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Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

Three Times Seven

Age 11 to 14 Challenge Level:

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.