# Find the sum of the remainders obtained when a number n is divided by 9 and 7 successively given that n is the smallest number that leaves remainders of 4,6, and 9 when divided successively by 13, 11 and 15. Option 1) Option 2) Option 3) Option 4) Option 5)

So, in order to get the number we have to fullfil the above mentioned numbers one by one....
The number with 15 divided 9 as remainder
The number with 11 divided 6 as remainder

The number with 13 divided 4 as remainder (Thus the general form of the required number)

In order to get the lowest put

sum of the remainders obtained when 3514 is divided by 9 and 7 successively=9

First Find N,

Divisors      = 13, 11, 15

Remainders= 4, 6, 9

N = {[9(11)+6]*13}+4

N = 1369

Now, if we check 1369 divided by 13 Remainder 4 and Quotient is 105

Successive division - 105 divided by 11 Remainder 6 and Qotient is 9

9 divided by 15 Remainder 9

As such, N = 1369

1369 divided by 9, 152 Quotient and 1 remainder

152 divided by 7, 21 Quor]tient and 5 remainder

Sum of remainders is 5+1=6

Option 5

If we do as @rishi.raj answer then In order to get the lowest put x=0 thus 1369+2145(0) = 1369

then N=1369

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