Smallest = 105, largest = 994. Use AP: a=105, d=7, last term=994. Find n.

How Many Three-Digit Numbers Are Divisible by 7?

Question

How many three-digit numbers are divisible by 7?

This is a classic CBSE board question — appeared in CBSE 2024 — and it tests whether you can recognize an AP and apply the nth term formula correctly.

Solution — Step by Step

The smallest three-digit number is 100. Divide: $100 \div 7 = 14.28...$ , so 100 itself isn’t divisible by 7.

The next multiple is $7 \times 15 = 105$ . So our first term is $a = 105$ .

The largest three-digit number is 999. Divide: $999 \div 7 = 142.71...$ , so 999 doesn’t work.

Go down: $7 \times 142 = 994$ . So our last term is $l = 994$ .

All multiples of 7 form an AP with common difference $d = 7$ . Our sequence is:

105, 112, 119, \ldots, 994

We need $n$ , the number of terms.

The formula for the last term of an AP is:

l = a + (n-1)d

Substituting:

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

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

889 = (n-1) \times 7

n - 1 = \frac{889}{7} = 127

n = 128

There are 128 three-digit numbers divisible by 7.

Why This Works

Multiples of any fixed integer $k$ always form an AP with $d = k$ . Once we spot that, the problem reduces to: given first term, last term, and common difference — find the count.

The nth term formula $l = a + (n-1)d$ is doing the counting for us. It’s asking: “how many steps of size $d$ does it take to get from $a$ to $l$ ?” That’s exactly what $n-1$ represents.

This same logic applies to any “how many integers between X and Y are divisible by k?” question. Just find the first and last valid terms, set up the AP, and solve. It’s a one-minute question once the pattern is clear.

Alternative Method

You can get $n$ directly using a division formula — no AP formula needed:

n = \left\lfloor \frac{999}{7} \right\rfloor - \left\lfloor \frac{99}{7} \right\rfloor

n = 142 - 14 = 128

This floor division method is faster in MCQ settings. $\lfloor 999/7 \rfloor$ counts all positive integers up to 999 divisible by 7. Subtract $\lfloor 99/7 \rfloor$ to remove the one- and two-digit ones. Same answer, fewer steps.

For board exams, show the AP method with working — it’s what the marking scheme expects. Use the floor method to verify quickly.

Common Mistake

The most common error is taking $a = 100$ or $l = 999$ without checking divisibility. Students assume the boundary values are automatically included. Always verify: $100 \div 7$ is not an integer, and $999 \div 7$ is not an integer. Starting with the wrong first or last term gives $n = 129$ or $n = 127$ — both wrong, and both a guaranteed mark loss in boards.

A secondary mistake: after getting $n - 1 = 127$ , some students write $n = 127$ and forget the $+1$ . Write the step $n = 127 + 1$ explicitly in your answer — don’t do it in your head.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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