# How to calculate 15! without using calculator

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

• "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? – Arthur Jan 30 at 11:02
• Are there two blanks there? Are the two blanks the same value? – Michael Burr Jan 30 at 11:03
• @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 $9$ – Henry Jan 30 at 11:10
• – NoChance Jan 30 at 20:42

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.

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 $$x=11-8=3$$.

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.