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integers

by Ankitaverma » Mon Dec 09, 2013 10:04 am
If n and k are integers whose product is 400, which of the following statements must be true?

(A) n + k > 0
(B) n≠k
(C) Either n or k is a multiple of 10.
(D) If n is even, then k is odd.
(E) If n is odd, then k is even.

Q/a-e can someone explain
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by Patrick_GMATFix » Mon Dec 09, 2013 11:47 am
two numbers have an even product if and only if at least one of them is even. So if one of the numbers is odd, the other one must be even.

That is why E must be true.

A would be false if both numbers were negative.
B would be false if both numbers were equal to square root of 400.
C would be false if n=16 and k=25. To find these numbers, break 400 into its prime factors. 400=2^4 * 5^2. To make sure that neither n, nor k is a multiple of 10, separate the 2s from the 5s. So n=2^4, k=5^2
D would be false if n and k were both even (eg: 2 and 200).
E must be true because if n is odd, we would have odd * k = even.

Hope that helps,
-Patrick
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by Brent@GMATPrepNow » Mon Dec 09, 2013 12:45 pm
Ankitaverma wrote:If n and k are integers whose product is 400, which of the following statements must be true?

(A) n + k > 0
(B) n≠k
(C) Either n or k is a multiple of 10.
(D) If n is even, then k is odd.
(E) If n is odd, then k is even.
Patrick's approach is fantastic.
Here's another (backup approach):

It may be useful to find the prime factorization of 400
400 = (2)(2)(2)(2)(5)(5)

Now let's examine answer choice E.
If n is ODD, there are only a few possible values for n.
n = 1, in which case k = 400 (in which case, k is EVEN)
n = 5, in which case k = 80 (in which case, k is EVEN)
n = 25, in which case k = 16 (in which case, k is EVEN)
n = -1, in which case k = -400 (in which case, k is EVEN)
n = -5, in which case k = -80 (in which case, k is EVEN)
n = -25, in which case k = -16 (in which case, k is EVEN)
So, for every possible ODD value of n, k is always EVEN

So, answer choice E must be true.

Cheers,
Brent
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by theCodeToGMAT » Mon Dec 09, 2013 9:35 pm
(n)(k) = 400

(A) n + k > 0
(-ve) * (-ve) = +ve
(+ve) * (+ve) = +ve
NO

(B) n≠k
(20)(20) = 400
NO

(C) Either n or k is a multiple of 10.
(20)(20)
(25)(16) = 400
NO


(D) If n is even, then k is odd.
(16)(25)
(20)(20)
NO


(E) If n is odd, then k is even.
(ODD)*(EVEN) == (EVEN; 400)
(ODD)*(ODD) == ODD ≠ 400 NOT VALID
TRUE

Answer [spoiler]{E}[/spoiler]
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by madhavanjc » Mon Dec 16, 2013 11:05 pm
The last two answers 'n' and 'k' are interchangeable.
I am still unable to get why Option D is incorrect.
Can anyone pls explain.
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by Uva@90 » Tue Dec 17, 2013 7:47 am
madhavanjc wrote:The last two answers 'n' and 'k' are interchangeable.
I am still unable to get why Option D is incorrect.
Can anyone pls explain.
Hi Madhavanjc,

Since you have issue with option D and E we will take that alone.

we know that 400 = n*k

Option D: If n is even, then k is odd.
Here first we are selecting Even Number,n.
so select even number for N,
Let, N= 2 then K must be 200 => K is Even
Now select N = 16 then K must be 25 => K is Odd
Hence Insufficient.

Option E : If n is odd, then k is even.
Here first we are selecting Odd Number,n.
So, select Odd number for N
Let N= 5 then k must be 80 => K is Even
Let N =25 then K must be 16 => K is Even
Hence Sufficient.

Answer is E

Hope it helps you.

Regards,
Uva.
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