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venstein
Could you, please, elaborate on where 1*56 and 56*1 pairs for n*p are coming from? To me m can be either 1 or 3, if it is 3 then n*p=5*7=35.If m is 1 then n*p=2,835. I don't see how n*p can equal 56...Thanks


I agree-Answer should be A-Please confirm
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hazelnut
If m, n, and p are positive integers, mnp= ?

1) \(m^4*np=2,835\)

2) p=7

E is correct. Consider the following:

m = 1, n = 3^4*5, p = 7 --> mnp = 2,835;
m = 3, n = 5, p = 7 --> mnp = 105.

Answer: E.
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Pay attention to special case in which m = 1!
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hazelnut
If m, n, and p are positive integers, mnp= ?

1) \(m^4*np=2,835\)

2) p=7

The answer is E

Factorization of 2835 =3^4*5*7

But we are not told that n and p have to be prime so following cases are possible
1*3
35*1
5*7
7*5
So insufficient
p=7
it does not provide information for other variable so insufficient

Together we know p=7 then n=5
Hence together they are sufficient
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hazelnut
If m, n, and p are positive integers, mnp= ?

1) \(m^4*np=2,835\)

2) p=7

When the question says m,n and p are positive integers, the confusion lies in wondering whether m, n and p are single digit integers or not.
If they are single digit then answer is A, if not then answer is E
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Statement I can be re-written as: m^3(mnp)=2835--- Insufficient to solve.
Statement II says p=7 which is not sufficient to put into any equation and solve for mnp.

Hence the answer is E.
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