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√n differ from √100

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by sanju09 » Mon Mar 07, 2011 1:35 am
For how many different positive integers n does √n differ from √100 by less than 1?
(A) 38
(B) 39
(C) 40
(D) 41
(E) 42
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by Anurag@Gurome » Mon Mar 07, 2011 2:06 am
sanju09 wrote:For how many different positive integers n does √n differ from √100 by less than 1?
(A) 38
(B) 39
(C) 40
(D) 41
(E) 42
Now 9 and 11 differs exactly by 1 from √100 = 10
Hence, √n must lie between 9 and 11.

Therefore, n must lie between 9² = 81 and 11² = 121

Hence, number of such positive integers = (121 - 81 - 1) = 39

The correct answer is B.
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by amit31 » Tue Mar 08, 2011 8:54 am
Anurag@Gurome wrote:
sanju09 wrote:For how many different positive integers n does √n differ from √100 by less than 1?
(A) 38
(B) 39
(C) 40
(D) 41
(E) 42
Now 9 and 11 differs exactly by 1 from √100 = 10
Hence, √n must lie between 9 and 11.

Therefore, n must lie between 9² = 81 and 11² = 121

Hence, number of such positive integers = (121 - 81 - 1) = 39

The correct answer is B.
But shouldn't we exclude the 81 and 121 since question says less than 1. If we include both 81 and 121 it would mean they differ by 1 and not by less than 1
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by rros0770 » Tue Mar 08, 2011 3:58 pm
Hi Amit-

Anurag did exclude 121 and 81

By subtracting 1 from the difference between 121 - 81, we are excluding the two extremes from the total count
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