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8712 and 9801 are the only 4-digit numbers which are multiples of their reversals:

8712 = 4*2178 and 9801 = 9*1089.

Without using a calculator/computer, can you find two 5-digit numbers with this property? Note that palindromes and multiples of 10 are not allowed.

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  • $\begingroup$ The reversal of a multiple of ten wouldn’t have the same number of digits, so these don’t have to be explicitly excluded. $\endgroup$ – Lawrence Jun 5 at 4:56
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They are:

10989 (ie. 98901), 21978 (ie. 87912)

Reasoning.

I'll rule out 1xxx2, 2xxx4, 3xxx6, 4xxx8, 1xxx3, 2xxx6, 3xxx9, 1xxx4, 1xxx5, 1xxx6, 1xxx7, 1xxx8 because it's impossible for exact multiples due to the possible quotients and units digits.

Now,

That leaves 2xxx8 and 1xxx9. For 1xxx9, it has to be 10xx9 because otherwise if the second digit were 1, it would need to be 11x99 which is clearly impossible. (More than 1 would lead to 6 digits) From here if we create an equation and simplify, we see the fourth digit would be 8. From there we can check 10989 (the only option where multiplying by 9 yields 98xxx), which works.

For the second case:

2abc8 * 4 = 8cba2. a has to be less than 3 so the 10000-s digit of the product is 8. 8cb02 and 8cb22 cannot be multiples of 4, leaving 21bc8*4=8cb12. Again we can write equations, getting c=2 or 7. 21b28 can't work since 21028*4>82000. So, we just need to check b=7,8,9 (again so multiplying by for 21b78 to get 21978 which works.

Sorry if my writing is bad, I'm new.

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  • $\begingroup$ Great start. Your second answer is correct, but it should be reversed. The first answer answer does not work. $\endgroup$ – Dmitry Kamenetsky Jun 5 at 3:07
  • $\begingroup$ They’ve got the right answer in the second spoiler box, @DmitryKamenetsky $\endgroup$ – El-Guest Jun 5 at 3:31
  • $\begingroup$ (I’ve edited the answer to reflect this) $\endgroup$ – El-Guest Jun 5 at 3:33
  • $\begingroup$ Sorry, it was a typo, and I somewhat misunderstood the question (I thought it was the other way around) $\endgroup$ – Forgotten_User Jun 5 at 3:39
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    $\begingroup$ Thanks @El-Guest for fixing it $\endgroup$ – Forgotten_User Jun 5 at 3:39

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