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by goyalsau » Tue Oct 05, 2010 3:12 am
If y is an integer, then the least possible value of ||23 - 5y|| is
3
1
2
4
5


[spoiler]OA is 3
I think it should be 2, 23-25 = 2
[/spoiler]
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by mj78ind » Tue Oct 05, 2010 3:44 am
I agree, 2......
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by goyalsau » Tue Oct 05, 2010 3:51 am
mj78ind wrote:I agree, 2......
Source is authentic
and it has happened with me a lot of times whenever i thought that question is wrong. somebody comes up and gives the reason behind it. :P
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by shovan85 » Tue Oct 05, 2010 5:31 am
goyalsau wrote:If y is an integer, then the least possible value of ||23 - 5y|| is
3
1
2
4
5


[spoiler]OA is 3
I think it should be 2, 23-25 = 2
[/spoiler]
Awesome!! Great Question. I would have gone for 2 at first glance. If the source is authentic then here are my 2 cents to match the answer ;):

What is |x|?

When x < 0: |x| = -x
When x > 0: |x| = x

So Break it and take the first absolute

when (23-5y) < 0: |23 - 5y| = -(23-5y) Now if u take y = 5 u will get 4 and if you take y = 4 u will get -3 ....(a)
when (23-5y) > 0: |23 - 5y| = (23-5y) Now if u take y = 5 u will get -2 and if you take y = 4 u will get 3 .....(b)

We cannot take 4 as an answer in (a) as (23-5y) < 0. So -3.
We cannot take -2 as an answer in (b) as (23-5y) > 0. So 3.

Now try the second absolute value on -3 or 3. Either case the least is 3.

I think the key here is not to consolidate both ABS at a time :)

Hope I am correct....:)
Last edited by shovan85 on Tue Oct 05, 2010 5:54 am, edited 1 time in total.
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by GMATGuruNY » Tue Oct 05, 2010 5:47 am
goyalsau wrote:If y is an integer, then the least possible value of ||23 - 5y|| is
3
1
2
4
5


[spoiler]OA is 3
I think it should be 2, 23-25 = 2
[/spoiler]
If y=5, then ⎮23-5*5⎮ = ⎮-2⎮ = 2. There is no way to make ⎮23-5y⎮=1 if y is an integer, so the correct answer is 2.

This is Q50 from OG12, in which the OA is 2.
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by shovan85 » Tue Oct 05, 2010 5:57 am
GMATGuruNY wrote:
If y=5, then ⎮23-5*5⎮ = ⎮-2⎮ = 2. There is no way to make ⎮23-5y⎮=1 if y is an integer, so the correct answer is 2.

This is Q50 from OG12, in which the OA is 2.
So Actual Question contains only one ABSOLUTE symbol?
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by GMATGuruNY » Tue Oct 05, 2010 6:03 am
shovan85 wrote:
GMATGuruNY wrote:
If y=5, then ⎮23-5*5⎮ = ⎮-2⎮ = 2. There is no way to make ⎮23-5y⎮=1 if y is an integer, so the correct answer is 2.

This is Q50 from OG12, in which the OA is 2.
So Actual Question contains only one ABSOLUTE symbol?
Yes, the question in the OG12 has only set of absolute value symbols, as written in my reply.
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by shovan85 » Tue Oct 05, 2010 6:07 am
GMATGuruNY wrote:
shovan85 wrote:
GMATGuruNY wrote:
If y=5, then ⎮23-5*5⎮ = ⎮-2⎮ = 2. There is no way to make ⎮23-5y⎮=1 if y is an integer, so the correct answer is 2.

This is Q50 from OG12, in which the OA is 2.
So Actual Question contains only one ABSOLUTE symbol?
Yes, the question in the OG12 has only set of absolute value symbols, as written in my reply.
Thanks a lot.

If the Question were true (||23 - 5y|| two absolute marks) then also it will be 2 right?
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by GMATMadeEasy » Tue Oct 05, 2010 6:23 am
@shovan85 : think what || does actually ? Just make values positive. Once internal one makes value positive, it is done. f you put thousands or two even, it is same.
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