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Bunuel
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Rabab36
gmatophobia can you pls explain statement 2

y=x-1

x=4,y=3
x=5,y=4
x=6,y=5
x=7,y=6

remainder will be same for x=4,5,6
not for x=7,8


therefore option c

Rabab36

We know that y (denominator) is one less than x (numerator). Hence except for when the denominator = 1, if we divide the numerator by denominator the remainder will always be equal to 1.

In this case the denominator cannot be equal to 1 as the question states that both the numerator and denominator are integers greater than 2. Hence the remainder will always be one.

To the address the second point -

x = 7 & y = 6, the remainder will when \(\frac{7}{6}\) will be 1.

Hence \(1^{356}\) = 1; hence 1 is the remainder.

Quote:
and i think you have taken x^365

actually as per q is x^356

Thanks for noticing. Me and my typo :facepalm_man:

This wouldn't change the solution because \(1^n\) is 1 (n can be any number).

Hope this helps.
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