14!=871a82b1200 without working out 14!, find a and b

I think it has something to do with maths rules regarding 9 or 3 (the digits adding up to either of those numbers) but not entirely sure!

  • $\begingroup$ That's a good guess. Have you tried pursuing it and seeing where it leads you? $\endgroup$ – MJD Feb 24 '15 at 18:43
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    $\begingroup$ You can also use a rule related to 11, I think. $\endgroup$ – user84413 Feb 24 '15 at 18:44
  • $\begingroup$ May be if you divide by 10 or 20 or a number judisiously chose and after you make a system of two equations with two unknown quantities, but I think you need a computer.:) $\endgroup$ – Hexacoordinate-C Feb 24 '15 at 18:51
  • $\begingroup$ @Shadock: You don't need a computer. $\endgroup$ – Jim Feb 24 '15 at 18:51
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    $\begingroup$ Also divisible by $7\times 11 \times 13=1001$ in case that helps any $\endgroup$ – Mark Bennet Feb 24 '15 at 19:08

The rule for finding the remainder when dividing by $3$ is to sum up the digits and divide THAT number by $3$, the remainders will be the same. As $14!$ is divisible by $3$ the remainder should be zero, so $$8 + 7 + 1 + a + 8 + 2 + b + 1 + 2 = 29 + a + b$$ should be divisible by $3$. To make it easier we can take factors of $3$ out of the $29$ and conclude that $2 + a + b$ should be divisible by $3$.

The rule for remainders when dividing by $9$ is the same, sum up the digits and divide THAT by $9$. There's also a rule for $11$ involving the alternating sum of digits. All these give you equations like the one I got above. Try and write down those equations and see if you can do the last step of solving them on your own.

  • $\begingroup$ Checking for divisibility by 3 is a waste of time; it can't tell you anything that the divisibility-by-9 test doesn't also tell you. $\endgroup$ – MJD Feb 24 '15 at 18:48
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    $\begingroup$ @MJD: What checking divisibility by $3$ does is let me explain the process explicitly without giving away the answer :) $\endgroup$ – Jim Feb 24 '15 at 18:53
  • $\begingroup$ I agree completely! Thanks for explaining. $\endgroup$ – MJD Feb 24 '15 at 22:16

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