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For nonnegative integers x and y, what is the remainder when [#permalink]

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15 Dec 2012, 15:53

good one... do you know the explanation? given x,y \geq 0; what is R in x = Qy + R 1) x=13y + 0.8y in this case we need value of 0.8y but we don't know values of x,y; Not Sufficient 2) x+y < \(10^4\); Not Sufficient

1+2 from 1) x=13.8y so from 2) 14.8y < \(10^4\) so y < \frac{\(10^4\)}{14.8} there exists multiple integer values of y less than 10^4/14.8 for which x is integer... So my answer is E.... Since OA is different, i think my thinking in the last part is wrong.... Not sure how to proceed from here....

Bunuel.... I need your presence here..... I am becoming reckless to get a solution for this.....

Last edited by Amateur on 15 Dec 2012, 18:12, edited 1 time in total.

Re: For nonnegative integers x and y, what is the remainder when [#permalink]

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16 Dec 2012, 06:57

shanmugamgsn wrote:

BangOn wrote:

Amateur wrote:

Bunuel.... I need your presence here..... I am becoming reckless to get a solution for this.....

Let me try

X/Y = 13+ .8 X/Y = 13 + 4/5

X and Y are integers... Possibilities of Remainder/Divisor = 4/5, 8/10, 12/15 if 4/5 X = 69 Y = 5 If 8/10 X=138 Y =10 If 12/15 X= 207 Y =15

B says we should have the total digits of X and Y less than 5 only 4/5 suffice. Hence OA.

B says we should have the total digits of X and Y less than 5 only 4/5 suffice.

Didnt this mean total no of digits less than 5? Can u explain ur soln Bangon

Yeah Sure. Total Digits in both X and Y should be less than 5. if 4/5 X = 69 Y = 5 No. Of digits in X and Y is 3 If 8/10 X=138 Y =10 No. Of digits in X and Y is 5 If 12/15 X= 207 Y =15 No. Of digits in X and Y is 5

So from C, we have only one possible solution X = 69 Y = 5
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Re: For nonnegative integers x and y, what is the remainder when [#permalink]

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16 Dec 2012, 14:42

JJ2014 wrote:

For nonnegative integers x and y, what is the remainder when x is divided by y?

(1) x/y = 13.8 (2) The numbers x and y have a combined total of less than 5 digits

BangOn wrote:

Amateur wrote:

Bunuel.... I need your presence here..... I am becoming reckless to get a solution for this.....

B says we should have the total digitsof X and Y less than 5 only 4/5 suffice. Hence OA.

BangOn wrote:

thank you for the explanation, statement B makes sense now. one thing- how did you get the possibilities of remainder/divisor?[/quote]

Possibilities by just multiplying the fraction .8 = 4/5 1) 4/5*1/1 = 4/5 2) 4/5 * 2/2 = 8/10 . . similarly[/quote] estimating reminders is fine.... but calculating x and y values when you have remainders, I think it is time consuming process for a 600-700 problem..... Also, B didnot say we should have total digits less than 5, b said x and y have a combined total of less than 5 digits.... if it did say the number of digits of x and y have a combined total of less than 5 you must have been correct.... but since it said less than 5 digits.... the equation will be x+y<10^4

Re: For nonnegative integers x and y, what is the remainder when [#permalink]

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16 Dec 2012, 20:12

Statement -1 : X= 13 Y + .8 Y As X & Y are Non negative integers ,hence .8Y integer Possible values of Y for which R =.8 Y is an integers : 5,10,15,... As there are multiple values of Y for which R is an integers- Not sufficient

Statement 2 : Insufficient

Together : As ,Total digits ( X & Y ) less than 5 : this implies Y cant be in double digit ,so Y=5 Hence C is sufficient

As I saw the explanations and the question stem, it revealed two flaws.

(1).Non - negative integer x,y '0' is also non-negative integer.

(2). Combined total of digits of x,y mean the sum total of x and y or as stated in the explanation

Rgds, TGC !
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Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: For nonnegative integers x and y, what is the remainder when [#permalink]

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10 Aug 2013, 08:00

2

This post received KUDOS

semwal wrote:

Could somebody help with this tough question:-

If zt < -3, is z < 4? E a. z < 9 b. t < -4

thanks

From your post I perceive that you are a new bee in this forum.You have to post your question at the right place and in a right way where in you provide OA/source/difficulty level.

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: For nonnegative integers x and y, what is the remainder when [#permalink]

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17 Jun 2014, 03:45

I was having a lot of trouble with this one.

I could cross of the first statement as both 138/10 and 1380/10 would give the same answer, but the remainder would've been different.

For the second one I got that we have a max of 4 numbers. I could not really solve it, but I figured it was possible given the information so I picked C. (However, it was impossible with just the information from B, I sometimes just jump ahead of time-.-)

what does combined total mean? That is a redundant usage...question not well formed

Source please

This means that the number of digits in x plus the number of digits in y is less than 5.

For nonnegative integers x and y, what is the remainder when x is divided by y?

(1) x/y = 13.8 --> x/y = 69/5 = 138/10 = 207/15 = ... You'll get different remainders for different x and y. Not sufficient.

(2) The numbers x and y have a combined total of less than 5 digits. Clearly insufficient.

(1)+(2) From (2) we know that the number of digits in x plus the number of digits in y is less than 5, so x and y could only be 69 and 5 (combined total of 3 digits). Sufficient.

Re: For nonnegative integers x and y, what is the remainder when [#permalink]

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04 Oct 2015, 08:24

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

For nonnegative integers x and y, what is the remainder when x is divided by y?

(1) x/y = 13.8 (2) The numbers x and y have a combined total of less than 5 digits

There are 2 variables (x,y) and 2 equations, so there is high chance that (C) will be our answer. Looking at the conditions together, we will substitute in values directly, as it is the fastest way to solve remainder problems. From x=yQ+r(Q=integer, 0<=r<y), x/y=(yQ+r)/y=Q+(r/y)=13+0.8 r/y=0.8에서 r=0.8y r is an integer, so y=5,10,15... . Then we can see that (x,y,r)=(69,5,4),(138,10,8),(207,15,12)... But the only set of values that suits conditions 2 is (x,y,r)=(69,5,4). The conditions give a unique value, so they are sufficient and the answer becomes (C).

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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