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)

    4

  • Option 2)

    5

  • Option 3)

    9

  • Option 4)

    7

  • Option 5)

    6

Answers (2)

The\: n \: can\: be\: written \: as,
n = 13k + 4
n = 11k' + 6
n = 15k" + 9

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= 15x+9
The number with 11 divided 6 as remainder = 11*(15x+9) + 6

The number with 13 divided 4 as remainder= 13*[11*(15x+9) + 6] +4 = 2145x +1369 (Thus the general form of the required number) 

In order to get the lowest put x= 1 \: thus\: 3514

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

Answer is 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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