10 & 11
Because
10^(2) = 100
11^(2) = 121
And 103 fall between 100 & 121
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The square root of 103 is roughly 10.14889157, so it is between root 100 (10) and root 121 (11).
They are; 8 and 9 but the square root of 68 is about 8.246211251
-5 and -4 or 4 and 5
Call the integer square roots of the specified pair of numbers l and g for lesser and greater respectively. Then, from the problem statement, g2 - l2 = 105. Possible values for l2 are successively 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, .... , and corresponding values for g2 are 106, 109, 114, 121, ... etc. The first of the latter series that is a perfect square number is 121. Therefore, the square numbers are 121 and 16.
sqrt(89) = -9.4 or +9.4 So one possible pair of consecutive integers between the square roots of 89 are 2 and 3.
Ordered pair (s)