A certain number, when successively divided by 8 and 11, leaves remain

IMO:

x = 8*k + 3 = 11, 19, 27, 35, ..., 51, ....
x = 11*k' + 7 = 7, 18, 28, 40, 51, ....

x = 88*k'' + R = 51 -----> if k''=0 ------> R = 51

Answer: C

The number in question has the form \(x = r +k\cdot 88\) where \(r\) is one of the answer choices. Since \(88\) is a multiple of \(8\) and \(11\), \(r\) and \(x\) have the same remainder when divided by \(11\) (or \(8\)).
Only answer choice (C) yields \(7\) when divided by \(11\).

Reading rajatchopra1994 I realize that I most likely misunderstood the question (lol).

we are "successively dividing"

so we take the Number = N

(1st) Dividing N by 8 -------> get a quotient = A and a Remainder = 3


(2nd) Dividing the quotient of A we got above by 11 ---------> get another quotient = Q and a Remainder = 7



Starting from the 2nd Division and moving to the 1st Division -----

(Quotient = A) / 11 = Q + (7/11)

A = 11Q + 7 ------ (equation 1)



the 1st Division involving the Number = N -----

(N / 8) = A + (3 / 8)

N = 8A + 3

---substituting in (equation 1) for the Quotient Variable of A-----

N = 8(11Q + 7) + 3

N = 88Q + 56 + 3

N = 88Q + 59


When we Divide this Original Number of N by 88, it will leave a Remainder = 59

-D-

As per question, they are asking for a certain number, when successively divided by 8 and 11, leaves remainders of 3 and 7 respectively. This means that

N = 8*x + 3, where x = 11y + 7

So we put x value in equation 1.

N = 8 * (11y + 7) + 3
N = 88 y + 56 + 3
N = 88 y + 59

So if we divide 88y + 59 by product of 8 & 11 the remainder will be 59

So Option D is answer