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Math: Number Theory

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Re: Math: Number Theory [#permalink]

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New post 14 Sep 2016, 06:53
Quote:
Similarly, all prime numbers above 3 are of the form 6n−16n−1 or 6n+16n+1, because all other numbers are divisible by 2 or 3.
In the prime number section it says prime numbers greater than 3 are of the form 6n-1 or 6n+1. However, this is not necessarily true. eg: n=36, then 6n-1 is 215 and 6n+1 is 217, divisible by 5 and 7 respectively.

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Re: Math: Number Theory [#permalink]

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New post 14 Sep 2016, 07:02
ks4196 wrote:
Quote:
Similarly, all prime numbers above 3 are of the form 6n−16n−1 or 6n+16n+1, because all other numbers are divisible by 2 or 3.
In the prime number section it says prime numbers greater than 3 are of the form 6n-1 or 6n+1. However, this is not necessarily true. eg: n=36, then 6n-1 is 215 and 6n+1 is 217, divisible by 5 and 7 respectively.


That's not what is said there.

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 \(p=6n-1\), 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 vise-versa 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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New post 27 Oct 2016, 02:30
• Any positive divisor of n is a product of prime divisors of n raised to some power.



pls someone explain with example

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New post 27 Oct 2016, 03:59
sanaexam wrote:
• Any positive divisor of n is a product of prime divisors of n raised to some power.



pls someone explain with example


For example, say n = 72. Consider it's factor 36 --> \(36 = 2^2*3^2\) --> 36 = product of prime divisors of n raised to some power.
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New post 27 Oct 2016, 05:27
Thank you Bunuel,.1 more doubt...Is it necessary that the prime factors should the prime factors of 72.?
For ex:35.when factorized....5^1*7^1
5 is factor of 35...so 5 when factorized...5^1.
So does the property states....that the divisors should have the primes which are within the number n.?

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New post 30 Jan 2017, 23:45
Thank you Bunuel, bb, and walker with kudos for the short notes of various math topics.
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New post 29 Mar 2017, 23:52
Great post!!
Thanks for sharing. It really cleared the very basic doubts especially the exponent rules.
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New post 27 Apr 2017, 18:01
@ Bunnet ,

Thank you so much for this consolidated Number Theory notes! It really helped me to boost my GMAT quant score from Q42 to Q49!

-Jyoti

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Math: Number Theory [#permalink]

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New post 29 Apr 2017, 07:26
Bunuel bb walker
What is the possibility that questions will come from concepts like "Finding the number of trailing zeros or powers of a prime/ non-prime number p, in the n!" on GMAT?

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New post 23 May 2017, 16:33
Bunuel wrote:
NUMBER THEORY

Division of two integers can result into an even/odd integer or a fraction.



Division of two integers (except the case where denominator is zero) can result into an even/odd integer or a fraction.

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New post 23 May 2017, 16:47
Bunuel wrote:
NUMBER THEORY

Fractions (also known as rational numbers) can be written as terminating (ending) or repeating decimals (such as 0.5, 0.76, or 0.333333....).

--------------------------------------------------------


0.333333... looks like a recurring decimal and is not a rational number. This can be better modified as 0.33333 etc so it is less confusing in my opinion.

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New post 24 May 2017, 04:02
workout wrote:
Bunuel wrote:
NUMBER THEORY

Fractions (also known as rational numbers) can be written as terminating (ending) or repeating decimals (such as 0.5, 0.76, or 0.333333....).

--------------------------------------------------------


0.333333... looks like a recurring decimal and is not a rational number. This can be better modified as 0.33333 etc so it is less confusing in my opinion.


0.333333... IS a rational number because it equals to the ratio of two integers: 1/3 = 0.3333......
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New post 24 May 2017, 11:00
Bunuel wrote:
NUMBER THEORY

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: \(\frac{(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)}{(a-1)*(b-1)*(c-1)}\)

--------------------------------------------------------


Does anyone have an opinion/suggestion if it helps(speeds up problem solving) to memorize these type of formulae for the exam ?

