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St 1: Since (p-1)(r+1) is Odd, it means both (p-1) and (r+1) are odd, which also means that both p and r are even (since they are integers).

A product of 3 integers would be even if atleast one of them is even. Hence pqr is even. Sufficient.

St 2: (q-r)^2 is odd.

This suggests that the difference between q and r is also odd (since square of odd integer is odd). Since the difference is odd, one of them has to be an Even integer. E.g. 6-3 = 3. Both cannot be odd or both cannot be even. E.g. 9-3=6 and 8-4=4.

Again, a product of 3 integers would be even if atleast one of them is even. Hence pqr is even. Sufficient.

indeed the question is if the product is even. my contention is what if r= 0? in that case, 1. (p-1)*1 = odd this p is even .. but pqr =0 2. (q-0)^2 = odd ..i.e. q = odd and pqr = 0

as 1. as explained above. Sufficient 2. Sufficient as (q-r)(q-r) is odd that is q <> r; q - r is not even => exactly one of them is even and the other is odd. so pqr is even = due to that even _________________

1. Sufficient Only odd x odd can equal an odd number. Hence, P and R are both even. Any number multiplied by an even number is even. As such, PQR is even.

2. Sufficient Just like #1, odd x odd is the only way to get odd. As such, Q - R must be odd. Either Q or R are even/odd or odd/even. Either way, one of the numbers in the set is even, so the product of PQR must be even.

Booyah! I would never have been able to figure that out without MGMAT!

hi all, i am of diff opinion. I am sure its C both statements are required to answer. 1 gives info of only p & r, while 2 gives info of q. We need to know the product value of p,q & r. so both statements are essentially needed. _________________

"When the going gets tough, the tough gets going!"

Is pqr even? This is asking if either p, q or r is even. If one is even, then Yes!

1) (p-1)(r+1) is odd

RULE ====================== ODD x EVEN results to EVEN ODD x ODD results to ODD ====================== Using that rule, (p-1) is ODD and (r+1) is ODD. This means p (the number after p-1) is EVEN. We can stop there and realize that pxqxr is EVEN. Sufficient!

2) (q-r)(q-r) ois ODD

This means q-r is ODD.

RULE ======================== ODD +/- ODD = EVEN EVEN +- ODD = ODD ======================== From the rule, one of q and r is EVEN. Then, it's SUFFICIENT to say that pxqxr is EVEN.

hi all, i am of diff opinion. I am sure its C both statements are required to answer. 1 gives info of only p & r, while 2 gives info of q. We need to know the product value of p,q & r. so both statements are essentially needed.

To choose (C) means that statement 1 nor statement 2 is sufficient. But in the sample above, Statement 1 and 2 can both answer the question without the need of each other.

1. Sufficient Only odd x odd can equal an odd number. Hence, P and R are both even. Any number multiplied by an even number is even. As such, PQR is even.

2. Sufficient Just like #1, odd x odd is the only way to get odd. As such, Q - R must be odd. Either Q or R are even/odd or odd/even. Either way, one of the numbers in the set is even, so the product of PQR must be even.

Booyah! I would never have been able to figure that out without MGMAT!

I do agree. MGMAT is really good in breaking these things down into easier to comprehend pieces.

indeed the question is if the product is even. my contention is what if r= 0? in that case, 1. (p-1)*1 = odd this p is even .. but pqr =0 2. (q-0)^2 = odd ..i.e. q = odd and pqr = 0

This is actually a pertinent point, I think. Question might need to be reworded to "positive integers" or "non-zero integers" instead of just "integers".

Zero is neither even nor odd (as a helpful forummer once pointed to to me).

indeed the question is if the product is even. my contention is what if r= 0? in that case, 1. (p-1)*1 = odd this p is even .. but pqr =0 2. (q-0)^2 = odd ..i.e. q = odd and pqr = 0

This is actually a pertinent point, I think. Question might need to be reworded to "positive integers" or "non-zero integers" instead of just "integers".

Zero is neither even nor odd (as a helpful forummer once pointed to to me).

"0" is definitely EVEN.

"0" is neither POSITIVE nor NEGATIVE. _________________

Zero is neither even nor odd (as a helpful forummer once pointed to to me).

In some exams, even numbers refer to positive even numbers only (e.g. in CAT for IIMs - if you don't know which exam I am talking about, just ignore it) but as far as GMAT is concerned, an even number is the one which is divisible by 2. Hence 0, -4 and 68 are all even numbers. (This is also the more widely accepted definition, I think)

My guess is someone might have told you that '1 is neither prime nor composite.' (which is true) _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...