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Given log₅(N) = I1 + f1 and log₃(N) = I2 + f2, where I1 and I2 are non-negative integers and f1, f2 are in [0, 1). If I1 = 3 and I2 = 4, find the number of possible integral values of N.

1. 119
2. 116
3. 117
4. 118

Correct answer: 118

Solution

log₅(N) = I1 + f1 with I1=3 and f1 in [0,1) means 3 <= log₅(N) < 4, so 5³ <= N < 5⁴, i.e., 125 <= N <= 624. log₃(N) = I2 + f2 with I2=4 and f2 in [0,1) means 4 <= log₃(N) < 5, so 3⁴ <= N < 3⁵, i.e., 81 <= N <= 242. Intersection: max(125,81) <= N <= min(624,242), i.e., 125 <= N <= 242. Count = 242 - 125 + 1 = 118.

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