The question is given from ncert Book of class 10th Chapter no. 5 Ex. 5.2 Q. 13. In the following question you have to find the number of three digit numbers divisible by 7. Give the solution of the above question.

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Solution:The 1st Three Digit number that is divisible by seven is,

First no. = 105

Second no. = 105+7 = 112

Third no.= 112+7 =119

Therefore, 105, 112, 119, …

All are three digit numbers are divisible by 7 and thus, all these are terms of an Arithmetic Progression having 1st term as 105 and c.d as 7.

The largest possible three-digit number is 999.

When we divide 999 by 7, 5 will be the remainder.

i.e, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.

Now the series is:

105, 112, 119, …, 994

Let 994 be the nth term of this Arithmetic Progression of this A.P.

a = 105

d = 7

a

_{n}= 994n = ?

As we know that;

a

_{n}= a+(n−1)d994 = 105+(n−1)7

889 = (n−1)7

(n−1) = 127

n = 128

i.e 128 three digit numbers are divisible by 7.