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Remainder Dilemma

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by zagcollins » Mon Jul 21, 2008 8:37 pm
How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

A.15
B.16
C.17
D.18
E.19

I'm getting an answer of 16...IS that correct or is the answer 17?
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by Vignesh.4384 » Mon Jul 21, 2008 9:31 pm
i think it is 16.

48/3 = 16
so 49 is also within range
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Re: Remainder Dilemma

by Stuart@KaplanGMAT » Mon Jul 21, 2008 9:49 pm
zagcollins wrote:How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

A.15
B.16
C.17
D.18
E.19

I'm getting an answer of 16...IS that correct or is the answer 17?
It's 17.

There are 51 integers from 0 to 50, inclusive.

If you divide by 3, there are 3 possible remainders: 0, 1 and 2.

So, if there are 51 consecutive integers and 3 possible remainders, there will be 51/3 = 17 of each.

Since 51 is evenly divisble by 3, we don't even have to worry about where in the cycle we started (which we would have had to do if the number of integers weren't divisible by 3).
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by zagcollins » Mon Jul 21, 2008 10:05 pm
thanks stuart...good logic...tell me something...is OG an accurate representation of the quant section of GMAT? if it isnt, what is?
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by Stuart@KaplanGMAT » Mon Jul 21, 2008 10:12 pm
zagcollins wrote:thanks stuart...good logic...tell me something...is OG an accurate representation of the quant section of GMAT? if it isnt, what is?
The majority of questions in the OG are mid range, i.e. 420-580 level. So, if you're scoring in that range it's great. If you're aiming for a high score, however, most of the questions that you see during the GMAT are tougher than what you see in the OG.

For the most part, the OG presents questions in order of difficulty, so the questions near the end of each section are the tougher ones. There are certainly some very hard questions in the OG, just not that many of them.

If you're looking for higher end questions (and strategies for dealing with them), I recommend Kaplan's 800 book for the GMAT. However, it's very important to be great at the mid-level questions as well. In the past I've helped some students who focused only on the high end questions and ended up getting crushed because they weren't familiar with the questions near the beginning of the exam; remember, you only get to the tough ones if you master the medium ones.
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by zagcollins » Mon Jul 21, 2008 11:32 pm
exactly...i made tht mistake in the past when i simply concentrated on tuff problems and thn i cudnt solve both, tuff and medium range..now i have started concentrating on both...i have used the kaplan 800 book already..their approach is quite different from the OG...but i like it...makes certain problems much simpler than they actually are...I have bn asking every1 the same question: Should I buy the manhattan series or stick to the books that I have, i.e. OG (all three books) and Kaplan 800??
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by pepeprepa » Tue Jul 22, 2008 1:18 am
Personnally, I find 16
I searched to maximize k in this inequality: 3k+1<=50 with k>0
I found k=16 so the last one wich has a remainder of 1 when divided by 3 is "3*16+1=49"
The first one is when k=1 so "3*1+1=4"
And from 1 to 16 we have ... 16 possible k terms


After that, I noticed some said 16 and other said 17 so I took time to list them :roll:
4 7 10 13 16 19 22 25 28 31 34 37 (40 instead of 39) 43 46 49

Explain to me if I forget one of them. Thks
Last edited by pepeprepa on Tue Jul 22, 2008 3:14 am, edited 1 time in total.
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by Suyog » Tue Jul 22, 2008 3:11 am
40 instead of 39... and u r missing 1 in ur list..
makes it 17. Choose (c)
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by pepeprepa » Tue Jul 22, 2008 3:19 am
Ok so 1 has a remainder of 1 when divided by 3
and 2 has a remainder of 2 when divided by 3
I know that now: "Every n divided by k with n<k has a remainder of n"

Thank you Suyog.

So it's 17
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by Ian Stewart » Tue Jul 22, 2008 10:16 am
pepeprepa wrote: I know that now: "Every n divided by k with n<k has a remainder of n"
That's almost true- it's definitely true if n is greater than or equal to zero, and less than k. If n can be negative, then the above is not true.

That said, it's pretty unlikely you'll see a question on the GMAT that asks about the remainder when you divide a negative number by a positive number. Still, it might be a good check to see if you fully understand remainders: "What is the remainder when -7 is divided by 5?" I've never seen a GMAT question that tests this, but I wouldn't be surprised if it showed up at some point.
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by pepeprepa » Tue Jul 22, 2008 10:39 am
"What is the remainder when -7 is divided by 5?" -2?
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by Ian Stewart » Tue Jul 22, 2008 12:31 pm
pepeprepa wrote:"What is the remainder when -7 is divided by 5?" -2?
Remember that when you divide an integer by 5, the only possible remainders are 0, 1, 2, 3 and 4- so one of those is the correct answer. Think about how you find the remainder when you divide, say, 9 by 5, and try to do the same with -7. I'll leave the question open for a while longer, then post the answer.
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by pepeprepa » Tue Jul 22, 2008 1:01 pm
"What is the remainder when -7 is divided by 5?"
3
Because -7=5x(-2)+3
It would mean you go on the other way but anyway I do not master the "concept".
If you have examples to master it would be cool.
Thank you for the help to understand.
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by Ian Stewart » Tue Jul 22, 2008 2:59 pm
pepeprepa wrote:"What is the remainder when -7 is divided by 5?"
3
Because -7=5x(-2)+3
It would mean you go on the other way but anyway I do not master the "concept".
If you have examples to master it would be cool.
Thank you for the help to understand.
Yes, 3 is the answer. When we find the remainder when we divide, say, 23 by 5, we find the nearest multiple of 5 which is lower than 23, and work out how far away we are. The remainder is 3 because 23 is 3 more than 20, the nearest (smaller) multiple of 5. When we divide -7 by 5, the remainder is also 3, because the nearest multiple of 5 that is smaller than -7 is -10, and -7 is 3 more than -10.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by parallel_chase » Tue Jul 22, 2008 3:22 pm
Ian Stewart wrote:
pepeprepa wrote:"What is the remainder when -7 is divided by 5?"
3
Because -7=5x(-2)+3
It would mean you go on the other way but anyway I do not master the "concept".
If you have examples to master it would be cool.
Thank you for the help to understand.
Yes, 3 is the answer. When we find the remainder when we divide, say, 23 by 5, we find the nearest multiple of 5 which is lower than 23, and work out how far away we are. The remainder is 3 because 23 is 3 more than 20, the nearest (smaller) multiple of 5. When we divide -7 by 5, the remainder is also 3, because the nearest multiple of 5 that is smaller than -7 is -10, and -7 is 3 more than -10.
Thanks Ian that clears everything.
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