Fun way to learn some basic number theory !!!

A test of divisibility by 3, 7 and 19:

The product of the prime numbers 3, 7 and 19 is 399. If a number 100a+b (where b is a two-digit number and a is any positive integer) is  divisible by 399 or by any of its divisors, then a+4b is divisible by the same number.

On your own, can u prove this? (Hint: Use 400a+4b as a link). Can you formulate and prove its converse?

Devise a simple test of divisibility by 3, 7 and 19.

More recreational problems in number theory are coming soon !!

Nalin Pithwa

 

 

One Comment

  1. Anubhav C. Singh
    Posted September 5, 2016 at 7:24 am | Permalink | Reply

    [If a number 10a+b when divided by c leaves remainder 0, number x(10a+b) will also be divisible by c.] So 100a+b and 400a+4b will be equally divisible by 3, 7 or 19.
    Now if 399a is subtracted from 400a+4b, the remainder still says the same as 399 is divisible by 3, 7 and 19. [26 leaves remainder 2 when divided by 3. When subtracted by a multiple of 3(15), no. becomes 11 which still leaves 2 as remainder.] So a+4b is as divisible as 10a+b by a divisor c (here 399 or any factors.)

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