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marcodonzelli
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abhijit_sen
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marcodonzelli
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pmenon
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I get C.

stat 1 says that x/4 has remainder 3. thats fine, so x can be 7, 11, etc. but stat 1 doesnt say anything about y.

stat 2 is the reverse. we know about x, but not about y. insuff.

together, we know that x=7,11,15... and y=6,10,14...

combining any of the two values, youll see that the product always has a remainder of 2. try it out.
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neelesh
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From 1 -> \(x = 4k_1 + 3\) == INSUFFICIENT

From 2 -> \(x = 4m_1 + 2\) == INSUFFICIENT

From 1 & 2

\((4k_1 + 3)(4m_1 + 2) = (4j_1 + remainder)\)

-> \(4(4k_1m_1 + 2k_1 + 3m_1) + 6 = (4j_1+ remainder)\)

-> \(4*(4k_1m_1 + 2 + 3m_1 + 1) + 2 = (4j_1+ remainder)\)

Hence remainder should be 2.



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