Primary & Middle School Mathematics · Grades 1–9

37Whole Numbers

Everything in mathematics starts with counting. This chapter reviews how whole numbers are written, compared and combined — and takes a careful look at division, the operation that gives both a quotient and a remainder.

37.1 Writing and comparing whole numbers

Definition 37.1 (Place value)

Our way of writing numbers uses ten digits ( $0$ to $9$ ), and the value of a digit depends on its place: in $5\,208$ , the digit $5$ counts thousands, the $2$ counts hundreds, the $0$ counts tens (there are none) and the $8$ counts units:

5\,208 = 5 \times 1000 + 2 \times 100 + 0 \times 10 + 8 .

The place-value table of 5\,208. The 0 matters: without it, the 5 and the 2 would slide into the wrong columns and the number would read 528. — The place-value table of $5\,208$ . The $0$ matters: without it, the $5$ and the $2$ would slide into the wrong columns and the number would read $528$ .

Method 37.2 (Comparing two whole numbers)

The number with more digits is the larger one ( $1\,203 > 989$ ).
If they have the same number of digits, compare digit by digit from the left; the first difference decides: $6\,742 > 6\,715$ because at the tens place, $4 > 1$ .

The symbols are $<$ (“less than”) and $>$ (“greater than”); the small end of the symbol points at the small number.

Example 37.3

Order from smallest to largest: $908$ ; $89$ ; $1\,001$ ; $980$ . First by digit count: $89$ (two digits) comes first, $1\,001$ (four digits) last. Between $908$ and $980$ : same hundreds digit $9$ , then tens $0 < 8$ , so $908 < 980$ . Final answer:

89 < 908 < 980 < 1\,001 .

Numbers live on a line: smaller means further left. Each tick here is worth 100. — Numbers live on a line: smaller means further left. Each tick here is worth $100$ .

37.2 Addition, subtraction, multiplication

Example 37.4 (Column computations)

To add or subtract, align the units under the units, the tens under the tens, and work from right to left, carrying when needed:

\begin{array}{r} 4\,5\,7 \\ +\ 2\,8\,6 \\ \hline 7\,4\,3 \end{array} \qquad\qquad \begin{array}{r} 6\,0\,3 \\ -\ 1\,4\,7 \\ \hline 4\,5\,6 \end{array}

In the addition: $7 + 6 = 13$ , write $3$ , carry $1$ ; then $5 + 8 + 1 = 14$ , write $4$ , carry $1$ ; then $4 + 2 + 1 = 7$ . Check the subtraction by adding back: $456 + 147 = 603$ .

Remark 37.5 (Vocabulary)

The result of an addition is a sum; of a subtraction, a difference; of a multiplication, a product; of a division, a quotient. The numbers being combined are the terms (for $+$ , $-$ ) or factors (for $\times$ ).

Example 37.6 (Smart computation)

Reordering terms or factors often saves work:

17 + 58 + 3 = (17 + 3) + 58 = 20 + 58 = 78 ,

25 \times 17 \times 4 = (25 \times 4) \times 17 = 100 \times 17 = 1700 .

37.3 Division

Theorem 37.7 (Euclidean division)

Given two whole numbers $a$ (the dividend) and $b \neq 0$ (the divisor), there is exactly one way to write

a = b \times q + r \qquad\text{with}\qquad 0 \leq r < b .

The number $q$ is the quotient and $r$ the remainder of the division of $a$ by $b$ .

Proof. Admitted at this level. ∎

Example 37.8

Divide $137$ by $4$ , step by step.

How many times does $4$ go into $13$ ? Three times ( $4 \times 3 = 12$ ), remainder $13 - 12 = 1$ .
Bring down the $7$ : how many times does $4$ go into $17$ ? Four times ( $4 \times 4 = 16$ ), remainder $1$ .

So $q = 34$ and $r = 1$ :

137 = 4 \times 34 + 1, \qquad\text{and indeed } 0 \leq 1 < 4 .

Check: $4 \times 34 = 136$ , and $136 + 1 = 137$ .

Dividing 137 by 4: you can fit 34 full groups of 4 (reaching 136), and 1 is left over — the remainder is always smaller than the divisor. — Dividing $137$ by $4$ : you can fit $34$ full groups of $4$ (reaching $136$ ), and $1$ is left over — the remainder is always smaller than the divisor.

Definition 37.9 (Divisible)

When the remainder is $0$ , we say that $a$ is divisible by $b$ : for instance $135$ is divisible by $5$ since $135 = 5 \times 27 + 0$ .

