No because 137 is a Prime number whose only factors are itself and one
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The square root of 137 is not an integer and so does not.
11 & 12... the square root of 133 is 11.532562594670795889354183238818
2 x 6 = 12 2 and 12 are factors of 12 2 and 6 are a factor pair of 12
It is: 137/12 = 11 with a remainder of 5
10, which has the factors 1,2,5 and 10. If the trivial pair 1 and the number itself are disallowed, the answer is 12 with factors 2,3,4 and 6.