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Time SUCKER- consecutive integers

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Time SUCKER- consecutive integers

by kanha81 » Fri Jul 03, 2009 7:43 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

x = w
x > w
x/y is an integer
w/z is an integer
x/z is an integer

How do you solve such problems within 2 minute time limit? This was such a time sucker. I ended up guessing and got wrong. I could have actually got it right if I had time on my hands :evil:

Any suggestions?
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Re: Time SUCKER- consecutive integers

by Ian Stewart » Fri Jul 03, 2009 8:22 am
kanha81 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

x = w
x > w
x/y is an integer
w/z is an integer
x/z is an integer

How do you solve such problems within 2 minute time limit? This was such a time sucker. I ended up guessing and got wrong. I could have actually got it right if I had time on my hands :evil:

Any suggestions?
You'd want to rule out, as quickly as possible, as many answer choices as possible by using extremely simple numbers. So, for example, z could certainly be equal to 1 here, which means w/z and x/z can certainly be integers. That leaves us with A, B and C. And we can continue to use z = 1; it might be that w is the sum of one number, say 3, while x is the sum of two numbers, say 1 and 2, and w can certainly equal z, and A is out. Further, if x is the sum of very large integers, and w is not, then x can easily be larger than w. So B is out. That leaves C.

Now, you might want to know why C cannot be true. If we take a list of an even number of consecutive integers, where S is the smallest, and L the largest, we have:

S, S+1, S+2, ..., L-2, L-1, L

If we have an even number of terms here, then S+L will certainly be odd. And since we have an even number of terms, we can add them by grouping the largest and smallest, second largest and second smallest, and so on:

x = (S + L) + (S + 1 + L - 1) + (S + 2 + L - 2) + ...
x = (S+L) + (S+L) + (S+L) + ...
x = (y/2)*(S+L)

because we have y/2 pairs of numbers, and each pair adds to S + L. Now, we want to know if x/y could be an integer; that is, we want to know if:

[(y/2)*(S+L)]/y = (S+L)/2

could be an integer. Since S+L is odd, it cannot.

Of course, there are other ways to prove this - using the formula for summing consecutive integers should also work, for example.
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by ghacker » Fri Jul 03, 2009 8:50 am
with out any fancy work you can solve it in under 2 min

The question asks what could be not what must be

There is a big difference in could be and in must be ,

For example one can use POE to solve this in under 2 min
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by vinayakdl » Fri Jul 03, 2009 9:26 am
Sum of N consecutive integers divided by N yields and integer when N is odd while it does not yield an integer if it is even. Since y=2z, y is even and hence C cannot be true.

Vinayak
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