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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.
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
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.