Whole Numbers & Operations

In our number system, the place of a digit tells you its value. Moving left, each place is worth ten times more than the one to its right. Reading right to left the places are: ones, tens, hundreds, then thousands, ten-thousands, hundred-thousands, then millions, ten-millions, hundred-millions.

We group digits into periods of three (ones, thousands, millions) separated by commas. Read each period, then say its group name.

To round to a place value: look at the digit just to the right of that place.

• If it is $5$ or more, round up (add one to the rounding digit).
• If it is $4$ or less, round down (keep the rounding digit).

Then replace every digit to the right with zeros. Estimation means rounding first, then computing, to get a quick, close answer.

Line up numbers by place value (ones under ones). Add from right to left; when a column totals $10$ or more, carry the extra ten to the next column. Subtract from right to left; when the top digit is too small, borrow (regroup) one unit from the next place to the left.

To multiply multi-digit numbers, multiply by each digit of the bottom number one at a time, shifting each partial product one place left (because each new digit is worth ten times more). Then add the partial products.

Division splits a total into equal groups. In $a \div b$, $a$ is the dividend, $b$ is the divisor, and the result is the quotient. Long division repeats four steps: Divide, Multiply, Subtract, Bring down. Whatever is left at the end that will not divide evenly is the remainder.

When an expression has several operations, do them in this order:

• Parentheses (grouping symbols) first.
• Exponents next.
• Multiplication and Division, left to right.
• Addition and Subtraction, left to right.

Multiplication and division rank equally, so you handle them in reading order. The same is true of addition and subtraction.

An exponent tells how many times to use the base as a factor. In $b^n$, $b$ is the base and $n$ is the exponent: $b^n = \underbrace{b \times b \times \cdots \times b}_{n\text{ times}}$. We read $5^2$ as “five squared” and $5^3$ as “five cubed.” A perfect square is a whole number times itself, like $1, 4, 9, 16, 25, \ldots$

To solve a word problem: (1) read carefully and find what is asked, (2) decide which operation(s) fit — combining suggests addition, taking away suggests subtraction, equal groups suggest multiplication, and splitting into equal parts suggests division, (3) compute in the right order, and (4) check that your answer makes sense.

Every long-division problem is really a statement about whole numbers. For any dividend $a$ and any divisor $b > 0$, there is exactly one pair of whole numbers — a quotient $q$ and a remainder $r$ — such that

This is called the Division Algorithm. The condition $0 \le r < b$ is what forces a single correct answer: the remainder is never negative and never as large as the divisor. This one equation is the foundation for fractions, modular arithmetic, and much of the number theory you will meet later.

Often we care only about the remainder of a division, not the quotient. We write

meaning $r$ is the remainder when $a$ is divided by $b$. A clock is the everyday example: it works “$\bmod\ 12$.” If it is $9$ o'clock and $7$ hours pass, we do not say $16$ o'clock — we compute $16 \bmod 12 = 4$, so it is $4$ o'clock. Numbers that share the same remainder are called congruent; for example $16$ and $4$ are congruent mod $12$ because both leave remainder $4$.

The quick tests for divisibility are not magic — each one comes from place value.

• By $2$, $5$, or $10$: only the ones digit matters, because $10$, $100$, $1000, \ldots$ are all divisible by $2$, $5$, and $10$. So the whole number is divisible exactly when its last digit is.
• By $3$ or $9$: add the digits. Since $10 = 9 + 1$, $100 = 99 + 1$, and so on, each place is a multiple of $9$ plus $1$. Subtracting off the multiples of $9$ leaves just the sum of the digits, so the number is divisible by $9$ (or $3$) exactly when that digit sum is.
• By $4$: check the last two digits, because $100$ is divisible by $4$, so everything above the hundreds place already is.

Our usual system is base ten: each place is a power of $10$ ($\ldots, 1000, 100, 10, 1$). Computers use base two (binary), where each place is a power of $2$ ($\ldots, 8, 4, 2, 1$) and the only digits are $0$ and $1$. To convert a whole number to binary, repeatedly divide by $2$ and record the remainders from bottom to top.

The triangular numbers count dots stacked in a growing triangle: $1, 3, 6, 10, 15, \ldots$, where the $n$th one is the sum $1 + 2 + 3 + \cdots + n$. Instead of adding term by term, there is a beautiful shortcut:

Why it works (Gauss's trick): write the sum forwards and backwards and add the two lines in columns:

Every column adds to $n+1$, and there are $n$ columns, giving $n(n+1)$ for twice the sum — so the sum itself is $\dfrac{n(n+1)}{2}$.

A few reliable shortcuts:

• Multiply by $5$: multiply by $10$ and halve. So $48 \times 5 = 480 \div 2 = 240$.
• Square a number ending in $5$: for a two-digit number $\underline{d}5$, compute $d \times (d{+}1)$ and attach “$25$.” So $35^2$: $3 \times 4 = 12$, giving $\mathbf{1225}$.
• Front-end estimation: add just the leading digits for a fast lower estimate, then refine. For $612 + 289$: $600 + 200 = 800$, plus the tens gives about $900$.