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Re: Math: Number Theory [#permalink]

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New post 24 May 2017, 11:54
workout wrote:
Bunuel wrote:
NUMBER THEORY

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: \(\frac{(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)}{(a-1)*(b-1)*(c-1)}\)

--------------------------------------------------------


Does anyone have an opinion/suggestion if it helps(speeds up problem solving) to memorize these type of formulae for the exam ?


It helps knowing the rule, although questions based solely on this will be rare. However another variation of this formula based on number of factors is really useful.

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: \((p+1)*(q+1)*(r+1)\)

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Re: Math: Number Theory [#permalink]

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New post 04 Aug 2017, 07:39
Bunuel, Is there any GCD between 0 & 1?
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Math: Number Theory [#permalink]

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New post 19 Aug 2017, 19:04
Bunuel wrote:
NUMBER THEORY

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, walker

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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.


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 non-terminating, non-repeating decimals are non-rational, 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 twenty-six 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 \(6n-1\) 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 non-zero 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.


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: \(\frac{(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)}{(a-1)*(b-1)*(c-1)}\)

Example: Finding the sum of all factors of 450: \(450=2^1*3^2*5^2\)

The sum of all factors of 450 is \(\frac{(2^{1+1}-1)*(3^{2+1}-1)*(5^{2+1}-1)}{(2-1)*(3-1)*(5-1)}\)\(=\frac{3*26*124}{1*2*4}=1209\)


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 non-zero integers, is the largest positive integer that divides the numbers without a remainder.

To find the GCF, you will need to do prime-factorization. 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 prime-factorization. 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 Odd-factors, and EVEN number of Even-factors.
• 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 non-negative 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 non-prime 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:
\(a_ n=a_1+d(n-1)\)

• 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, \(Sum=\frac{2a_1+d(n-1)}{2}*n\)

• 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: \(a_n=2n-1\). 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 part-to-whole 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 non-negative 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 mixed-recurring decimal to fraction:
1. Write down the number consisting with non-repeating digits and repeating digits.
2. Subtract non-repeating number from above.
3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's.

Example #2: Convert \(0.2512(12)\) to a fraction.
1. The number consisting with non-repeating digits and repeating digits is 2512;
2. Subtract 25 (non-repeating number) from above: 2512-25=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 non-zero 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

\(\frac{a-b}{b}=\frac{c-d}{d}\) - dividendo

\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\) - 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)
\(a^n*a^m=a^{n+m}\)

\(\frac{a^n}{a^m}=a^{n-m}\)

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:
\(a^n-b^n\) is ALWAYS divisible by \(a-b\).
\(a^n-b^n\) 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 lower-precedence 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 \((a^m)^n\)


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hi

gone through the excellent learning experience....thanks

Anyway, I have 2 questions....

I quote you as under:

"""""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 =2^2∗3, so we should calculate how many 2-s and 3-s are in 18!

Calculating 2-s: 18/2+18/2^2+18/2^3+18/2^4=9+4+2+1=16. So the power of 2 (the highest power) in prime factorization of 18! is 16.

Calculating 3-s: 18/3+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=2^2∗3 we need twice as many 2-s as 3-s. 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..""""""

So, here, there are 16 2s and 8 3s. Although I am okay with your arithmetic, can it be seen as such that 3 is a limiting factor here, as, 16 2s cannot make 12 without the same number of 3s in the prime factorization ....here we can see 8 3s, so the highest power 12 can have in 18! is 8, isn't it ..?

what is more, the prime factorization of 12 says that there are 2, 2s and 1, 3, so virtually there are 8 4s also...

please correct me if I am missing something ...

Another question...

when to find the trailing zeros, you have asked to find the factors of 5 only, citing the reason that there are at least as many 2s....
Why should I assume that there are at lease as many 2s ...?

5 and 1 are also factors of 5, but they don't have any 2s into them ....

thanks in advance .....

