2018 in all bases

For how many values of b N b \in \mathbb{N} does 201 8 10 2018_{10} end with 202 \ldots 202 when written in base b b ?


The answer is 4.

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2 solutions

Otto Bretscher
Oct 28, 2018

We seek positive integers a , b a,b , with b > 2 b>2 , such that a b 3 + 2 b 2 + 2 = 2018 ab^3+2b^2+2=2018 or a b 3 + 2 b 2 = 2016 = 2 5 3 2 7 ab^3+2b^2=2016=2^5 3^2 7 . Thus b 2 b^2 must divide 2016 2016 ; the only candidates are b = 3 , 4 , 6 , 12 b=3,4,6,12 . We check that all of those "work"; thus the answer is 4 \boxed{4} .

Henry U
Oct 28, 2018

We can check in small bases and find that

201 8 10 = 220220 2 3 = 13320 2 4 = 1320 2 6 = 120 2 12 2018_{10} = 2202202_3 = 133202_4 = 13202_6 = 1202_{12}

Since the numbers keep getting shorter with every base, after 1202 1202 there is only the possibility 202 202 left. This means that

2 b 2 + 0 b 1 + 2 b 0 = 2018 b = ± 7 12 N 2 \cdot b^2 + 0 \cdot b^1 + 2 \cdot b^0 = 2018 \Leftrightarrow b = \pm 7 \sqrt{12} \notin \mathbb{N}

So, b = 3 , 4 , 6 , 12 b = 3,4,6,12 are the 4 only possibilities.

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