Finding Exponent
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Professor Calculus gives the following problem to his students: given two integers X (≥ 2) and Y (≥ 2), find the smallest positive integral exponent E such that the decimal expansion of XE begins with Y. For example, if X = 8 and Y = 51, then X3 = 512 begins with Y = 51, so E = 3. It is easy to see that neither E = 1 nor E = 2 will satisfy this property, so 3 is the smallest possible value of E in
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solution.cppC++17
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