Today I noticed that $294$ is a multiple of $49$, which is the last two digits of $294$ reversed.

How many other numbers have this property? That is, how many three-digit numbers have a factor which is given by taking the last two digits and reversing them?


  • I don't know the answer to this question, but I imagine it must be doable in a relatively neat way without too much case-bashing. No programming answers, please.
  • Leading zeros are not allowed. When I say "three-digit numbers" it excludes one- and two- digit numbers, and the two-digit factor shouldn't be a one-digit number.
  • $\begingroup$ The programming answer is actually rather fun, but unfortunately out of scope. $\endgroup$ Aug 13, 2020 at 16:00
  • $\begingroup$ I know it says no-computers but I couldn't resist: (Python) sorted(set([n if n%10!=0 and n%int(str(n)[:0:-1])==0 else 0 for n in range(100,1000)]))[1:] $\endgroup$
    – Vepir
    Aug 15, 2020 at 13:31

1 Answer 1


There are


My process:

Eliminate the possibilities with a zero in the 2nd or 3rd digit, as we can't have a 1-digit factor, and having a zero as the last digit of the factor would require the result also end in zero.
Eliminate those where the last two digits are the same.
Eliminate the numbers where the 2nd digit is even, and the last digit is odd, as the an odd number is never divisible by an even number.
Eliminate the numbers where the 2nd digit is 5, as multiplying by 5 will make the last digit either 5 or 0, and we already know the last digit can't be zero.

At this point,

we really just need the last two digits. My idea was to multiply the reverse of the last two digits, to get a 3-digit number. There are 44 2-digit numbers to check, which is somewhat trivial to verify, as there are at most 2 numbers, that when multiplied by the last digit, will get the first digit as the final digit of the total.

The numbers, and divisors are:

713 / 31 = 23
819 / 91 = 9
924 / 42 = 22
328 / 82 = 4
962 / 26 = 37
972 / 27 = 36
679 / 97 = 7
384 / 48 = 8
294 / 49 = 6
295 / 59 = 5

  • $\begingroup$ Good answer and I've upvoted, but I wonder if there's a way of reducing the amount of case checking still further? Quite possibly there isn't, as I didn't have a solution in mind before posting, but intuitively it feels like modular arithmetic should help. $\endgroup$ Aug 13, 2020 at 18:19
  • 1
    $\begingroup$ You missed one... $\endgroup$ Aug 13, 2020 at 22:59
  • 1
    $\begingroup$ @DanielMathias found and added it. $\endgroup$
    – Herb
    Aug 14, 2020 at 0:42

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