Question

vah kaun si sankhya hai jiska 7,9,11 se bhagh dene per sesh 1,2,3 buche​

Answer 1

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