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Picking Numbers on DS

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Picking Numbers on DS

by doclkk » Wed Jun 24, 2009 5:10 am
I've ran into many questions number properties questions and I was wondering what all of your approaches were.

Example:

If x and y are nonzero integers, what is the remainder when x is divided by y ?

(1) When x is divided by 2y, the remainder is 4.

(2) When x + y is divided by y, the remainder is 4.

I look at this question and I think - can I rephrase. No

Number properties... hmm, not too scary. I think. Let's begin.

Statement 1.

X / 2y R 4

HMMM ... What divided by 2 of something gives a remainder of 4.

Ah Ha

10/2(3) R4.

10/6

so now I think

10/3 = R1

R1 for 10/3

Next Number hmmm. Crap, I can't think of something else. 11 and what.

Bingo 11/7! Crap ... 11 Doesn't work

12/8 - Huzzah - works

12/2(4) = R4

12/4 = R0.

HUZZAH!

Statement 1 = insufficient

Statement 2

Which numbers to pick ... Hmm, I don't know. How about the numbers I picked for the first one?

10+3 / 3 = 13/3 = R1

12+4 / 3= 16/3 = R1

Crap that's two.

Hmmm what new number to pick.

13+3 / 3 = 16/3 = R1.

3 Numbers work. Crap - 1:50 spent on question. Try another one ? um um um B?

This happens to be B - but 1:50 and a shaky B at best.

What are your thoughts?
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by david4431 » Wed Jun 24, 2009 5:43 am
Hi doc,

Here is my opinion. S1 is probably the trickier of the 2 statements, so let's start with S2.

The key to S2 is to realize that you can break the equation up into 2 separate equations. (x+y)/y can be broken down into x/y + y/y. This equals x/y + 1. Since you're looking for the remainder of x/y and you're told what it is in S2, it is sufficient.

I actually started with S1, and my approach was to factor out the 1/2 so that I was left with (1/2)(x/y). On hindsight, all you had to do in S1 is multiply by 2 to get ride of the fraction. The remainder would also double. The question is: what is the denominator? If the denominator was 5, you would have a remainder of 1. If it was 9, you would have a remainder of 8.

Having said all this, the key for me was after solving S2, I realized that the correct answer for x/y was remainder 4. I looked at S1 and didn't see how I could manipulate a remained 4 out of that answer, so my answer was B.



But your real question is on the issue of time and speed while testing - I think this is an issue that many of us have. I personally don't see any shortcuts here. Here is an idea for you though: https://www.beatthegmat.com/inequality-d ... tml#162331

Read deltaforce's post in regards to having a set of numbers that you try out. Your speed should increase if you already have a predetermined set of numbers. The rest of your speed increase will come from practice, problem recognition and improved understanding of the quant material being tested. Good luck.
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by abhinav85 » Wed Jun 24, 2009 7:55 am
IMO A

What is the OA???
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Re: Picking Numbers on DS

by Ian Stewart » Wed Jun 24, 2009 9:35 am
doclkk wrote:I've ran into many questions number properties questions and I was wondering what all of your approaches were.

Example:

If x and y are nonzero integers, what is the remainder when x is divided by y ?

(1) When x is divided by 2y, the remainder is 4.

(2) When x + y is divided by y, the remainder is 4.
There are a couple of principles that you might find useful here:

First, if you're dividing by d, then ..., x - 2d, x - d, x, x+d, x+2d, x+3d, etc all give the same remainder. From this fact, Statement 2 is immediately sufficient.