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Re: Math: Number Theory [#permalink]

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New post 27 Aug 2017, 10:57
Hi Bunuel,

Thanks a lot for such an excellent post. But I have doubt on the following point

"Finding the Number of Factors of an Integer

First make prime factorization of an integer n=ap∗bq∗crn=ap∗bq∗cr, where aa, bb, and cc are prime factors of nn and pp, qq, and rr are their powers.

The number of factors of nn will be expressed by the formula (p+1)(q+1)(r+1)(p+1)(q+1)(r+1). NOTE: this will include 1 and n itself."

I think the above list contains only the positive factors.
Do we need to consider the negative factors as well by multiplying 2?
Please give your valuable opinion and guide me further.
Thanks in advance.
_________________

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Re: Math: Number Theory [#permalink]

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New post 27 Aug 2017, 10:59
buan15 wrote:
Hi Bunuel,

Thanks a lot for such an excellent post. But I have doubt on the following point

"Finding the Number of Factors of an Integer

First make prime factorization of an integer n=ap∗bq∗crn=ap∗bq∗cr, where aa, bb, and cc are prime factors of nn and pp, qq, and rr are their powers.

The number of factors of nn will be expressed by the formula (p+1)(q+1)(r+1)(p+1)(q+1)(r+1). NOTE: this will include 1 and n itself."

I think the above list contains only the positive factors.
Do we need to consider the negative factors as well by multiplying 2?
Please give your valuable opinion and guide me further.
Thanks in advance.


On the GMAT a factor is a positive divisor.
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Re: Math: Number Theory [#permalink]

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New post 29 Aug 2017, 20:32
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KUDOS
Bunuel wrote:
buan15 wrote:
Hi Bunuel,

Thanks a lot for such an excellent post. But I have doubt on the following point

"Finding the Number of Factors of an Integer

First make prime factorization of an integer n=ap∗bq∗crn=ap∗bq∗cr, where aa, bb, and cc are prime factors of nn and pp, qq, and rr are their powers.

The number of factors of nn will be expressed by the formula (p+1)(q+1)(r+1)(p+1)(q+1)(r+1). NOTE: this will include 1 and n itself."

I think the above list contains only the positive factors.
Do we need to consider the negative factors as well by multiplying 2?
Please give your valuable opinion and guide me further.
Thanks in advance.


On the GMAT a factor is a positive divisor.


Thanks for your reply.
Can you please generalize the cases involving number system [specifically DS] problems in which we need to consider negative numbers as well while solving a problem.
Also mention the cases where we need not think about negative numbers at all.
I am taking longer time for calculating the possibilities and request you to provide some clues to solve first.
Thanks.
_________________

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Re: Math: Number Theory [#permalink]

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New post 09 Sep 2017, 08:51
buan15 wrote:
Bunuel wrote:
buan15 wrote:
Hi Bunuel,

Thanks a lot for such an excellent post. But I have doubt on the following point

"Finding the Number of Factors of an Integer

First make prime factorization of an integer n=ap∗bq∗crn=ap∗bq∗cr, where aa, bb, and cc are prime factors of nn and pp, qq, and rr are their powers.

The number of factors of nn will be expressed by the formula (p+1)(q+1)(r+1)(p+1)(q+1)(r+1). NOTE: this will include 1 and n itself."

I think the above list contains only the positive factors.
Do we need to consider the negative factors as well by multiplying 2?
Please give your valuable opinion and guide me further.
Thanks in advance.


On the GMAT a factor is a positive divisor.


Thanks for your reply.
Can you please generalize the cases involving number system [specifically DS] problems in which we need to consider negative numbers as well while solving a problem.
Also mention the cases where we need not think about negative numbers at all.
I am taking longer time for calculating the possibilities and request you to provide some clues to solve first.
Thanks.


I also have same doubt, can anyone please help with the test cases?
Thanks.

Kudos [?]: [0], given: 3

Re: Math: Number Theory   [#permalink] 09 Sep 2017, 08:51

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