Someone has recently tweeted:

Last year I had my 50th. The year before was my 100th. This year it's my 30th. #NotAllNumbersAreDecimal

What is he turning (base 10) this year?

There's nothing in the tweet to suggest there is only one solution. How many are there?


Let $x$ be his age on his birthday this year. Then $x-1$ is 50 in some base, i.e. a multiple of 5; $x-2$ is 100 in some base, i.e. a perfect square; and $x$ is 30 in some base, i.e. a multiple of 3.

By the Chinese remainder theorem, being a multiple of 3 and one more than a multiple of 5 is exactly equivalent to being 6 more than a multiple of 15 (congruent to 6 mod 15). So $x$ must be one of 6, 21, 36, 51, 66, 81, 96, 111, 126, 141 (assuming $x<150$, which seems reasonable). Subtract 2 from each of these to see whether we get a square, and we find that the only possible values of $x$ are

6, 51, and 66.

Let's check each of these individually. 6 turns out to be too small since 5 isn't written as 50 in any base (thanks @KateGregory!), but both the other two seem to be eligible.

  • 1
    $\begingroup$ 6 doesn't work. The base is too small - you can't write 50 to represent it. $\endgroup$ – Kate Gregory Jun 21 '15 at 14:44
  • $\begingroup$ @KateGregory Thanks - I've fixed that. Is it plausible that there's not a unique solution? Nice puzzle btw :-) $\endgroup$ – Rand al'Thor Jun 21 '15 at 14:47
  • $\begingroup$ I'm not going to change the question on you, but is there a "neat" number he could tack onto the series that would eliminate one of the two? $\endgroup$ – Kate Gregory Jun 21 '15 at 14:56
  • $\begingroup$ @KateGregory Simply by adding another case you'll limit the other possibility. Convert 63 (or 48) to another base and say (the year before the year before last year was X). You can also do the next year to be less confusing. $\endgroup$ – Quark Jun 21 '15 at 14:59
  • 1
    $\begingroup$ @KateGregory "Next year I'll be 40" would restrict it to 51. "Two years ago I was also 1000" would restrict it to 66, but doesn't look so nice. $\endgroup$ – Rand al'Thor Jun 21 '15 at 15:11

There are two potential answer:

If 50 was base 10, 100 was base 7 (49), and 30 was base 17 (51), it would work. So 51 would work. The other possible answer is that if 30 was base 22 (66), 50 was base 13 (65), and 100 was base 8 (64), so the answer would be 66. Credit to rand al'thor for finding the other possible answer.

  • $\begingroup$ is that the only combination that works? $\endgroup$ – Kate Gregory Jun 21 '15 at 14:44
  • $\begingroup$ @KateGregory I agree with rand that there are two answers, I solved for the bases they'd require but his answer shows why there aren't others. $\endgroup$ – Quark Jun 21 '15 at 14:54

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