Find the smallest number that if divided by 7,9,11 leaves a remainder of 1,2, and 3 respectively

Question

Find the smallest number that if divided by 7,9,11 leaves a remainder of 1,2, and 3 respectively

Answer 1

Hi Yuri,

 

This is a tricky one but lets give it a try.

 

N/7 =1

 

N/9 =2

 

N/11 = 3

 

So the following must be true

 

2N/7 = 2 remainder

 

2N/9 = 4 remainder

 

2N/11 = 6 remainder

 

But if we add five to each of the equations they will have a remainder of zero

 

so

 

(2n+5)/7 = 0 remainder

 

(2n+5)/9 = 0 remainder

 

(2n+5)/11 = 0 reminder

 

therefore

 

(2n+5)  must be a multiple of 7*9*11 =693

 

(2n+5) = 693

 

2n = 688

 

N = 344

 

N/7 = 49 Remainder 1

 

N/9 = 38 Reminder 2

 

N/11 = 31 Remainder 3

 

So the answer is 344

 

Hope this helps