Missing digits in 19!

Number Theory Level 1

$\large 19! = \overline{12164510{\color{#3D99F6}a}4{\color{#3D99F6}a}8832000}$

Determine the digit ${\color{#3D99F6}a}$ .

Note: $19 ! = 19 \times 18 \times 17 \times \cdots \times 2 \times 1$

8 4 7 5 0

2 solutions

Yatin Khanna
Feb 27, 2017

Clearly, the number is divisible by 11.
Now, by using the divisibility test for 11; the difference between the sums of the digits at odd and even places should be a multiple of 11.
Therefore, 1+1+4+1+a+a+8+2+0 - (2+6+5+0+4+8+3+0+0) is divisible by 11.
That is 2a-11 is divisible by 11. Or 2a is divisible by 11 therefore the only possible digit value of a is 0.

One can also use a calculator!

Dale Gray - 4 years, 2 months ago

Lol; well, you aren't expected to use it for this.

Syed Hamza Khalid - 2 years, 9 months ago

Larry Cunningham
Mar 24, 2017

Since 19! is, by definition, evenly divisible by 3 (one of the factors in 19!) then the sum of the digits of 19! must also be evenly divisible by 3.

So using the divisibility rule for 3 the sum of the digits ... 1+2+1+6+4+5+1+0+a+4+a+8+8+3+2+0+0 = 45+2a ... must also be evenly divisible by 3. Therefore 2a must be evenly divisible by 3. Of the choices listed for the digit "a" only 0 satisfies this (2 x 0 = 0 which, of course, is evenly divisible by 3).

Obviously if the choices for "a" listed included any or all of 3, 6 or 9 as well then an additional test would have to be made to determine what the correct answer is but of the choices listed only 0 works.

Missing digits in 19!

2 solutions

0 pending reports