + n/100 + n/1000  THIS IS WRONG !!!



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27 Oct 2012, 02:00
I shall like to add one trick to find square of number :
To find square of a number \((ab)^2\) where a is tens digit and b is units digit =>
1. Multiply tens digit a by (a+1) i.e. a * (a+1) = A 2. Suffix 25 to the above derived result A and the new number will be A25
Example: what is \(35^2\)? Solution: 3 * (3+1) = 3*4 = 12 and so answer is 1225



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27 Oct 2012, 02:34
About Exponents and divisibility:
\((a + b)^2 = a^2+ 2ab + b^2\) Square of a Sum \((a  b)^2 = a^2  2ab + b^2\) Square of a Difference
\(a^n  b^n\) is always divisble by ab i.e. irrespective of n being odd or even Proof: \(a^2  b^2 = (ab)(a+b)\) \(a^3  b^3 = (ab)(a^2+ab+b^2)\)
Thus divisible by a b in both cases where n = 2 i.e. even and 3 i.e. odd
\(a^n + b^n\) is divisble by a+b i.e. only if n = odd Proof: \(a^3  b^3 = (a+b)(a^2ab+b^2)\) Thus divisible by a + b as n = 3 i.e. odd



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12 Jun 2015, 11:27
Bunuel wrote: • Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m<\sqrt{n}\). Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime. A minor point, but the inequalities here should not be strict. If you want to test if some large integer n is prime, then you need to try dividing by numbers up to and including \(\sqrt{n}\). We must include \(\sqrt{n}\), in case our number is equal to the square of a prime. And it might be worth mentioning that it is only necessary to try dividing by prime numbers up to \(\sqrt{n}\), since if n has any divisors at all (besides 1 and n), then it must have a prime divisor. It's very rare, though, that one needs to test if a number is prime on the GMAT. It is, computationally, extremely timeconsuming to test if a large number is prime, so the GMAT cannot ask you to do that. If a GMAT question asks if a large number is prime, the answer really must be 'no', because while you can often quickly prove a large number is not prime (for example, 1,000,011 is not prime because it is divisible by 3, as we see by summing digits), you cannot quickly prove that a large number is prime.
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Re: Math: Number Theory
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31 Jan 2010, 11:34
ariel wrote: Appreciate the very prompt response, walker. To your point re divisibility by 7: I'm having a hard time proving this algebraically, is it a fair statement to say that the only nonprime numbers of the form 6n1 and 6n+1 are the ones that are divisible by 7?
If so, a quick way to check whether a big number is prime would be to: 1) check whether it's of the form 6n1 or 6n+1 2) check whether it's divisible by 7
Is this correct? Not so. Divisibility by 7 does not check whether the number is prime or not. Actually this issue is covered in the post. First you should know that all prime numbers except 2 and 5 end in 1, 3, 7 or 9. So if it ends in some other digit it's not prime. Next, if the above didn't help (meaning that number ends in 1, 3, 7 or 9) there is a way to check whether the number is prime or not. Walker gave an example how to do this, but here it is again: Verifying the primality of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m<\sqrt{n}\). Examples: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\). We should check \(161\) on divisibility by numbers from 2 to 13. From integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime. Verifying the primality of \(149\): \(\sqrt{149}\) is little more than \(12\). We should check \(149\) on divisibility by numbers from 2 to 12, inclusive. \(149\) is not divisible by any of the integers from \(2\) to \(12\), hence \(149\) is prime. Verifying the primality of \(73\): \(\sqrt{73}\) is little less than \(9\). We should check \(73\) on divisibility by numbers from 2 to 9. \(73\) is not divisible by any of the integers from \(2\) to \(9\), hence \(149\) is prime. Hope it helps.
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Re: Math: Number Theory
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20 Sep 2010, 00:27
7  Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.
Is this the only way to check divisibility by 7? For huge numbers there is no big difference to divide the number directly by 7 or to use the algorithm above.



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Re: Math: Number Theory
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19 Nov 2010, 01:53
Araj wrote: Hello Bunuel  thank you so much for this fantastic post! with regards to checking for primality: Quote: Verifying the primality (checking whether the number is a prime) of a given number can be done by trial division, that is to say dividing by all integer numbers smaller than , thereby checking whether is a multiple of . Example: Verifying the primality of : is little less than , from integers from to , is divisible by , hence is not prime.
Would it be accurate to say that a number is prime ONLY if it gives a remainder of 1 or 5 when divided by 6? i.e, for eg. 10973/6 gives a remainder of 5, so it has to be prime... i found the reasoning behind this in one of the OG solutions: prime numbers always take the form: 6n+1 or 6n+5 .... the only possible remainders when any number is divided by 6 are [0,1,2,3,4,5] ... A prime number always gives a remainder of 1 or 5, because: a) if the remainder is 2 or 4, then the number must be even b) if the remainder is 3, then it is divisible by 3 ... hence, if a number divided by 6 yields 1 or 5 as its remainder, then it must be prime ...? Raj First of all there is no known formula of prime numbers.Next:Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3). So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) or \(p=6n1\), where n is an integer >1. But:Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so viseversa of above property is not correct. For example 25 yields a remainder of 1 upon division be 6 and it's not a prime number. Hope it's clear.
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Re: Math: Number Theory
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06 Dec 2010, 01:08
shrive555 wrote: \((a^m)^n=a^{mn}\) 1
\((2^2)^2 = 2^2*^2 =2^4\)
\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) 2
\(2^2^2 = 2^(2^2) = 2^4\)
If above example is correct then whats the difference 1 & 2. Please clarify thanks If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\). So: \((a^m)^n=a^{mn}\); \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\). Now, there are some specific values of \(a\), \(m\) and \(n\) for which \(a^m^n\) equals to \(a^{mn}\). For example: \(a=1\): \(1^{m^n}=1=1^{mn}\); \(m=0\): \(a^0^n=a^0=1\) and \(a^{0*n}=a^0=1\); \(m=2\) and \(n=2\) > \(a^{2^2}=a^4\) and \(a^{2*2}=a^4\); \(m=4\) and \(n=\frac{1}{2}\) > \(a^{4^{\frac{1}{2}}}=a^2\) and \(a^{4*{\frac{1}{2}}}=a^2\); ... So, generally \(a^m^n\) does not equal to \((a^m)^n\), but for specific values of given variables it does. shrive555 wrote: In question would that be given explicitly ... i mean the Brackets ( ) \(a^m^n\) ALWAYS means \(a^{(m^n)}\), so no brackets are needed. For example \(2^{3^4}=2^{(3^4)}=2^{81}\); If GMAT wants the order of operation to be different then the necessary brackets will be put. For example: \((2^3)^4=2^{(3*4)}=2^{12}\). Hope it's clear.
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Re: Math: Number Theory
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03 Jan 2011, 09:52
resh924 wrote: Bunuel,
For determining last digit of a power for numbers 0, 1, 5, and 6, I am not clear on how to determine the last digit.
Your post says: • Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
What is the last digit of 345^27 is the last digit 5? What is the last digit of 216^32is the last digit 6? What is the last digit of 111^56is the last digit 1?
Any clarification would be helpful.
Thanks for all your help. First of all: last digit of 345^27 is the same as that of 5^27 (the same for 216^32 and 111^56); Next: 1 in any integer power is 1; 5^1= 5, 5^2=2 5, 5^3=12 5, ... 6^1= 6, 6^2=3 6, 5^3=21 6, ... So yes, integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base: thus 0, 1, 5, and 6 respectively. Hope it's clear.
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Re: Math: Number Theory
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