#math trivia for #May9: #130 can be expressed with two 1s and the rest (if any) all 0s in what bases?
Answer. This is equivalent to asking for which bases b can 130 be expressed as
130 = b^x + b^y
for distinct x, y. Without loss of generality, let y be the smaller of the two exponents. Then b^y must divide into 130, which means that the possible values for b^y are 1, 2, 5, 10, 13, 26 and 65.
Now consider each value of b^y in turn:
- 1: y = 0 and b^x = 129, so b = 129, x = 1, y = 0. 130 is 11 base 129.
- 2: y = 1, b = 2, and b^x = 128, so x = 7. 130 is 10000010 base 2.
- 5: y = 1, b = 5, and b^x = 125, so x = 3. 130 is 1010 base 5.
- 10: y = 1, b = 10, and b^x = 120; no solution for x.
- 13: y = 1, b = 13, and b^x = 117; no solution for x
- 26: y = 1, b = 26, and b^x = 104; again no solution.
- 65: y = 1, b = 65, and b^x = 65; no solution because x would be the same as y.
There are therefore three bases that work: 129, 2 and 5.
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