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Emdad
When x is divided by 13, the answer is y with a remainder of 3. When x is divided by 7, the answer is z with a remainder of 3. If x, y, and z are all possible integers, what is the remainder of yz/13?

(A) 0
(B) 3
(C) 4
(D) 7
(E) None of these

From the text you can reason that:

x= 13y + 3 (when x is divided by 13 it gives a quotient of y a remainder of 3)
x= 7z + 3 (when x is divided by 7 it gives a quotient of z and it gives a remainder of 3)

Now reverse the equations to find y and z

y=(x-3)/13
z= (x-3)/7

You know that x, y and z are INTEGERS thus the product YZ is divisible by (13*7)=81 because it's a multiple of it.

But 81 is divisible by 13 thus the remainder is 0.
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x/13 = y + 3/13 and x/7 = z + 3/7

Multiply these 2 equations by 13 and 7 respectively

We get

x = 13y + 3 and x = 7z + 3

13y = 7z

y = 7z/13

The question asks remainder when yz is divided by 13

taking the value of y above we get = 7z/13*z = 7z^2/13, the remainder is 0 since z is an integer.
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