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Problem 2 (due Monday, March 2)
Recall that the smallest integer greater or equal than a given real number x is denoted by ⌈x⌉ and called the ceiling of x. Let p be a prime number and 1≤a<p an integer. Prove that the number
⌈(ap−1−1)pp−1⌉ is divisible by p. What can you say when a=p?
This problem was solved by only one participant: Yuqiao Huang. The submitted solution does not discuss the case when a=p. For 1≤a≤p−1, the solution claims correctly that ⌈(ap−1−1)pp−1⌉=ap−a, hence the result follows from Fermat's Little Theorem. In order to justify the claim, the solver proves that ap−a−1<(ap−1−1)pp−1≤ap−a. The proof of the right hand side inequality is fairly simple; the submitted proof of the left hand side inequality is rather long and complicated, and we will not reproduce it here. To see a detailed solution click the following link Solution