GMAT

Integers, parity and signs

The integers are the whole numbers with their negatives: $\dots, -2, -1, 0, 1, 2, \dots$. Zero is an integer, it is even, and it is neither positive nor negative. An integer is even if it is a multiple of 2 and odd otherwise.

Parity follows fixed rules, and they are quicker than trying numbers.

For signs, a product or quotient of two numbers is positive when they have the same sign and negative otherwise. A product of several non-zero numbers is negative exactly when it has an odd number of negative factors. A number raised to an even power is never negative, so $(-3)^4 = 81$ while $(-3)^3 = -27$.

Divisibility

An integer $a$ is divisible by a non-zero integer $b$ when $a \div b$ is an integer, and then $b$ is a factor (divisor) of $a$ and $a$ is a multiple of $b$. Every integer divides 0.

For 11, add and subtract the digits alternately from the right. For 2,728: $8 - 2 + 7 - 2 = 11$, so 2,728 is divisible by 11, and indeed $2{,}728 = 11 \times 248$.

Two facts about divisibility are used constantly. If $a$ divides $b$ and $a$ divides $c$, then $a$ divides $b + c$ and $b - c$. If $a$ and $b$ are coprime (share no factor above 1) and both divide $n$, then $ab$ divides $n$; this is why divisibility by 6 needs both 2 and 3, but divisibility by 12 cannot be checked with 2 and 6, which are not coprime.

Primes and factorisation

A prime is an integer greater than 1 whose only positive factors are 1 and itself. The smallest prime is 2, and it is the only even prime. The primes below 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. The number 1 is not prime.

Every integer above 1 has exactly one prime factorisation, found by dividing out the smallest primes repeatedly. For example $360 = 2^3 \times 3^2 \times 5$.

If $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ with distinct primes $p_i$, then the number of positive factors of $n$ is

For 360 this gives $4 \times 3 \times 2 = 24$ factors. A factor is built by choosing an exponent from $0$ to $a_i$ for each prime, which is where the formula comes from. The odd factors use only the odd primes, so 360 has $3 \times 2 = 6$ odd factors. A positive integer has an odd number of factors exactly when it is a perfect square, because every exponent is then even and every bracket is odd.

Greatest common factor and least common multiple

The greatest common factor (gcf, also called the highest common factor) of two integers is the largest integer dividing both, and the least common multiple (lcm) is the smallest positive integer that both divide. From the factorisations, the gcf takes each prime to its smaller power and the lcm takes each prime to its larger power.

Take $84 = 2^2 \times 3 \times 7$ and $126 = 2 \times 3^2 \times 7$:

For any two positive integers, $\text{gcf} \times \text{lcm} = a \times b$. Here $42 \times 252 = 10{,}584 = 84 \times 126$. The lcm is the tool for events that repeat together: two lights flashing every 12 and 18 seconds coincide every $\text{lcm}(12, 18) = 36$ seconds.

Remainders

When a positive integer $n$ is divided by a positive integer $d$, there is a unique quotient $q$ and remainder $r$ with

The remainder is always smaller than the divisor. A number with remainder 0 is divisible by $d$, and a number $n$ with remainder $r$ on division by $d$ can be written $n = qd + r$ for some integer $q$, which is how to reason about it in general.

Remainders combine like the numbers themselves, and the result is reduced again at the end. If $n$ leaves remainder 5 and $m$ leaves remainder 7 on division by 9, then $n + m$ leaves the remainder of $5 + 7 = 12$, which is 3, and $nm$ leaves the remainder of $5 \times 7 = 35$, which is 8.

To find a number with two remainders, list the numbers that meet the larger divisor's condition and test them. If $n$ leaves remainder 3 on division by 5 and remainder 2 on division by 7, the numbers $3, 8, 13, 18, 23, \dots$ satisfy the first condition, and 23 is the first that leaves remainder 2 when divided by 7. The solutions are then $23 + 35k$, because 35 is the lcm of 5 and 7 and adding it changes neither remainder.

Units digits and cycles

The last digit of a power repeats in a cycle. Powers of 7 end in 7, 9, 3, 1, 7, 9, 3, 1 and so on, a cycle of length 4. To find the last digit of $7^{53}$, take $53$ divided by 4, which leaves remainder 1, so the last digit is the first in the cycle, 7. Powers of 2 and 3 also cycle with length 4, powers of 4 and 9 with length 2, and powers of 0, 1, 5 and 6 never change their last digit.

Consecutive integers

The product of any $k$ consecutive integers is divisible by $k!$. So any two consecutive integers multiply to an even number, any three multiply to a multiple of 6, and any four multiply to a multiple of 24. The sum of $k$ consecutive integers is divisible by $k$ when $k$ is odd, since the sum is $k$ times the middle term. For even $k$ the sum is not divisible by $k$; four consecutive integers starting at 1 sum to 10.

The sum of the integers from 1 to $n$ is $\dfrac{n(n+1)}{2}$, so $1 + 2 + \dots + 50 = \dfrac{50 \times 51}{2} = 1{,}275$.

Place value and decimals

Each digit's value is the digit times a power of 10 set by its position, so $4{,}072.35 = 4 \times 10^3 + 0 \times 10^2 + 7 \times 10 + 2 + 3 \times 10^{-1} + 5 \times 10^{-2}$. A two-digit number with tens digit $t$ and units digit $u$ is $10t + u$, and swapping the digits gives $10u + t$. The difference between a two-digit number and its reversal is $9(t - u)$, so it is always a multiple of 9.