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by ankita1709 » Wed May 30, 2012 2:58 am
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:
1. x = w
2. x > w
3. x/y is an integer
4. w/z is an integer
5. x/z is an integer

I am really confused
neither x= can be true nor x/y is an integer can be true.. what should be the answer
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by Anurag@Gurome » Wed May 30, 2012 4:07 am
ankita1709 wrote: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:
Say, n consecutive integers are a, (a + 1), (a + 2),..., and (a + n - 1)
Hence, sum of this n integers = [na + n(n - 1)/2] = n*[a + (n - 1)/2]

Now, if n is odd, then (n - 1) is even --> (n - 1)/2 is an integer --> The sum is a multiple of n
And, if n is even, then (n - 1) is odd --> (n - 1)/2 is not an integer --> The sum is not a multiple of n

From the question, y = 2z = an even integer.
Hence, sum of y consecutive integers, x will not be a multiple of y.
Hence, x/y is not an integer.

The correct answer is C.
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by Anurag@Gurome » Wed May 30, 2012 4:21 am
The above method is not likely that one will try if (s)he doesn't know of it. But it is the easiest and least time consuming one.

We can solve this by plugging number, but to prove that something is always false (or always true) by plugging numbers is a tough job. Hence, let us try to plug numbers in such a way that we can prove other options can be true for at least one case.
  • 1. Say, z = 1 --> y = 2. Then take x = (1 + 2) = 3 and w = 3 --> x = w --> Possible
    2. Say, z = 1 --> y = 2. Then take x = (1 + 2) = 3 and w = 2 --> x > w --> Possible
    3. (Try this for some time and then leave it for now and move on)
    4. Take any of the first two examples. Both 3/1 and 2/1 are integers --> Possible
    5. Take any of the first two examples. 3/1 is an integer --> Possible
Hence, all but C could be true.

The correct answer is C.
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by ankita1709 » Wed May 30, 2012 7:07 am
Anurag@Gurome wrote: 1. Say, z = 1 --> y = 2. Then take x = (1 + 2) = 3 and w = 3 --> x = w --> Possible

The correct answer is C.
How did you get x=w
lets say z=3--> y=6
x=6(6+1)/2 = 21
w=3(3+1)/2 = 6
When can x=w

Pardon me if I am misinterpreting something
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by Anurag@Gurome » Wed May 30, 2012 9:22 am
ankita1709 wrote: 1. Say, z = 1 --> y = 2. Then take x = (1 + 2) = 3 and w = 3 --> x = w --> Possible

...

How did you get x=w
I assumed z = 1. Hence, y = 2
Now, I assumed y (= 2) consecutive integers are 1 and 2. Hence, x = (1 + 2) = 3
And, I assumed z (= 1) consecutive integer 3. Hence, w = 3

Hence, x = w
ankita1709 wrote:lets say z=3--> y=6
x=6(6+1)/2 = 21
w=3(3+1)/2 = 6
When can x=w
The question asked ...each of the following could be true... not always true. Hence, there may be infinite number of examples where x ≠ w, BUT there are infinite number of examples where x = w.

Also if I'm not wrong you are mixing up consecutive integers with first n natural numbers. I'm not sure, but your example points in that direction.
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by aneesh.kg » Wed May 30, 2012 12:18 pm
ankita1709 wrote: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:
1. x = w
2. x > w
3. x/y is an integer
4. w/z is an integer
5. x/z is an integer

I am really confused
neither x= can be true nor x/y is an integer can be true.. what should be the answer
Let's look at Option No. 3:
x/y is the average of y consecutive integers
If y = 2z, then y is an even integer.

Important: For every Arithmetic Progression, Median = Mean.

A set of consecutive integers is an Arithmetic Progression with constant difference = 1.
A set of consecutive integers which has even number of terms will have two central terms, both of which will be consecutive integers. The Median (= Mean) is equal the average of these two consecutive integers, which CANNOT be a fraction.
If the two central terms are x and (x + 1), then the Median (= Mean) = x + 0.5, which is not an integer.

Option 3 CANNOT be true.

[spoiler](C)[/spoiler] is correct.

If you're wondering when is Mean = Median, read this post:
https://www.beatthegmat.com/when-is-the- ... tml#475351
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by ankita1709 » Thu May 31, 2012 2:26 am
Anurag@Gurome wrote: Also if I'm not wrong you are mixing up consecutive integers with first n natural numbers. I'm not sure, but your example points in that direction.
Ohhhhh .. Thanks so much Anurag. I was exactly messing up with that thing..
Thanks a lot
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