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
an = 994
n = ?
As we know that;
an = a+(n−1)d
994 = 105+(n−1)7
889 = (n−1)7
(n−1) = 127
n = 128
i.e 128 three digit numbers are divisible by 7.