What is the remainder when the positive integer x is divided

[RonPurewal]

Hey Ron,

I understand it is easy to do these kind of questions by just substituting a number. But it would be really nice if you could explain to me the method for:

i really don't know -- maybe i'm just really tired today, but i honestly can't follow the proof/argument above.
however, i can really, truly, honestly tell you that any piece of number-theory argumentation that is this long and convoluted ... is not going to help you on test day. quite the opposite, actually.

the purpose of many problems on this test is to be outrageously difficult, or even impossible, to solve by "textbook" methods.
the POINT of these problems is the mental flexibility -- i.e., the problem is not so much intended to test remainders as it's intended to test your ability to quit using approaches that aren't working and start using backup approaches.
by continuing to hammer your head against the proverbial brick wall, you are totally walking right into the trap that the test makers are setting for you.

for a whole hour and a half of problems on which traditional algebra doesn't work at all, see the NOVEMBER 4 lecture at the following address:
http://www.manhattangmat.com/thursdays-with-ron.cfm

[rachelhong2012]

Hi,

I know we can plug #'s to test the choices. But I wonder if this theoretical approach is also correct:

What is the remainder when positive integer x is divided by 3?

Looking at the question, I generated 3 possibilities:

multiple of 3 + remainder of 0
multiple of 3 + remainder of 1
multiple of 3 + remainder of 2
The rephrased question: which possiblity is it?

a. When x is divided by 6 remainder is 2

x = multiple of 6 + remainder of 2
x = multiple of 3 + remainder of 2

sufficient
b. When x is divided by 15, remainder is 2

x = multiple of 15 + remainder of 2
x = multiple of 3 + remainder of 2

sufficient


Am I right?
Thanks!

[tim]

yes, but this only works because 3 is a factor of both 6 and 15..

[rachelhong2012]

yes, but this only works because 3 is a factor of both 6 and 15..

I see, thanks Tim!

[tim]

no problem!

[rkafc81]

another brute force approach - generate a chart of a bunch of ascending numbers (starting from anywhere, but easier to start from 0 as it is a bit easier to work out the remainder of numbers like say 0,1,2,3,4 than 45,46,47,48) and applyting the constraints given:

(Statement 1)
............ REMAINDER WHEN DIVIDED BY:
NUMBER /3 /2 /6
-------------------
0 3 2 6
1 1 1 1
2 2 0 2
3 0 1 3
4 1 0 4
5 2 1 5
6 0 1 6
7 1 1 1
8 2 0 2
9 0 1 3
10
11
12
13
14
15
16
17
18
19
20


etc. I can't be bothered to type it all out sorry -- try it and you'll know what I mean -- and eventually you'll see that all the numbers where both given conditions match (i.e. 3,9,15,21 etc.) all give a remainder of 3 when divided by 6... SUFFICIENT

[tim]

that'll work in a pinch. i do want to point out that your first row is incorrect though - 0 has a remainder of 0 no matter what you divide it by. the corollary to that is that EVERY integer is a factor of 0. (if you are skeptical, go back to the definitions!)

[Khush]

Yeah, that works.

This is also one of those problems where 'decoding' the statements by translating them into numbers works wonders: just sort through numbers by trial and error, and find the list of numbers that works for each statement.

(1) The numbers that work for this statement are 3, 9, 15, 21, etc. adding 6 each time.
(2) The numbers that work for this statement are 3, 15, 27, 39, etc. adding 12 each time.

If you make these lists, you'll find that either statement is sufficient on its own, so, D.

Incidentally, you should at least be able to translate '1 is the remainder upon dividing X by 2' into 'X is odd', IMMEDIATELY.

Hi Ron,

Thank You!

Plugging in numbers is really working great with these question types.
I do have few questions regarding this method in certain circumstances..

If in a question, combination of two statements gives values of x such as x= 13, 41, 69, etc.
If the question asks "what is the remainder when x is divided by 3", then in the above case the division yields different remainders.

Is it possible that in an official question, we will see different remainders for a pattern of values of x?

In general, how many values of x should be divided by the given divisor to check for the value of the remainder. I usually check for only one value of x.

Please advise.

[RonPurewal]

Is it possible that in an official question, we will see different remainders for a pattern of values of x?

If the question is multiple-choice, then it will have only one correct answer. So, if the answer choices are numbers, you'll only see one number.
In that case, checking a single case will suffice.

If the question is data sufficiency, then the whole point is to determine whether there's only one possible number ("sufficient") or more than one ("not sufficient"). So, for DS, you'll have to check multiple cases, until one of the following happens:
* You get 2 different answers, in which case you're done and it's "not sufficient"
* You keep getting the same 1 answer over and over, until you're thoroughly convinced of the pattern ("sufficient")

[Khush]

Thank you so much Ron!

This was helpful.......

[RonPurewal]

You're welcome.