 |
| Author |
Message |
|
Guest
|
Post subject: x is the sum of y consecutive integers Posted: Thu May 08, 2008 1:04 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
1. x = w
2. x > w
3. x/y is an integer
4. w/z is an integer
5. x/z is an integer
Explanation given:
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.
Question:
Can you please explain another way to solve this problem? Is there another way to do it instead of plugging in numbers (which would take a long time in this case)?
|
|
 |
|
 |
|
Guest
|
Post subject: Re: x is the sum of y consecutive integers Posted: Wed May 14, 2008 12:20 am |
|
|
|
Anonymous 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
Explanation given: 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.
Question: Can you please explain another way to solve this problem? Is there another way to do it instead of plugging in numbers (which would take a long time in this case)?
y = 2z
if z = 1, y 2
if z = 2, y = 4
if z = 3, y = 6
if z = 4, y = 8
so y is always even and average of even consecutive integers is always not an integer.
lets write down some even consecutive integers. suppose y = 4:
so the integers are: a, a+1, a+2, a+3
x = a + a+1 + a+2 + a+3
x = 4a + 6
so 4a+6 is not evenly divisible by 4. therefore x/y is not an integer...
|
|
 |
|
 |
|
UPA
|
Post subject: Re: x is the sum of y consecutive integers Posted: Wed May 14, 2008 11:36 pm |
|
|
|
Anonymous wrote: Anonymous 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
Explanation given: 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.
Question: Can you please explain another way to solve this problem? Is there another way to do it instead of plugging in numbers (which would take a long time in this case)? y = 2z if z = 1, y 2 if z = 2, y = 4 if z = 3, y = 6 if z = 4, y = 8 so y is always even and average of even consecutive integers is always not an integer. lets write down some even consecutive integers. suppose y = 4: so the integers are: a, a+1, a+2, a+3 x = a + a+1 + a+2 + a+3 x = 4a + 6 so 4a+6 is not evenly divisible by 4. therefore x/y is not an integer...
wow i forgot to put my username.
|
|
 |
|
 |
|
rfernandez
|
Post subject: Posted: Fri May 30, 2008 5:34 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 386
|
|
UPA's solution is akin to how I would solve it as well. I usually find it helpful to express consecutive sets using variables like a, a+1, a+2, etc. and seeing what insights I can draw from that analysis.
Rey
|
|
 |
|
 |
|
viv09
|
Post subject: Re: x is the sum of y consecutive integers Posted: Sun Sep 11, 2011 11:51 pm |
|
 |
| Students |
|
|
Posts: 12
|
|
average of even consecutive integers is always not an integer.
Is this a rule?
eg. 2+4+6/3 = 12/3 = 4
Can some one please explain the statement given above.
Thanks
|
|
 |
|
 |
|
saptadeepc
|
Post subject: Re: x is the sum of y consecutive integers Posted: Mon Sep 12, 2011 12:26 am |
|
 |
| Students |
|
|
Posts: 55
|
viv09 wrote: average of even consecutive integers is always not an integer.
Is this a rule?
eg. 2+4+6/3 = 12/3 = 4
Can some one please explain the statement given above.
Thanks what he meant is average of even "numbers" of consecutive numbers is not an interger for example 1 + 2 / 2 OR 1 + 2 + 3 + 4 / 4 OR 1 + 2 + 3 + 4 + 5 + 6 / 6 are never integers
|
|
 |
|
 |
|
viv09
|
Post subject: Re: x is the sum of y consecutive integers Posted: Mon Sep 12, 2011 9:42 pm |
|
 |
| Students |
|
|
Posts: 12
|
|
oh, I think I got caught up in the words... Thanks!!
|
|
 |
|
 |
|
jnelson0612
|
Post subject: Re: x is the sum of y consecutive integers Posted: Mon Oct 17, 2011 12:31 am |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 1857
|
|
Great! :-)
_________________ Jamie Nelson ManhattanGMAT Instructor
|
|
 |
|
 |
|
Users browsing this forum: No registered users and 0 guests |
| |
|
|
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum
|
|
|
 |