[and if you want to know why, you can either think of some actual numbers -- when dividing by 7, for example, the numbers 3, 10, 17, 24, 31, 38, etc all give the same remainder -- or you can prove this abstractly. Say the remainder is r when x is divided by d. Then,

x = qd + r

(x is r greater than a multiple of d). Now add d to both sides:

x + d = qd + r + d = (q + 1)d + r

so x + d is also r greater than a multiple of d, and the remainder is r when x + d is divided by d. ]

Second, if you want to pick numbers in a remainder question, the easiest number is usually the remainder itself. Remember that when dividing, say, 4 by 7, the remainder is 4, and the quotient is zero (if you have four apples, and seven children, you can give each child zero apples, so zero is the quotient, and you have four left over - that's the remainder). So, when looking at Statement 1, there's no reason not to use x = 4, just as long as 2y is greater than 4. If you let x = 4 and y = 3, or let x = 4 and y = 4, you get two different answers to the original question, and S1 is insufficient.
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by abhinav85 » Wed Jun 24, 2009 9:43 am
Hey Ian,

I used this logic to make A sufficeint.
If its not suff. as you mentioned can you tell what i am missing???

the question is basically asking what is the remainder when x/y.

From statement 1 we get,

x = 2y + 4, so if remainder is 4 y has to be more then that.
so the next number will be 5.

14 = (2 x 5) + 4

This is my reasoning for A.
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by Ian Stewart » Wed Jun 24, 2009 10:39 am
abhinav85 wrote:
I used this logic to make A sufficeint.
If its not suff. as you mentioned can you tell what i am missing???

the question is basically asking what is the remainder when x/y.

From statement 1 we get,

x = 2y + 4, so if remainder is 4 y has to be more then that.
so the next number will be 5.

14 = (2 x 5) + 4

This is my reasoning for A.
I'm not sure I follow the progression here, but there's certainly a mistake when you've translated the Statement into algebra. Remember, if the remainder is r when n is divided by d, then we know that n = qd + r. So here, if the remainder is 4 when x is divided by 2y, then we know that 2y is greater than 4 (and not, as you say above, that y is greater than 4) and that:

x = q*2y + 4

You left out the quotient q. From this, we can see that x is 4 more than a multiple of y. *If* y is greater than 4, this guarantees that the remainder will be 4 when x is divided by y. However, y can be 3 or 4 here, in which case the remainder cannot be 4 when x is divided by y, which is why the statement is insufficient.
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by doclkk » Wed Jun 24, 2009 12:48 pm
abhinav85 wrote:IMO A

What is the OA???
It's B
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Re: Picking Numbers on DS

by doclkk » Wed Jun 24, 2009 1:07 pm
Ian Stewart wrote:
doclkk wrote:I've ran into many questions number properties questions and I was wondering what all of your approaches were.

Example:

If x and y are nonzero integers, what is the remainder when x is divided by y ?

(1) When x is divided by 2y, the remainder is 4.

(2) When x + y is divided by y, the remainder is 4.
There are a couple of principles that you might find useful here:

First, if you're dividing by d, then ..., x - 2d, x - d, x, x+d, x+2d, x+3d, etc all give the same remainder. From this fact, Statement 2 is immediately sufficient.

[and if you want to know why, you can either think of some actual numbers -- when dividing by 7, for example, the numbers 3, 10, 17, 24, 31, 38, etc all give the same remainder -- or you can prove this abstractly. Say the remainder is r when x is divided by d. Then,

x = qd + r

(x is r greater than a multiple of d). Now add d to both sides:

x + d = qd + r + d = (q + 1)d + r

so x + d is also r greater than a multiple of d, and the remainder is r when x + d is divided by d. ]

Second, if you want to pick numbers in a remainder question, the easiest number is usually the remainder itself. Remember that when dividing, say, 4 by 7, the remainder is 4, and the quotient is zero (if you have four apples, and seven children, you can give each child zero apples, so zero is the quotient, and you have four left over - that's the remainder). So, when looking at Statement 1, there's no reason not to use x = 4, just as long as 2y is greater than 4. If you let x = 4 and y = 3, or let x = 4 and y = 4, you get two different answers to the original question, and S1 is insufficient.
Wow - this is definitely different thinking. I typically don't think of a remainder with a smaller numerator. so you would argue these numbers are easier to work with?
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by rah_pandey » Wed Jun 24, 2009 10:35 pm
Just to add

by first statement
x=2*k*y+4

x/y=2k+4/y

therefore remainder depends on 4/y
since we are not told that y>4 therefore possible remainders are 0,1,4

hence insufficient
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