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Now 9 and 11 differs exactly by 1 from √100 = 10sanju09 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
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 1Anurag@Gurome wrote:Now 9 and 11 differs exactly by 1 from √100 = 10sanju09 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
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.
