When the positive integer x is divided by 4

[Anne1276]

For (1) I understand why x must be an odd number but what information leads us to the fact that it must be a multiple of 3?


THANKS!

[GMATPaduan]

Anne,

It would be helpful if you posted the entire question, so that everyone on the forum could benefit from the advice given by the instructors.

I also do not know if the instructors will answer a question, where the entire question is not writtten out and the source cited - they don't always have a OG book in front of them at all times.

[Anne1276]

Here is the original problem:

When the positive integer x is divided by 4, is the remainder equal to 3?

(1) When x/3 is divided by 2, the remainder is 1.

(2) x is divisible by 5.

[StaceyKoprince]

Yes, we do need the entire question (and answer choices) as well as the source before we can answer the question. Anne, you list the source as CAT#1. Is this an MGMAT CAT or GMATPrep or some other source?

[Anne1276]

Sorry, it's my first time posting.

This is MGMAT CAT #1.

[StaceyKoprince]

No problem! If you look at the top of the list of threads in each forum, there's a sticky explaining how to post - take a look when you have a minute.

Statement 1 is a little tricky - it's got lots of info layered in there.

The statement tells us that when the combo (x/3) is divided by 2, the remainder is 1. Since the remainder is an integer, the number (x/3) itself must also be an integer. In order for (x/3) to be an integer, x must be divisible by 3. Otherwise, we would have a fraction or decimal amount for x/3.

For example 9 is divisible by 3, but 10 and 11 are not. (9/3) = 3 and 3/2 = 1 r.1. 10/3 = 3.333 so I can't get an integer remainder when dividing by 2.

Hope this helps!

[unique]

Can someone solve this using Algebra please. I made a mistake interpreting 1.

I treated x/3 divided by 2 as x/6 -- how can I avoid such mistakes?

[StaceyKoprince]

You just need to study how to do this math so you know the problem is a remainder problem and you know how best to approach remainder problems. You could technically approach this with algebra, but that would make it an even harder problem than it already is.

Technically, if I say "when x/3 is divided by 2" ultimately I really am taking x and dividing it by 6 - so there isn't an error there to call it x/6. You just have to know how to continue to handle the information.

If I write it that way, though, I may not notice the key piece of info that x is divisible by 3 - so I just have to be careful. I should also notice that, if x has a remainder of 1 when I divide it by 2, then it must be odd.

So, I know from (1) that x is divisible by 3 and it is odd. So x could be, say, 3 or 9, but it can't be 6. If x = 9, the remainder is 1 when divided by 4. If x = 15 (the next largest possibility), the remainder is 3 when divided by 4. So I can't answer the question definitively yes or no. Insufficient.

(2) x is div. by 5. So x could be 5, 10, 15, etc. If x=5, the remainder is 1 when divided by 4. If x=15, the remainder is 3 when divided by 4. Once again, I can't answer definitively - insufficient.

(1) AND (2). Must be a multiple of 5 and 3, and be odd. Possibilities: 15, 45, 75, etc. I already know from previous work that the remainder is 3 when divided by 4. Let's try another. When 45 is divided by 4, the remainder is 1. I have "yes" and "no" again, so insufficient.

Answer is E.

[DCE]

Now Let's see how do we represent the first equation in Algebra

Whenever you are dividing consequently, pick the second half first i.e dividing by 2 leaves a remainder 1
This gives us: 2k+1

Now this has to be divisible by 3, so mutliply it by 3. Now we get: 3(2k +1). This is what it boils down to 6k +3.

Hope that this helps.

Thanks,
DCE

[RonPurewal]

Now Let's see how do we represent the first equation in Algebra

Whenever you are dividing consequently, pick the second half first i.e dividing by 2 leaves a remainder 1
This gives us: 2k+1

Now this has to be divisible by 3, so mutliply it by 3. Now we get: 3(2k +1). This is what it boils down to 6k +3.

Hope that this helps.

Thanks,
DCE

correct.

it may be more straightforward if you actually include x/3 in this calculation, because the problem is written with reference to the quantity x/3.

therefore:
"if you divide x/3 by 2, the remainder is 1"
--> x/3 = 2k + 1
multiply by 3 --> x = 6k + 3