How many three-digit numbers are divisible by ? (Hint: All three-digit numbers divisible by form an AP. Find the smallest and largest such three-digit numbers.)

Question

Homi · Accepted Answer

Understand the Question

The three-digit numbers range from $100$ to $999$ .
All three-digit numbers divisible by $7$ form an Arithmetic Progression (AP) with a common difference of $d = 7$ .
We find the first term ( $a$ ) and the last term ( $a_n$ ) in this range, then use the $n$ -th term formula $a_n = a + (n - 1)d$ to find the total number of terms ( $n$ ).

Step 1 · Find the First and Last Three-Digit Multiples of 7

The smallest three-digit number is $100$ . Dividing $100$ by $7$ : $100 = 7 \times 14 + 2$

To find the first three-digit number divisible by $7$ : $100 + (7 - 2) = 105$

The largest three-digit number is $999$ . Dividing $999$ by $7$ : $999 = 7 \times 142 + 5$

To find the last three-digit number divisible by $7$ : $999 - 5 = 994$

Thus, the sequence of three-digit numbers divisible by $7$ is: $105, 112, 119, \dots, 994$

Step 2 · Calculate the Number of Terms

This sequence forms an AP where:

First term, $a = 105$
Common difference, $d = 7$
Last term, $a_n = 994$

Using the formula for the $n^{\text{th}}$ term of an AP: $a_n = a + (n - 1)d$

Substitute the values:

\begin{aligned} 994 &= 105 + (n - 1) \times 7 \\[0.6em] 994 - 105 &= 7(n - 1) \\[0.6em] 889 &= 7(n - 1) \\[0.6em] n - 1 &= \dfrac{889}{7} \\[0.6em] n - 1 &= 127 \\[0.6em] n &= 128 \end{aligned}

Answer

$128$

Common Mistakes

First Term Calculation Error: Subtracting the remainder from $100$ gives $98$ (a two-digit number). Always add $(7 - \text{remainder})$ to get the smallest three-digit multiple: $100 + (7 - 2) = 105$ .
Off-by-One Error: Forgetting to add $1$ after dividing: $n - 1 = 127 \implies n = 128$ , not $127$ .

Question 3

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