*Spoiler for New Scientist Enigma 1742: “Clive’s number” *(Follow the link to see the puzzle.)

I think the obvious thing to start with is that, since FIVE is divisible by 5 and no letter is associated with 0, E must be 5. That means THREE and NINE both are divisible by 5, so they are not powers of 3 or 9 respectively. The only 4-digit powers of 4 are 1024 and 4096, both of which contain 0 so cannot be FOUR, and the only 5-digit power of 7 is 16807 which does not match SEVEN. This then means the consecutive digits whose corresponding words are powers of those digits must be 5 and 6; the only 4-digit power of 5 is 5^{5} = 3125 and the only 3-digit power of 6 is 6^{3} = 216. So F is 3, I is 1, S is 2, V is 2, and X is 6.

Then SEVEN = 2525N and we need N=6 to get a multiple of 7.

Then NINE = 6165 which is indeed a multiple of 9. (Use the usual test for divisibility by 9: 6+1+6+5 = 18 which is a multiple of 9, therefore 6165 is too.)

Now it’s tempting to try to figure out what TWO, THREE, FOUR, and EIGHT are, but you don’t need to — indeed, there’s no unique answer. Instead go ahead with SCIENTIST. Use the test for divisibility by 11: S–C+I–E+N–T+I–S+T must be 0 (mod 11). The Ss and Ts cancel and we can fill in values for I, E, and N: –C+1–5+6+1 = –C+3 = 0 so C = 3.

Now for divisibility of CLIVE by 11: C–L+I–V+E = 3–L+1–2+5 = –L+7 = 0 so L = 7.

Then CLIVE = 37125.

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