NUMBER THEORY
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 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.

Note that you can perform this operation number of times. Also you won't need to check divisibility by 7 for huge numbers on GMAT.

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:.
• Perfect square always has even number of powers of prime factors.

Hello Bunuel - thank you so much for this fantastic post!

with regards to checking for primality:



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

I think its just that you have taken a bad example here. Consider a=2, m=3, b=2

(a^m)^n=(2^3)^2=8^2=64
a^{(m^n)}=2^{(3^2)}=2^9=512