For how many values of b ∈ N does 2 0 1 8 1 0 end with … 2 0 2 when written in base b ?
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We can check in small bases and find that
2 0 1 8 1 0 = 2 2 0 2 2 0 2 3 = 1 3 3 2 0 2 4 = 1 3 2 0 2 6 = 1 2 0 2 1 2
Since the numbers keep getting shorter with every base, after 1 2 0 2 there is only the possibility 2 0 2 left. This means that
2 ⋅ b 2 + 0 ⋅ b 1 + 2 ⋅ b 0 = 2 0 1 8 ⇔ b = ± 7 1 2 ∈ / N
So, b = 3 , 4 , 6 , 1 2 are the 4 only possibilities.
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We seek positive integers a , b , with b > 2 , such that a b 3 + 2 b 2 + 2 = 2 0 1 8 or a b 3 + 2 b 2 = 2 0 1 6 = 2 5 3 2 7 . Thus b 2 must divide 2 0 1 6 ; the only candidates are b = 3 , 4 , 6 , 1 2 . We check that all of those "work"; thus the answer is 4 .