15!

Number Theory Level 1

$\large 1 \times 2 \times \cdots \times 15 = \overline{13076 7{ \color{#D61F06} a }368 000}$

What is the missing digit $a$ ?

0 2 4 6 8

4 solutions

Saya Suka
Dec 3, 2016

The divisibility test for 9 is applicable here (which is glaringly obvious since $1 \leq 9 \leq 15$ ).
The total sum of the digits will be equal to some multiple of 9.

We have $1 \times ( 1 + 8 + a ) + 2 \times ( 3 + 6 + 7) + 4 \times 0$ is a multiple of 9.
This is equal to $41 + a$ .
Since $0 \leq a \leq 9$ , hence we must have $a = 4$ to give $41 + 4 = 45$ as the multiple of 9.

Note:

$15!$ has $\lfloor \frac{15}{3} \rfloor + \lfloor \frac{15}{3^2} \rfloor = 5 + 1 = 6$ factors of 3, so $15!$ is a multiple of $3^6$ . This approach shows us that even $6!, 7!, 8!$ are multiples of 9 even though 9 doesn't appear in the expanded multiplication.
Note: If the digit sum without $a$ is a multiple of 9, then $a$ could only be 0 or 9. We will need to do more work.

"Note: If the digit sum without is a multiple of 9, then could only be 0 or 9. We will need to do more work."

This gives me an inspiration !

Christopher Boo - 4 years, 6 months ago

I like that question!

Eric's approach would work better for this problem, as we will always get just 1 answer. If more digits were hidden, then we will need more tests.

Chung Kevin - 4 years, 6 months ago

That's essentially the logic.

Can you complete the calculation to explain how to arrive at the answer? Thanks!

Calvin Lin Staff - 4 years, 6 months ago

Thanks. Great explanations and tips that you pointed out.

I've edited the presentation order so that it's easier to read the solution.

Calvin Lin Staff - 4 years, 6 months ago

$15!$ has $\lfloor \frac{15}{2} \rfloor + \lfloor \frac{15}{2^2} \rfloor + \lfloor \frac{15}{2^3} \rfloor = 7 + 3 + 1 = 11$ factors of $2$ . You can be sure that the last n digits is divisible by $2^n$ ; for example $a368000$ is divisible by $2^7$ or $a368$ is divisible by $2^4$ , if the double possibility of $0$ or $9$ arises.

Saya Suka - 3 months, 1 week ago

Eric Kim
Dec 6, 2016

Similar to Saya Suka's solution, but with 11.

Let each digits in 130767a368000 have indexes, where the first digit (1) has an index of 1, the second digit (3) has an index of 2, and etc.

The divisibility test for 11 is that the sum of all the digits with even indexes differ from the sum of all the digits with odd indexes by a multiple of 11 (including 0).

Sum of digits with odd digits: 1 + 0 + 6 + a + 6 + 0 + 0 = 13 + a Sum of digits with even digits: 3 + 7 + 7 + 3 + 8 + 0 = 28

13 + a = 28 + 11n
a= 15 + 11n
Since a is a digit, 0<= a <= 9.
Thus, n = -1 and a = 4

Nice! The benefit of using the rule of 11 is that we're guaranteed a unique answer no matter which digit is hidden (assuming that all the other digits are correct).

Calvin Lin Staff - 4 years, 6 months ago

11 might be a bit overkill for this problem, but is the best method in this variation !

Christopher Boo - 4 years, 6 months ago

I don't think it's an overkill because of how simple the rule is. If we tried to apply divisibility by 17 which doesn't have a real simple use, then that would be an overkill.

Calvin Lin Staff - 4 years, 6 months ago

Same way :) (+1)

Yatin Khanna - 4 years, 6 months ago

Ashish Kedia
Jan 17, 2017

Find sum of all digits except a

thats 41 but it should be a multiple of 3... so 4 is the only possible option

Also, $1$ and $7$ make the number divisible by $3$ .

Jesse Nieminen - 4 years, 3 months ago

B D
Aug 31, 2018

Use a calculator.

15!

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