How many three-digit numbers are divisible by 7?

Question

Homi · Accepted Answer

We will find the first and last three-digit numbers divisible by 7. Then we will use the arithmetic progression formula.

Step 1 — Find the first term

Let's find the smallest three-digit number. This number is 100. We divide 100 by 7.

$100 = 7 \times 14 + 2$

The remainder is 2. To make it divisible by 7, we add $(7 - 2)$ .

$100 + (7 - 2)$

$= 100 + 5$

$\boxed{105}$

Step 2 — Find the last term

Let's find the largest three-digit number. This number is 999. We divide 999 by 7.

$999 = 7 \times 142 + 5$

The remainder is 5. To make it divisible by 7, we subtract the remainder.

$999 - 5$

$\boxed{994}$

Step 3 — Count the numbers

The numbers form an arithmetic progression. The first term ( $a$ ) is 105. The common difference ( $d$ ) is 7. The last term ( $a_n$ ) is 994. We use the formula $a_n = a + (n-1)d$ .

$994 = 105 + (n-1) \times 7$

$994 - 105 = (n-1) \times 7$

$889 = (n-1) \times 7$

$\frac{889}{7} = n-1$

$127 = n-1$

$n = 127 + 1$

$\boxed{128}$

Answer

(i) There are 128 three-digit numbers divisible by 7.

Question 13

Answer

More questions in Exercise 5.2