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The product of $2×653×733×977$ is a number with nine digits and it contains every digit once except for one, which digit is that?
I noticed that it is a product of primes, but so far I cannot solve it without multiplying it out. I suspect I can solve this using modular arithmetic but I have no idea how. Any help?

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  • $\begingroup$ seven... wolframalpha.com/input/?i=+2%C3%97653%C3%97733%C3%97977 $\endgroup$ – janmarqz Oct 8 '15 at 0:19
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    $\begingroup$ Hint: Do it mod $9$ $\endgroup$ – Robert Israel Oct 8 '15 at 0:19
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    $\begingroup$ @janmarqz Do you really think that is helpful? Anybody can use a calculator. $\endgroup$ – Thomas Andrews Oct 8 '15 at 0:20
  • $\begingroup$ "Do you really think that is helpful? Anybody can use a calculator" Well, sure. But tell them not to. It's a puzzle and a challenge to was mathematical reasoning. You can always use a calculator to check yourself. $\endgroup$ – fleablood Oct 8 '15 at 1:26
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Calculate the number $\bmod 9$

Full solution:

$\bmod 9$ the number is $2$, if we had each digit once the number would be $0\bmod 9$. If the missing digit is $x$ we have $0-x\equiv 2\bmod 9\implies x+2\equiv 0\bmod 9\implies x\equiv -2\equiv 7\bmod 9$ so the digit is $7$.

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Hint: The number is $2\pmod 9$. Which digit is missing?

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In mod 9 we have

2*653*733*977 = 2*5*4*5 = 2 mod 9

Now we make the following table :

0+1+2+3+4+5+6+7+8+() = 36 = 0 mod 9

0+1+2+3+4+5+6+7+()+9 = 37 = 1 mod 9

0+1+2+3+4+5+6+()+8+9 = 38 = ...

0+1+2+3+4+5+()+7+8+9 = 39 = ...

0+1+2+3+4+()+6+7+8+9 = 40 = ...

0+1+2+3+()+5+6+7+8+9 = 41 = ...

0+1+2+()+4+5+6+7+8+9 = 42 = ...

0+1+()+3+4+5+6+7+8+9 = 43 = ...

0+()+2+3+4+5+6+7+8+9 = 44 = ...

()+1+2+3+4+5+6+7+8+9 = 45 = ...

After that, the missing number is .....

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