Method 37.10 (Word problems with division)

After computing $a = bq + r$ , always return to the question:

“How many full boxes/teams/pages?” — the answer is the quotient $q$ ;
“How many are left over?” — the answer is the remainder $r$ ;
“How many boxes are needed for everyone?” — if $r > 0$ , one more than the quotient: $q + 1$ .

Example 37.11

$137$ students go on a trip in buses of $40$ seats. Division: $137 = 40 \times 3 + 17$ . Three buses carry $120$ students, and $17$ students remain — they need a bus too! So $4$ buses are needed, even though the quotient is $3$ .

37.4 Exercises

Exercise 37.1 ★

Write in digits: “three thousand and forty-seven”; “twenty thousand five hundred”; “one hundred four”. Then write $60\,013$ in words.

Solution

Solution of Exercise 37.1.

$3\,047$ ; $20\,500$ ; $104$ . $60\,013$ : “sixty thousand and thirteen”.

Exercise 37.2 ★

In the number $47\,209$ : what is the digit of the hundreds? The digit of the units? What does the digit $4$ count? Write the number as a sum of multiples of $1$ , $10$ , $100$ , … as in Definition 37.1.

Solution

Solution of Exercise 37.2.

Hundreds digit: $2$ ; units digit: $9$ ; the $4$ counts the ten-thousands. Decomposition:

47\,209 = 4 \times 10\,000 + 7 \times 1000 + 2 \times 100 + 0 \times 10 + 9 .

Exercise 37.3 ★

Copy and complete with $<$ or $>$ :

507 \;?\; 570, \qquad 1\,099 \;?\; 989, \qquad 23\,456 \;?\; 23\,465, \qquad 9\,999 \;?\; 10\,000 .

Solution

Solution of Exercise 37.3.

$507 < 570$ (tens: $0 < 7$ ); $1\,099 > 989$ (four digits beat three); $23\,456 < 23\,465$ (tens: $5 < 6$ ); $9\,999 < 10\,000$ (four digits against five).

Exercise 37.4 ★

Order from smallest to largest: $2\,043$ ; $978$ ; $2\,304$ ; $2\,034$ ; $1\,001$ .

Solution

Solution of Exercise 37.4.

By digit count, $978$ and $1\,001$ come before the three four-digit numbers starting with $2$ ; among those, compare the tens and units: $2\,034 < 2\,043 < 2\,304$ . Final order:

978 < 1\,001 < 2\,034 < 2\,043 < 2\,304 .

Exercise 37.5 ★

Compute in columns: $348 + 576$ ; $805 - 268$ ; $307 \times 26$ . Check the subtraction with an addition.

Solution

Solution of Exercise 37.5.

$348 + 576 = 924$ ; $805 - 268 = 537$ (check: $537 + 268 = 805$ ); $307 \times 26 = 307 \times 20 + 307 \times 6 = 6\,140 + 1\,842 = 7\,982$ .

Exercise 37.6 ★

Compute smartly, grouping terms or factors:

38 + 45 + 12 + 5, \qquad 2 \times 83 \times 50, \qquad 4 \times 78 \times 25 .

Solution

Solution of Exercise 37.6.

$38 + 45 + 12 + 5 = (38 + 12) + (45 + 5) = 50 + 50 = 100$ .

$2 \times 83 \times 50 = (2 \times 50) \times 83 = 100 \times 83 = 8\,300$ .

$4 \times 78 \times 25 = (4 \times 25) \times 78 = 100 \times 78 = 7\,800$ .

Exercise 37.7 ★

For each division, give the quotient and the remainder, and write the equality $a = b \times q + r$ :

95 \div 7, \qquad 230 \div 6, \qquad 504 \div 8 .

Solution

Solution of Exercise 37.7.

$95 = 7 \times 13 + 4$ (quotient $13$ , remainder $4$ ).

$230 = 6 \times 38 + 2$ (quotient $38$ , remainder $2$ ).

$504 = 8 \times 63 + 0$ (quotient $63$ , remainder $0$ : $504$ is divisible by $8$ ).

Exercise 37.8 ★

Without dividing, explain why the equality $173 = 8 \times 20 + 13$ is not the Euclidean division of $173$ by $8$ , and find the correct quotient and remainder.

Solution

Solution of Exercise 37.8.

The equality is true, but the remainder $13$ is not smaller than the divisor $8$ , so it is not the Euclidean division. Take one more $8$ out of the remainder: $173 = 8 \times 21 + 5$ , with $0 \leq 5 < 8$ : quotient $21$ , remainder $5$ .

Exercise 37.9 ★★

Eggs are packed in boxes of $12$ .

A farm has $2\,000$ eggs. How many full boxes can it fill, and how many eggs are left over?
A bakery needs $150$ eggs. How many boxes must it buy?

