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
On
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 $x=11-8=3$.
On
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.
On
We assume that before entering the contest we've become an 'expert' at modulo $11$ calculations, so that we 'know' (or can quickly calculate) the following:
$\quad 5! \equiv 10 \pmod{11}$
$\quad 6! \equiv 5 \pmod{11}$
$\quad 7! \equiv 2 \pmod{11}$
$\quad 8! \equiv 5 \pmod{11}$
$\quad 9! \equiv 1 \pmod{11}$
$\quad 10! \equiv 10 \pmod{11}$
Problem: Find $a$ and $b$ where $15! = 1a0767436b000$.
For the problem at hand we don't need these shortcuts, since
$\quad 15! \equiv 0 \pmod{11}$
You first solve for $b$ in the manner suggested by Michael Burr, and then write out
$\tag 1 15! = 1a07674368000$
Although in this problem there are exactly three $0's$ to the right of $b$ and it to easy to factor out $10^3$, there are more complicated variations on these contest problems. In the $2^{nd}$ section of this answer a general method for finding that 'far right digit' $b$ by using modulo $10^5$ calculations is given; the contest tactic of using the fact that $10! \equiv 28800 \pmod{10^5}$ can be the starting line of attack for these problems.
Applying $(\text{mod } 11)$ to $\text{(1)}$ we can use the alternating sum of the digits technique,
$\quad 0 \equiv 0 - 0 + 0 - 8 + 6 - 3 + 4 - 7 + 6 - 7 + 0 - a + 1 \equiv -a -8 \pmod{11}$
and we conclude that $a = 3$,
$\tag{ANS.} 15! = 1307674368000$
'Contest' Theory
Let $n$ be any integer and let the expansion, $R$, of $n!$ be given with exactly one digit, $a$ missing. Let $p$ be any prime number greater than or equal to $11$. Then by applying $\text{mod } p$ to
$\quad n! = R$
you can solve for $a \text{ modulo } p$.
Since $0 \le a \le 9$, the modulo solution will be a 'spot on' solution.
Fortunately, there is no need to stray away from $p=11$ for these problems.
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.