6
$\begingroup$

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

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.

$\endgroup$
  • 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 at 21:36
  • $\begingroup$ Why is the sum of the digits divisible by 9? $\endgroup$ – Darkice Jan 2 at 21:37
  • 3
    $\begingroup$ @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 at 21:41
  • 1
    $\begingroup$ 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. $\endgroup$ – Darkice Jan 2 at 21:59
  • 6
    $\begingroup$ 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 at 22:09
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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 at 23:34
  • $\begingroup$ The order is also given by the divisibilty rule for 7. $\endgroup$ – Darkice Jan 3 at 8:54
1
$\begingroup$

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 $1a0767436....$

$\endgroup$
1
$\begingroup$

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 $d_1=3$.

$\endgroup$
1
$\begingroup$

$15!=2^{11}\cdot 5^3\cdot 7^2\cdot 11\cdot 13=(1000)X$ 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 $9$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.