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If m, n, p are positive integers, is (m2 + n)(2m + p) odd
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13 Jan 2014, 05:38
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77% (01:34) correct 23% (01:42) wrong based on 134 sessions
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If m, n, and p are positive integers, is (m^2 + n)(2m + p) an odd integer? (1) n is an odd integer. (2) p is an even integer. OE (1) For (m^2 + n). Since don't know if m is odd or even, don't know if m^2 is odd or even, which means adding odd, n could make whole statement odd or even. (odd + odd) = even, but (even + odd) = odd. So don't know if (m^2 + n) is odd or even. For (2m + p). Since m is integer, 2m must even. However, don't know if p is odd or even. So don't know if (2m + p) is odd or even. Since don't know if (m^2 + n) is odd or even and don't know if (2m + p) is odd or even, don't know if (m^2 + n)(2m + p) is odd or even. Insufficient (2) For (m^2 + n). Don't know if m is odd or even, so don't know if m^2 is odd or even. Don't know if n is odd or even. So don't know if (m^2 + n) is odd or even. For (2m + p). Since m is integer, 2m is even integer. Since p is even, (2m + p) = (even integer + even integer) = even. (2m + p) is even. Since (m^2 + n) is integer, (m^2 + n)(2m + p) = (integer x even integer) = even integer. Always No. Sufficient
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Re: If m, n, p are positive integers, is (m2 + n)(2m + p) odd
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13 Jan 2014, 06:37
goodyear2013 wrote: If m, n, and p are positive integers, is (m^2 + n)(2m + p) an odd integer? (1) n is an odd integer. (2) p is an even integer. OE (1) For (m^2 + n). Since don't know if m is odd or even, don't know if m^2 is odd or even, which means adding odd, n could make whole statement odd or even. (odd + odd) = even, but (even + odd) = odd. So don't know if (m^2 + n) is odd or even. For (2m + p). Since m is integer, 2m must even. However, don't know if p is odd or even. So don't know if (2m + p) is odd or even. Since don't know if (m^2 + n) is odd or even and don't know if (2m + p) is odd or even, don't know if (m^2 + n)(2m + p) is odd or even. Insufficient (2) For (m^2 + n). Don't know if m is odd or even, so don't know if m^2 is odd or even. Don't know if n is odd or even. So don't know if (m^2 + n) is odd or even. For (2m + p). Since m is integer, 2m is even integer. Since p is even, (2m + p) = (even integer + even integer) = even. (2m + p) is even. Since (m^2 + n) is integer, (m^2 + n)(2m + p) = (integer x even integer) = even integer. Always No. Sufficient If m, n, and p are positive integers, is (m^2 + n)(2m + p) an odd integer?(1) n is an odd integer > (m^2 + n)(2m + p) = (m^2 + odd)(even + p) > If m=odd, then m^2 + odd = odd + odd = even and the whole product is even but if m=even and p=odd, then both (m^2 + odd) and (even + p) are odd and the product is odd. Not sufficient. (2) p is an even integer > (m^2 + n)(2m + p) = (m^2 + n)(even + even) = (m^2 + n)*even = even. So, the product is NOT odd. Sufficient. Answer: B.
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Re: If m, n, p are positive integers, is (m2 + n)(2m + p) odd
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12 Oct 2018, 21:57
We want to determine whether or not there is sufficiency to say that \((m^2 + n)(2m + p)\) even or that \((m^2 + n)(2m + p)\) is not even. Statement (1) says that n is an odd integer. Let's consider \(m^2 + n\). Since we don't know if m is odd or even, we don't know if \(m^2\) is odd or even. We don't know if n is odd or even. So we don't know if \(m^2 + n\) is odd or even. Let's consider \(2m + p\). We see that since m is an integer, \(2m\) must be even. However, we don't know if p is odd or even. So we don't know if \(2m + p\) is odd or even. Since we don't know if \(m^2 + n\) is odd or even and we don't know if \(2m + p\) is odd or even, we don't know if \((m^2 + n)(2m + p)\) is odd or even. Statement (1) is insufficient. We can eliminate choices (A) and (D). Statement (2) says that p is an even integer. Let's look at \(m^2 + n\). We don't know if m is odd or even, so we don't know if \(m^2\) is odd or even. We don't know if n is odd or even. So we don't know if \(m^2 + n\) is odd or even. Now let's look at \(2m + p\). Since m is an integer, \(2m\) is an even integer. Since p is even, \(2m + p\) is the sum of an even integer and an even integer. An even plus an even is even, so \(2m + p\) is even. Since \((m^2 + n)\) is an integer, \((m^2 + n)(2m + p)\) is an integer times an even integer, which must be an even integer. The answer to the question is "no." Statement (2) is sufficient. Choice (B) is correct.
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Re: If m, n, p are positive integers, is (m2 + n)(2m + p) odd
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12 Oct 2018, 21:57






