Find the smallest integer that produces remainder of 2,4,6&1 when it is divided by 3,5,7&11 respectively?

1 Answer
May 2, 2017

419

Explanation:

Call the number n.

  • In order to give a remainder of 2 when divided by 3, n must be one less than a multiple of 3.

  • In order to give a remainder of 4 when divided by 5, n must be one less than a multiple of 5.

  • In order to give a remainder of 6 when divided by 7, n must be one less than a multiple of 7.

So n is one less than a multiple of the LCM of 3, 5 and 7, namely:

3*5*7 = 105

So:

n = 105k-1" " for some integer k.

In order to give a remainder of 1 when divided by 11, n must be of the form 11m+1 for some integer m.

So we have:

11m+1 = 105k-1

That is:

105k = 11m + 2

Now:

105/11 = 9" " with remainder 6

So what are the multiples of 6 modulo 11?

1*6 = 6 = 0*11+6

2*6 = 12 = 1*11 + 1

3*6 = 18 = 1*11 + 7

4*6 = 24 = 2*11 + 2

So the smallest possible positive value of k which gives us the required remainder is k=4

So:

n = 105k-1 = 105*4-1 = 419