Remainders, odd and even, and positive and negative

What a remainder really says

Dividing a positive integer $n$ by a positive integer $d$ gives a quotient $q$ and a remainder $r$ with

$$n = dq + r, \qquad 0 \le r < d$$

The condition on $r$ matters: a remainder is never negative and always smaller than the divisor. Dividing by 7 can only leave 0, 1, 2, 3, 4, 5 or 6, and a remainder of 0 means $d$ divides $n$.

The equation is the useful form. "When $n$ is divided by 7 the remainder is 5" becomes $n = 7q + 5$, and you can list the numbers that fit: 5, 12, 19, 26, 33, … Starting from the remainder itself, add the divisor again and again. Notice that the smallest such number is 5, with quotient 0; people often forget it.

Remainders of sums, products and powers

Remainders can be combined without finding $n$. If $a$ leaves remainder $r$ and $b$ leaves remainder $s$ when divided by $d$, then:

• $a + b$ leaves the same remainder as $r + s$;
• $ab$ leaves the same remainder as $rs$;
• $a^k$ leaves the same remainder as $r^k$.

Each time, reduce at the end: if $r + s$ or $rs$ is $d$ or more, divide it by $d$ again. Why does this work? Write $a = dq + r$. Every term in $a^2 = d^2q^2 + 2dqr + r^2$ except the last is a multiple of $d$, so only $r^2$ affects the remainder.

The units digit of a number is its remainder on division by 10, so the same rule finds units digits of big products and powers. Powers of 3 end in 3, 9, 7, 1 and then repeat; to find the units digit of $3^{50}$, divide 50 by the cycle length 4 (remainder 2), and take the second entry: 9.

Two conditions at once

When a number must leave one remainder for one divisor and another for a second divisor, list the candidates for the larger divisor and test each against the other condition. Once you find the first number that fits, the rest come every $\operatorname{lcm}$ of the two divisors.

Odd and even

An integer is even if it is divisible by 2 and odd otherwise. 0 is even, and negative integers are odd or even too: $-6$ is even, $-3$ is odd.

Operation	Rule
sum or difference	even if the two have the same parity; odd if they differ
product	even if at least one factor is even; odd only if all are odd
power $n^k$ ($k \ge 1$)	same parity as $n$

Two consequences come up all the time. A product of consecutive integers, such as $n(n+1)$, is always even, because one of any two consecutive integers is even. And if $a + b$ is odd, one of $a$ and $b$ is odd and the other even, so $ab$ is even.

Division has no parity rule: $\frac{12}{2} = 6$ is even while $\frac{6}{2} = 3$ is odd. Work with sums and products instead.

Positive and negative

The sign of a product depends only on how many negative factors it has: an even number of negatives gives a positive product, an odd number a negative one. Any factor of 0 makes the product 0, which is neither positive nor negative.

So an even power is never negative ($x^2 \ge 0$), and an odd power keeps the sign of its base. Given $xy > 0$, you know $x$ and $y$ have the same sign but not which; given $xy < 0$, they have opposite signs.

Pick numbers wisely

In a comparison question, variables can be negative, zero or fractions unless the stem says otherwise. Read every restriction ("integer", "positive", "odd"), then test values from each kind the stem allows. A claim that works for 2 and 3 may fail for $-1$ or 0.