Question

find the smallest number which when divided by 9 11 13 leaves remainder 1 2 3 respectively

Answer 1

Let the number be p
P/9 => remainder 1
 P/11 => remainder 2
P/13 => remainder 3

2P/9 => remainder 2
2P/11 => remainder 4
2P/13 => remainder 6

Now comes the tricky part.. Notice that if we add 7 to 2P , these equation will become completely divisible by 9,11 and 13 respectively, i.e. remainder becomes 0.

2P+7/9 => remainder 0
2P+7/11 => remainder 0
2P+7/13 => remainder 0
this means 2P+7 is divisible by 9,11 & 13. So it will be divisible by 9x11×13=1287
2P+7=1287
2P= 1280
P=640
So 640 is the answer
Hope this is useful.

Answer 2

Let the number be x
x/9=remainder 1
x/11=remainder 2
x/13=remainder3

2x/9=remainder 2
2x/11=remainder 4
2x/13=remainder 6

Add 7 to 2x
So it is completely divisible
2P+7/9 => remainder 0
2P+7/11 => remainder 0
2P+7/13 => remainder 0
this means 2P+7 is divisible by 9,11 & 13. So it will be divisible by 9x11×13=1287
2P+7=1287
2P= 1280
P=640
So 640 is the answer
Hope this is useful