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sum of consecutive integers

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by canuckclint » Fri Oct 31, 2008 5:04 pm
x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT
x = w
x > w
x/y is an integer


w/z is an integer
x/z is an integer
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Source: — Problem Solving |

by jayjk78 » Fri Oct 31, 2008 9:26 pm
My guess is I does not work with any comnibation

II - X>W is possible x= 10 (1,2,3,4), w=3 (1,2)
III x/w=INT is possible x= 10 (1,2,3,4), w=(2,3)
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by cramya » Fri Oct 31, 2008 9:53 pm
x=w is possible

Let z=1(the problem stem does not say z>1) y=2

w = {-3}
x{-2,-1}


It should be x/y
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by 4meonly » Sat Nov 01, 2008 1:43 am
I guess it would be C
but do not have exhaustive evidence
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by rohangupta83 » Sat Nov 01, 2008 6:41 am
let Y be a series of 5 consecutive positive integers

(1,2,3,4,5)

therefore, sum = 1+2+3+4+5 = 15

x/y = 15/5 = 3 (which is an integer)
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by cramya » Sat Nov 01, 2008 7:16 am
Rohan,
y cant be 5 since y is even.
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by rohangupta83 » Sat Nov 01, 2008 7:47 am
cramya wrote:Rohan,
y cant be 5 since y is even.
ahh, i see. That is the significance of y=2z

that y is even

C it is then
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by piyushdabomb » Sun Nov 02, 2008 10:06 am
Isn't there a full proof method on solving these sort of problems? There seems to be a lot of guessing and I really don't think that you're suppose to be picking and choosing numbers here (won't you lose time on the gmat?)

Can someone provide a theoretical approach to this?
-------------------
Sincerely,

Piyush A.
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by canuckclint » Mon Nov 03, 2008 9:31 pm
The official MGMAT answer:


For any set of consecutive integers with an odd number of terms, the sum of the integers is always a multiple of the number of terms. For example, the sum of 1, 2, and 3 (three consecutives -- an odd number) is 6, which is a multiple of 3. For any set of consecutive integers with an even number of terms, the sum of the integers is never a multiple of the number of terms. For example, the sum of 1, 2, 3, and 4 (four consecutives -- an even number) is 10, which is not a multiple of 4.

The question tells us that y = 2z, which allows us to deduce that y is even. Since y is even, then the sum of y integers, x, cannot be a multiple of y. Therefore, x/y cannot be an integer; choice C is the correct answer. We can verify this by showing that the other choices could indeed be true:

(A) The sum x can equal the sum w: 4 + 5 + 6 + 7 + 8 + 9 = 12 + 13 + 14 = 39, for example.

(B) The sum x can be greater than the sum w: 1 + 2 + 3 + 4 > 1 + 2, for example.

(D) z could be odd (the question does not restrict this), making the sum w a multiple of z. Thus, w/z could be an integer. For example, if z = 3, then we are dealing with three consecutive integers. We can choose any three: 2, 3, and 4, for example. 2 + 3 + 4 = 9 and 9/3 = 3, which is an integer.

(E) x/z could be an integer. If z = 2 and if x is an even sum, then x/z would be an integer. For example, if z = 2, then y = 4. We can choose any four consecutive integers: 1 + 2 + 3 + 4, for example. So the sum x of these four integers is 10. 10/2 = 5, which is an integer.

The correct answer is C.
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