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I am joining a maths competition and recently I am preparing for it. I came across a question that asks me to fill the blank of a number:

1_0767436_000

And this number is the product of $15!= 15\times 14\times 13\times 12...\times 1$.

The competition doesn't allow to use a calculator, so I am wondering how to multiply these without calculator. Can I have a solution that can solve a similar question and also the solution to this question? Tqvm

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    $\begingroup$ "solution that can solve similar question" is a very tall order. Questions of this type are highly individual, and there is no one solution to fit all. Especially if this is for a competition, where problems are usually explicitly constructed to not be too similar to earlier problems. What does help, though, is lots of experience and a healthy "try a bunch of things" attitude. So, with that in mind, have you tried a bunch of things? What did you try? How did it go? $\endgroup$ – Arthur Jan 30 at 11:02
  • $\begingroup$ Are there two blanks there? Are the two blanks the same value? $\endgroup$ – Michael Burr Jan 30 at 11:03
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    $\begingroup$ @MichaelBurr - they are in fact not the same value - the second is obviously even and the first will then be different to get all the digits to add up to a multiple of $9$ $\endgroup$ – Henry Jan 30 at 11:10
  • $\begingroup$ See: math.stackexchange.com/questions/1163638/… and quora.com/… $\endgroup$ – NoChance Jan 30 at 20:42
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As the comments above mention, these types of problems are usually ad hoc. For this one, for example, you can try the following:

Sketch: The lowest nonzero should be easy to calculate via $\mod{10}$ calculations (after dividing out by the three factors of $10$). Then, for the other number, observe that $15!$ is divisible by $9$ and use a divisiblity test.

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Let $x$ and $y$ be the left and right missing digits respectively. First apply the divisibility test for $9$, which demands that the sum of all digits be divisible by $9$ for $1×2×...×9×...×15$. Thereby

$x+y\in \{2,11\}$

Now count factors of $2$ and $5$ in the factorial. There is a factor of $5$ coming from each of $5,10,15$ so three factors of $5$. There are factors of $2$ coming from $2,4,...,14$, with each multiple of $4$ providing another factor of $2$ and $8$ providing an additional factor beyond those. Thus $11$ factors of $2$, eight more than the factors of $5$. So there are only three terminal zeroes and the remaining digits must be divisible by $2^8$, thus the last three digits before the zeroes must be a multiple of $8$.

Both $360$ and $368$ satisfy divisibility by $8$, but there cannot be a fourth terminal zero so $y=8$ forcing $x=11-8=3$.

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Usually math competitions are about ideas rather than calculations. I would be surprised if calculating 15! would be non-avoidable. It might be that considering properties of 15! Is helpful for some problems and when this is the case considering prime factors is usually a good idea. I recommend to read about Legendre’s Formula.

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