---
title: "Whole Numbers"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 37
exercises: 12
source: https://one-course.com/books/math/1/en/chapter/37-whole-numbers
---

# Chapter 37 — Whole 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) *and* a [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-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.](https://one-course.com/images/onecourse/chapters/math-1/g6-wholes/fig-b71a98be01c7.svg)

*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).**

1. The number with more digits is the larger one ( $1\,203 > 989$ ).
2. 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.](https://one-course.com/images/onecourse/chapters/math-1/g6-wholes/fig-9fc3a4fd334b.svg)

*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](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple)*), 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.

1. How many times does $4$ go into $13$ ? Three times ( $4 \times 3 = 12$ ), [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $13 - 12 = 1$ .
2. Bring down the $7$ : how many times does $4$ go into $17$ ? Four times ( $4 \times 4 = 16$ ), [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-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.](https://one-course.com/images/onecourse/chapters/math-1/g6-wholes/fig-dfef293fb775.svg)

*Dividing $137$ by $4$: you can fit $34$ full groups of $4$ (reaching $136$), and $1$ is left over — the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) is always smaller than the [divisor](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple).*

**Definition 37.9 (Divisible).**

When the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-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:

1. “How many full boxes/teams/pages?” — the answer is the [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $q$ ;
2. “How many are left over?” — the answer is the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $r$ ;
3. “How many boxes are needed for everyone?” — if $r > 0$ , one *more* than the [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) : $q + 1$ .

**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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) 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 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of [multiples](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) of $1$, $10$, $100$, … as in [Definition 37.1](#def-g6-wholes-place).

**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 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 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 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 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) and the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder), and write the equality $a = b \times q + r$:

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

**Solution of Exercise 37.7.**

$95 = 7 \times 13 + 4$ ([quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $13$, [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $4$).

$230 = 6 \times 38 + 2$ ([quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $38$, [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $2$).

$504 = 8 \times 63 + 0$ ([quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $63$, [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $0$: $504$ is [divisible](#def-g6-wholes-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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) and [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder).

**Solution of Exercise 37.8.**

The equality is true, but the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $13$ is *not* smaller than the [divisor](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) $8$, so it is not the Euclidean division. Take one more $8$ out of the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder): $173 = 8 \times 21 + 5$, with $0 \leq 5 < 8$: [quotient](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $21$, [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $5$.

**Exercise 37.9 ★★.**

Eggs are packed in boxes of $12$.

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

**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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder).)

**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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) is $56$ and the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) is the largest it can possibly be. What is the dividend?

**Solution of Exercise 37.11.**

The largest possible [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) in a division by $9$ is $8$ (the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) must be smaller than the [divisor](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple)). 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of my digits is $9$. Who am I? (Reason step by step, then check.)

**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](#ex-g6-wholes-smart)): 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) appear in daily life.

**Part I — Gauss’s pairs.**

1. Compute $1 + 2 + 3 + 4 + 5$ by grouping terms cleverly.
2. 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.
3. Same method for $1 + 2 + \dots + 20$ : what is each pair worth, how many pairs, what total?
4. Now the schoolmaster’s [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) , $1 + 2 + \dots + 100$ : 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 $50 + 51$ .
5. Write Gauss’s answer as a single multiplication, and compute it.

**Part II — A trick that never fails.**

6. 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) .
7. Here is a version of the trick that works whether the count of terms is [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) or [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd): write the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the two lines is $9 \times 10 = 90$, and conclude that $1 + 2 + \dots + 9 = 45$ — the same answer as in question 6.
8. Use the two-line trick to compute $1 + 2 + \dots + 50$ , then $1 + 2 + \dots + 200$ .
9. Compute the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the first hundred *[even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)* numbers, $2 + 4 + 6 + \dots + 200$ . (Compare each term with a term of Gauss’s [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) : no new pairing is needed.)
10. Compute $101 + 102 + \dots + 200$ , using two [sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) you already know and a subtraction.

**Part III — [Triangular numbers](#pb-g6-wholes-1) everywhere.** The totals $1$, $3$, $6$, $10$, $15, \dots$ of the [sums](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $1 + 2 + \dots + n$ are called *triangular numbers*, because they count objects stacked in a triangle.

11. 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.
12. A grocer builds a triangular [pyramid](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) 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.)
13. Draw the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $1 + 2 + 3 + 4$ 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.
14. 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?
15. A last one for Gauss: $5\,050$ seconds — is that more or less than an hour and a [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) ? Convert it into hours, minutes and seconds, using Euclidean division ( [Theorem 37.7](#thm-g6-wholes-division) ) twice.

**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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) written twice, so one copy is worth [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $50 \times 51 = 2\,550$, so the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) is $2\,550 \div 2 = 1\,275$. For $1$ to $200$: $200$ columns worth $201$, double [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $40\,200$, [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $20\,100$.

**9.** Every [even number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) is the double of a number of Gauss’s [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def): $2 = 2 \times 1$, $4 = 2 \times 2$, …, $200 = 2 \times 100$. So the whole [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $12 \times 13 = 156$, 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) ($1$ h $30$ min).