NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.2
Question 13

Question. 13

How many three-digit numbers are divisible by 7?

Answer:

128

Detailed Answer with Explanation:

Step 1: Identify the range of three-digit numbers.

The smallest three-digit number is 100 and the largest is 999.

Step 2: Find the smallest three-digit number divisible by 7.

Divide 100 by 7: \(100 \div 7 \approx 14.28\). The next whole number is 15.

So the first three-digit multiple of 7 is:

\[7 \times 15 = 105\]

Step 3: Find the largest three-digit number divisible by 7.

Divide 999 by 7: \(999 \div 7 \approx 142.71\). Take the whole number part 142.

So the last three-digit multiple of 7 is:

\[7 \times 142 = 994\]

Step 4: Observe that these form an AP.

The numbers divisible by 7 between 100 and 999 form an arithmetic progression:

\[105, 112, 119, \ldots, 994\]

Here, first term \(a = 105\), common difference \(d = 7\), last term \(l = 994\).

Step 5: Use the formula for the number of terms in an AP.

For an AP, the number of terms \(n\) is:

\[n = \frac{l - a}{d} + 1\]

Substitute the values:

\[n = \frac{994 - 105}{7} + 1\]

\[n = \frac{889}{7} + 1\]

\[n = 127 + 1 = 128\]

Conclusion: There are 128 three-digit numbers that are divisible by 7.

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