i think it is 16.

48/3 = 16
so 49 is also within range

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).

thanks stuart...good logic...tell me something...is OG an accurate representation of the quant section of GMAT? if it isnt, what is?

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.

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??

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

40 instead of 39... and u r missing 1 in ur list..
makes it 17. Choose (c)

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

"What is the remainder when -7 is divided by 5?" -2?

"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.

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.