$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$

Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.

How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation: $$15! \approx \sqrt{2\pi\cdot 15} \cdot \left(\frac{15}{e}\right)^{15}$$ which is not feasible to calculate by hand.

The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:

$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $ $-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $

The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.

  • 8
    $\begingroup$ Well, you know that the sum of the digits is divisible by $9$ and the alternating sum of the digits is divisible by $11$. That should tell you what the two digits are, though not their order. But you should be able to argue that the first non-zero digit from the right must be even. $\endgroup$ – lulu Jan 2 '19 at 21:36
  • $\begingroup$ Why is the sum of the digits divisible by 9? $\endgroup$ – Stephan Jan 2 '19 at 21:37
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    $\begingroup$ @Darkice because for every natural number $n\geq 6$ you have that $n!$ is a multiple of nine and a natural number is a multiple of nine if and only if the sum of its digits is a multiple of nine. $\endgroup$ – JMoravitz Jan 2 '19 at 21:41
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    $\begingroup$ Alright, now I know that the digits must add up to a multiple of 9 and must be divisible by 11. So the digits could be 7 and 4 or 3 and 8 which would both fulfil the requirements. $\endgroup$ – Stephan Jan 2 '19 at 21:59
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    $\begingroup$ For your specific case, one of the numbers you are looking to find is the final nonzero digit. As $15! = 2^{11}\cdot3^6\cdot5^3\cdot7^2\cdot11\cdot13$ the final nonzero digit will be $2^8\cdot3^6\cdot7^2\cdot11\cdot13\pmod{10}$ which is a much easier calculation to do modulo 10 than fully expanding out the product. $2^8\cdot3^6\cdot7^2\cdot11\cdot13\equiv 6\cdot 9\cdot 9\cdot 1\cdot 3\equiv 8\pmod{10}$. This would be admittedly more difficult if both of the numbers were located further in the middle and not next to the trailing zeroes. $\endgroup$ – JMoravitz Jan 2 '19 at 22:09

Another way to reason is to note that $15!$ is divisible by $2\cdot4\cdot2\cdot8\cdot2\cdot4\cdot2=2^{11}$, which means $1\square0767436\square$ is divisible by $2^8$. In particular, it's divisible by $8$. But since $8\mid1000$ and $8\mid360$, the final $\square$ must be either $0$ or $8$. But it can't be $0$, since $15!$ has only three powers of $5$ (from $5$, $10$, and $15$), and those were already accounted for in the final three $0$'s of the number. Thus the final $\square$ is an $8$. Casting out $9$'s now reveals that the first $\square$ is a $3$.

Remark: It wasn't strictly necessary to determine the exact power of $2$ (namely $2^{11}$) that divides $15!$, merely that $2^6$ divides it, but it wasn't that hard to do.


You can cast out $9$’s and $11$’s: \begin{align} 1+x+0+7+6+7+4+3+6+y+0+0+0&=x+y+34 \\ 1-x+0-7+6-7+4-3+6-y+0-0+0&=-x-y \end{align} Thus $x+y=11$ (it can't be $x=y=0$).

Then find the remainder modulo $10000$; since $$ 15!=2^{11}\cdot 3^6\cdot 5^3\cdot 7^2\cdot11\cdot13=1000\cdot 2^8\cdot3^6\cdot7^2\cdot 11\cdot 13 $$ this means finding the remainder modulo $10$ of $$ 2^8\cdot3^6\cdot7^2\cdot 11\cdot 13 $$ that gives $8$ with a short computation.


Using the divisibility rule for 7 the answer boils down to 3 and 8:

$-368+674+307+1 \mod 7 \equiv 0$

  • $\begingroup$ for the order problem you could remove three zeros from the Least significant digit side and say it still needs to be divisible by 2 or 4 cause there are lots of 2's and 4's in the original multiplication, so 3 couldn't be on the right side. $\endgroup$ – no0ob Jan 2 '19 at 23:34
  • $\begingroup$ The order is also given by the divisibilty rule for 7. $\endgroup$ – Stephan Jan 3 '19 at 8:54

Okay, $15! = 1*2*3..... *15=1a0767436b000$.

Why does it end with $000$? Well, obviously because $5,10,15$ all divide into it so $5^3$ divides into it and at least three copies of $2$ divid into it so $2^3*5^3 =1000$ divide into it.

If we divide $15!$ by $100 = 8*5^3$ we get

$1a0767436b = 1*2*3*4*6*7*9*2*11*12*13*14*3$

If we want to find the last digit of that we can do

$1a0767436b \equiv b \pmod {10}$ and

$1*2*3*4*6*7*9*2*11*12*13*14*3\equiv 2*3*4*(-4)*(-3)*(-1)*2*1*2*3*4*3\equiv$

$-2^9*3^4 \equiv -512*81\equiv -2 \equiv 8\pmod {10}$..

So $b = 8$.

But what is $a$?

Well, $11|1a0767436b$ and $9|1a0767436b$.

So $1+0+6+4+6 - a - 7-7-3-b = 11k$ for some integer $k$. And $1+a+0+7+6+7+4+3+6+b = 9j$ for some integer $j$.

So $-a -8 =11k$ so as $0\le a \le 9$ we have $a = 3$.

And that's that $15! = 1307674368000$..... IF we assume the person who asked this question was telling the truth.

We do know that $15!$ ends with $.... 8000$ but we are completely taking someone elses word for it that it begins with $1a0767436....$


Let $d_1$ and $d_2$ be the two unknown digits.

The number must be divisible by $8000$, because $15!$ contains $8$ and $1000$.

$d_2$ is non-zero number, because $15!$ contains only three $5$s. It implies $1d_10767436d_2$ must be divisible by $8$. It implies $36d_2$ is divisible by $8$. Hence, $d_2=8$.

Now you can use the divisibility by $9$ ($d_1+d_2=11$) and find $d_1=3$.


$15!=2^{11}\cdot 5^3\cdot 7^2\cdot 11\cdot 13=(1000)X$ where $X=2^8\cdot 3^6\cdot 7^2\cdot 11\cdot 13.$

The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$

Modulo $10$ we have $6\cdot 9\cdot 9 \cdot 1\cdot 3\equiv 6\cdot(-1)^2\cdot 3\equiv 18\equiv 8$. So the last digit of $X$ is an $8$.

Therefore the 2nd digit of $15!$ must be a $3$ in order for the sum of all its digits to be divisible by $9$.


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