Solution

Solution of Exercise 37.9.

1. $2\,000 = 12 \times 166 + 8$ : the farm fills $166$ full boxes and $8$ eggs are left over.

2. $150 = 12 \times 12 + 6$ : twelve boxes contain only $144$ eggs, not enough. The bakery must buy $13$ boxes.

Exercise 37.10 ★★

Today is a Tuesday. What day of the week will it be in $100$ days? (Weeks have $7$ days: divide and think about the remainder.)

Solution

Solution of Exercise 37.10.

$100 = 7 \times 14 + 2$ : one hundred days are $14$ full weeks and $2$ days more. Fourteen weeks later it is again a Tuesday; two days after a Tuesday is a Thursday.

Exercise 37.11 ★★

In a division by $9$ , the quotient is $56$ and the remainder is the largest it can possibly be. What is the dividend?

Solution

Solution of Exercise 37.11.

The largest possible remainder in a division by $9$ is $8$ (the remainder must be smaller than the divisor). So the dividend is

9 \times 56 + 8 = 504 + 8 = 512 .

Exercise 37.12 ★★★

I am a three-digit number. My hundreds digit is twice my units digit, my tens digit is $0$ , and the sum of my digits is $9$ . Who am I? (Reason step by step, then check.)

Solution

Solution of Exercise 37.12.

Let the digits be $h$ (hundreds), $0$ (tens), $u$ (units). We know $h = 2 \times u$ and $h + 0 + u = 9$ , so $2u + u = 9$ , giving $3u = 9$ , $u = 3$ , and $h = 6$ . The number is $603$ . Check: $6$ is twice $3$ , the tens digit is $0$ , and $6 + 0 + 3 = 9$ .

37.5 Problem: Little Gauss and the sum of the first hundred numbers

Problem 37.1

Weekend problem — adding $1 + 2 + 3 + \dots + 100$ in ten seconds, and the triangular numbers

A famous story: to keep his class busy, a schoolmaster asked his pupils to add up all the numbers from $1$ to $100$ . One boy wrote $5\,050$ on his slate almost at once — Carl Friedrich Gauss, nine years old, who would grow into one of the greatest mathematicians of all time. His secret was nothing but smart computation (Example 37.6): grouping the terms so that every group is easy. This problem rediscovers his trick, makes it work in every situation, and shows how often these sums appear in daily life.

Part I — Gauss’s pairs.

Compute $1 + 2 + 3 + 4 + 5$ by grouping terms cleverly.
For $1 + 2 + 3 + \dots + 10$ : pair the first term with the last ( $1 + 10$ ), the second with the next-to-last ( $2 + 9$ ), and so on. What is each pair worth? How many pairs are there? Finish the computation.
Same method for $1 + 2 + \dots + 20$ : what is each pair worth, how many pairs, what total?
Now the schoolmaster’s sum, $1 + 2 + \dots + 100$ : explain why the pairs $1 + 100$ , $2 + 99$ , $3 + 98, \dots$ are all worth the same, and why there are exactly $50$ of them, the last one being $50 + 51$ .
Write Gauss’s answer as a single multiplication, and compute it.

Part II — A trick that never fails.

Try the pairing on $1 + 2 + \dots + 9$ : this time one number is left without a partner. Which one? Pair around it and compute the sum.
Here is a version of the trick that works whether the count of terms is even or odd: write the sum twice, once forwards and once backwards, one line under the other,

$\begin{array}{ccccccc} 1 & + & 2 & + & \dots & + & 9 \\ 9 & + & 8 & + & \dots & + & 1 \end{array}$

and add column by column. What is each column worth? How many columns are there? Explain why the sum of the two lines is $9 \times 10 = 90$ , and conclude that $1 + 2 + \dots + 9 = 45$ — the same answer as in question 6.
Use the two-line trick to compute $1 + 2 + \dots + 50$ , then $1 + 2 + \dots + 200$ .
Compute the sum of the first hundred even numbers, $2 + 4 + 6 + \dots + 200$ . (Compare each term with a term of Gauss’s sum: no new pairing is needed.)
Compute $101 + 102 + \dots + 200$ , using two sums you already know and a subtraction.

Part III — Triangular numbers everywhere. The totals $1$ , $3$ , $6$ , $10$ , $15, \dots$ of the sums $1 + 2 + \dots + n$ are called triangular numbers, because they count objects stacked in a triangle.

