Question

What is the product of all possible digits $x$ such that the six-digit number $341,\!4x7$ is divisible by 3?

Answer 1

80

Explanation:

We have the number:

341,4x7

And we want this number to be divisible by 3.

A number is divisible by 3 if the sum of all it's digits is also divisible by 3, then:

3 + 4 + 1 + 4 + x + 7 = 19 + x

Then 19 + x needs to be divisible by 3.

We know that the multiples of 3 (near these values) are:

3*6 = 18

3*7 = 21

3*8 = 24

3*9 = 27

Then if we take x = 2, we have:

19 + x = 19 + 2 = 21

Then:

341,427 is divisible by 3.

if we take x = 5, then:

19 + x = 19 + 5 = 24

Then:

341,457 is divisible by 3.

if we take x = 8, then:

19 + x = 19 + 8 = 27

then:

341,487 is divisible by 3.

Then the values of x such that the number 341,4x7 is divisible by 3, are:

2, 5, and 8.

The product of these numbers is:

2*5*8 = 10* = 80