Concatenation Property?

Number Theory Level 1

14 , 21 , 28 , 35 , 42 , 49 , 56 14, \ \ \ 21, \ \ \ 28, \ \ \ 35, \ \ \ 42, \ \ \ 49, \ \ \ 56

The above shows a sequence of multiples of 7. Is it true that if I combine these numbers together by their numerals in that order, then it is also divisible by 7? In other words, is the large number

14212835424956 14 21 28 35 42 49 56

divisible by 7?

Yes, it is divisible by 7 No, it is not divisible by 7

Pi Han Goh by Pi Han Goh , 30, Malaysia

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2 solutions

It is divisible by 7. It will just be the quotient when each of the numbers is divided by 7 but with a 0 in between each one of them i.e. 2030405060708.

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That final 0 after 8, is not necessary.

Marcus Jeffries - 4 years, 11 months ago

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Edited thanks

A Former Brilliant Member - 4 years, 11 months ago
Anandmay Patel
Aug 3, 2016

14212835424956 can be written as 14x10^12+21x10^10+28x10^8+35x10^6+42x10^4+49x20^2+56x10^0 which when divided by 7,obviously results in an integer.So 7 divides the number.

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