Question 17 of 44intermediate🔧 ApplyShort Answer3 marks

How many three-digit numbers are divisible by 7?

Correct Answer

The three-digit numbers divisible by 7 form an AP: 105, 112, 119, . . . , 994. Here, the first term is a=105a = 105, the common difference is d=7d = 7, and the last term is an=994a_n = 994. We need to find the number of terms, nn.

Using the formula an=a+(n−1)da_n = a + (n - 1)d:

994=105+(n−1)×7994 = 105 + (n - 1) \times 7 889=(n−1)×7889 = (n - 1) \times 7 n−1=127n - 1 = 127 n=128n = 128

Therefore, there are 128 three-digit numbers divisible by 7.

Exercise: EXERCISE 5.2 | Q: 13 | (Page Page 14)
For More Understanding

Explanation

This question is solved exactly like Example 7 in the provided context. The student identifies the sequence of numbers as an Arithmetic Progression (AP). The first term 'a' is the smallest 3-digit multiple of 7 (105), and the last term 'an' is the largest 3-digit multiple of 7 (994). The common difference 'd' is 7. Applying the formula for the nth term allows calculation of the total count.

Solution Steps

  1. Step 1: Identify the AP. The sequence is 105, 112, 119, ..., 994. Here, a = 105, d = 7, an = 994.

  2. Step 2: Apply the nth term formula: an = a + (n - 1)d.

  3. Step 3: Substitute the values: 994 = 105 + (n - 1) * 7.

  4. Step 4: Solve for n: 889 = (n - 1) * 7, which gives n - 1 = 127, so n = 128.

PreviousNext