Question

Find the smallest positive integer which, when divided by 6, give a remainder of 1 and when divided by 11, gives a remainder of 6.

Answer 1

let N= 6x+1=11y+6

6x-6=11y-1

6(x-1)=11y-1

this means (11y-1) is perfectly divisible by 6

so it can be written as

5y-1 is divisible by 6 (11%6 is 5)

5y-1 = 6K ,K be any constant.

hence y = ((6K+1)/5)

Now

as N= 11y+6 , putting the above value of y we have,

N = 11((6K+1)/5)+6

putting K= 4 for integral solution

N=61 which is the answer.

Enjoy….

Answer 2

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