Three times Seven
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
Age
11 to 14
Challenge level
Image
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7.
Why?
Correct solutions were received from Andrei of School 205 Bucharest (whose solution forms the basis of the one below), Mary of Birchwood High and Chen from The Chinese High School, Singapore.
If two numbers are both divisible by 7, then their sum will always be divisible by 7.
Any 3-digit number could be written as:
abc = 100a + 10b + c
We must prove that if 2a + 3b + c is divisible by 7, then abc is also divisible by 7.
Subtracting (2a + 3b + c) [which is divisible by 7] from 100a + 10b + c [the value of the number abc], gives:
98a + 7b = 7 (14a + b)
This number is always divisible by 7.
This means that abc is divisible by 7 when (2a + 3b + c) is divisible by 7.