Question

The smallest positive integer, when it is divided by 7,9,11
leaves the remainder 1, 2, 3 respectively is-
​

Answer 1

Answer:

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

Answer 2

Answer:

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