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DefinitionIntegers 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 NumbersAn 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.
Prime NumbersA 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 \(m<\sqrt{n}\).
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.
• If \(n\) is a positive integer greater than 1, then there is always a prime number \(p\) with\(n < p < 2n\).
FactorsA 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 IntegerFirst 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 IntegerFirst 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 - LCMThe 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).
Divisibility Rules2 - 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.
FactorialsFactorial of a positive integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to n. For instance \(5!=1*2*3*4*5\).
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 \(5^k<n\).
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 \(p^x<n\)
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 IntegersConsecutive 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 SetEvenly 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 integers: \(1+2+...+n=\frac{1+n}{2}*n\)
Sum of n first 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 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.
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10 Mar 2010, 06:20
. Thank you. +1Kudos. 