Eight friends meet, and each one shakes hands exactly once with each of the others. Explain why the number of handshakes is $7 + 6 + 5 + 4 + 3 + 2 + 1$ , and compute it.
A grocer builds a triangular pyramid of cans: $12$ cans in the bottom row, $11$ in the next, and so on up to a single can on top. How many cans does she need? (Use the trick of question 7.)
Draw the sum $1 + 2 + 3 + 4$ as a staircase of squares: one square in the first column, two in the second, three in the third, four in the fourth. Show on your drawing that two such staircases, one of them turned upside down, fit together exactly into a rectangle of $4$ rows of $5$ squares — and explain why this is the two-line trick of question 7, drawn instead of written.
A clock strikes $1$ time at one o’clock, $2$ times at two o’clock, …, $12$ times at twelve o’clock. How many strikes is that over the twelve hours? And over a whole day?
A last one for Gauss: $5\,050$ seconds — is that more or less than an hour and a half? Convert it into hours, minutes and seconds, using Euclidean division (Theorem 37.7) twice.

Solution

Solution of Problem 37.1.

1. For instance $(1 + 4) + (2 + 3) + 5 = 5 + 5 + 5 = 15$ .

2. Each pair is worth $11$ : $1 + 10$ , $2 + 9$ , $3 + 8$ , $4 + 7$ , $5 + 6$ . That is $5$ pairs (ten numbers, two per pair), so the sum is $5 \times 11 = 55$ .

3. Pairs worth $21$ ( $1 + 20$ , $2 + 19, \dots$ ), and there are $10$ of them: $10 \times 21 = 210$ .

4. In each pair, when the first number grows by $1$ the second shrinks by $1$ , so the total never changes: $1 + 100 = 2 + 99 = 3 + 98 = \dots = 101$ . The hundred numbers make $50$ pairs (two numbers each), the innermost being $50 + 51 = 101$ .

5. $50$ pairs of $101$ :

1 + 2 + \dots + 100 = 50 \times 101 = 5\,050 .

6. With nine numbers, the middle one, $5$ , has no partner. Around it: $1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 10$ , four pairs. Total: $4 \times 10 + 5 = 45$ .

7. Each column is worth $1 + 9 = 2 + 8 = \dots = 9 + 1 = 10$ , and there are $9$ columns: the two lines together are worth $9 \times 10 = 90$ . But the two lines are the same sum written twice, so one copy is worth half: $1 + 2 + \dots + 9 = 90 \div 2 = 45$ . No number is ever left without a partner — its partner is just underneath it.

8. For $1$ to $50$ : $50$ columns worth $51$ , double sum $50 \times 51 = 2\,550$ , so the sum is $2\,550 \div 2 = 1\,275$ . For $1$ to $200$ : $200$ columns worth $201$ , double sum $40\,200$ , sum $20\,100$ .

9. Every even number is the double of a number of Gauss’s sum: $2 = 2 \times 1$ , $4 = 2 \times 2$ , …, $200 = 2 \times 100$ . So the whole sum is the double of Gauss’s:

2 + 4 + \dots + 200 = 2 \times 5\,050 = 10\,100 .

10. The numbers from $101$ to $200$ are the numbers from $1$ to $200$ with the numbers from $1$ to $100$ removed:

101 + 102 + \dots + 200 = 20\,100 - 5\,050 = 15\,050 .

11. Line up the friends. The first shakes $7$ hands (one with each of the others). The second has already greeted the first, so he shakes $6$ new hands; the third, $5$ ; and so on, the seventh shaking $1$ and the eighth none that is new. Every handshake is counted exactly once:

7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 \text{ handshakes}

(pair around the middle: $(7+1) + (6+2) + (5+3) + 4 = 8 + 8 + 8 + 4 = 28$ ).

12. $1 + 2 + \dots + 12$ : twelve columns worth $13$ , double sum $12 \times 13 = 156$ , so $156 \div 2 = 78$ cans.

13. The staircase of $1 + 2 + 3 + 4$ squares, plus the same staircase upside down, fill a rectangle of $4$ rows and $5$ columns exactly: each row of the rectangle is one column of the two-line trick (a step of the rising staircase completed by a step of the falling one, $1 + 4$ , $2 + 3$ , $3 + 2$ , $4 + 1$ — always $5$ ). So twice the sum is $4 \times 5 = 20$ squares, and $1 + 2 + 3 + 4 = 10$ : the picture is question 7.

14. $1 + 2 + \dots + 12 = 78$ strikes in twelve hours (question 12), hence $78 \times 2 = 156$ strikes in a whole day.

15. Divide by $60$ : $5\,050 = 60 \times 84 + 10$ , so $5\,050$ seconds is $84$ minutes and $10$ seconds. Divide the minutes by $60$ : $84 = 60 \times 1 + 24$ , so $84$ minutes is $1$ hour $24$ minutes. In all: $1$ h $24$ min $10$ s — a little less than an hour and a half ( $1$ h $30$ min).