Invisible Divisible
Number Theory
Level
3
How many three-digit positive integers divisible by 3 are there such that the digits are consecutive in some order and none of the digits are zero?
Example numbers: 465, 321.
The answer is 42.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Consider the consecutive integers x , x + 1 , and x + 2 . Their sum is divisible by 3: x + x + 1 + x + 2 = 3 x + 3 = 3 ( x + 1 ) .
Since a number is divisible by 3 when the digits add to a number divisible by 3, any 3-digit number with consecutive digits in some order is divisible by 3.
That means we now just need to count how many numbers like this there are. There are 7 consecutive sequences using the digits 1 to 9 (1-2-3, 2-3-4, 3-4-5, 4-5-6, 5-6-7, 6-7-8, 7-8-9) and each of those can be rearranged in 6 ways (for example, 1-2-3 can be 123, 132, 213, 312, 231, or 321), so that means there are 7 × 6 = 4 2 numbers that meet the conditions.