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!
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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.