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When positive integer x is divided by positive integer y, the result i
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28 Jan 2015, 06:40
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Re: When positive integer x is divided by positive integer y, the result i
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02 Feb 2015, 02:03
Bunuel wrote: When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2digit remainders for x/y?
A. 560 B. 616 C. 672 D. 728 E. 784
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:An important division concept that the GMAT likes to test is that the remainder divided by the divisor becomes the decimal places. (Try this with smaller numbers like 5 divided by 4. The remainder is 1, which when divided by 4 becomes .25. Those are the decimal places in the result 1.25.) Here you're given the decimal places as .32, which translates to "thirtytwo hundredths" or 32/100. Since 32/100 reduces to 8/25, you know that in order to have a twodigit remainder you must have a divisor that's a multiple of 25 and a remainder that's a multiple of 8 and that's a twodigit number. That leaves you with these possibilities: 16/50 24/75 32/100 40/125 etc. Which should tell you that the possible remainders are all multiples of 8 from 16 to 96. To sum those values, take the number of terms (there are 11 of them, since that's 2 * 8 through 12 * 8) and multiply by the middle value (56, the average of 16 and 96) and the answer is 616.
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Re: When positive integer x is divided by positive integer y, the result i
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28 Jan 2015, 08:59
ans B 616... remainders = .32=32/100=8/25=16/50 and so on.. so two digit remainders are 16+24+32+....+96.. =8(2+3+4....+12)=616
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Re: When positive integer x is divided by positive integer y, the result i
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28 Jan 2015, 15:58
chetan2u wrote: ans B 616... remainders = .32=32/100=8/25=16/50 and so on.. so two digit remainders are 16+24+32+....+96.. =8(2+3+4....+12)=616 Why would we not count the 8/25? I got 624. Wouldn't it be =8(1+2+3+4....12)? ****Nevermind reread question and see that it is only 2 digit answers that we are using so then B=616



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Re: When positive integer x is divided by positive integer y, the result i
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28 Jan 2015, 21:09
Here we go
Remainder/Divisor = decimal part
Say remainder = r
r/y = 0.32 > 32/100
so we can have 16+24+32+40+48+56+64+72+80+88+96 = 616
option B is correct



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Re: When positive integer x is divided by positive integer y, the result i
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04 Feb 2015, 00:50
\(\frac{32}{100} = \frac{16}{50} = \frac{24}{75} =\) ..... Two digit remainder addition = 16+24+32+40+48+56+64+72+80+88+96 = 8*2 + ................ + 8*12 = 8(2+.......+12) = 77*8 = 616 Answer = B
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Re: When positive integer x is divided by positive integer y, the result i
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10 Apr 2015, 01:32
x/y = 96.12 x/y ==> 9612/100 ==> x/y = 96 12/100 From that you can see that the reminder is 12. If you make the reminder 9, that means that you multiply 12 by 3/4 Multiply 100 also by 3/4 and you get 75.
Answer B



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When positive integer x is divided by positive integer y, the result i
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Updated on: 21 Aug 2017, 13:53
.32 reduces to 8/25, 16/50, 24/75, 32/100....96/300 the sum of all 2 digit multiples of 8≤96=11*(16+96)/2=616 B
Originally posted by gracie on 13 Jun 2016, 14:16.
Last edited by gracie on 21 Aug 2017, 13:53, edited 5 times in total.



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Re: When positive integer x is divided by positive integer y, the result i
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13 Jun 2016, 16:22
Bunuel wrote: When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2digit remainders for x/y?
A. 560 B. 616 C. 672 D. 728 E. 784
Kudos for a correct solution. Reminder of x/y= 32/100= 8/25 First two digits multiple of 8 = 16 Last two digit multiple of 8 = 96 (8*12) Total = Average * Total number of digits in the set Since its a uniformly arranged set, average = 1st digit+last digit/2= 16+96/2 Total number of digits= 11 Total= 16+96/2*11= 616 B is the answer
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Re: When positive integer x is divided by positive integer y, the result i
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17 Jun 2016, 05:46
Bunuel wrote: When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2digit remainders for x/y?
A. 560 B. 616 C. 672 D. 728 E. 784
Kudos for a correct solution. After I saw the solution..I got to know what exactly the question was asking for..somehow..I feel that this question should ask it in a better way..if thats not the case, then please help me comprehend such abridged questions.
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Re: When positive integer x is divided by positive integer y, the result i
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07 Jul 2016, 02:18
Hi everyone!
A question here...
I don't really get what the Q is asking, since 8/25, 16/50 etc all represent the same exact number! so, how come we are summing numbers from 8 to 96? for what? They would reduce to the simplest fraction 8/25, so how can those be different numbers (remainders in this case)?



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Re: When positive integer x is divided by positive integer y, the result i
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21 Aug 2017, 12:25
iliavko wrote: Hi everyone!
A question here...
I don't really get what the Q is asking, since 8/25, 16/50 etc all represent the same exact number! so, how come we are summing numbers from 8 to 96? for what? They would reduce to the simplest fraction 8/25, so how can those be different numbers (reminders in this case)? i found a better explanation 32/100 reduces down to 8/25 you can not use 8 because that is not a doubledigit number, but you want to look for everything that could l reduce down to 8/25 which represent all the other numbers that could be the remainers the easy way is to multiply by all numbers, 1* 8/ 25= 8/25. 2* 8/25 = 16/50 ...... 12* 8/25 = 96/ 300, and you can not use 13 because that is greater than two digits. so basically you then add up all those values in between



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Re: When positive integer x is divided by positive integer y, the result i
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16 Oct 2018, 19:31
Lowest fraction equivalent of the decimal part is 8/25. Now since we are considering only 2 digit remainders, only the numerator 16 onwards will be used and any fraction with numberator greater than 96 will not be considered as it will become more than 2 digits \(=\frac{8*2}{25*2}=\frac{16}{50}\) \(=\frac{8*3}{25*3}=\frac{24}{75}\) .. .. .. \(=\frac{8*12}{25*12}=\frac{96}{300}\) 2 digit remainders are 16,24,32,40,48,56,64,72,80,88,96 Sum of above series= \(\frac{n*(n1 + nl)}{2}\) where n= number of terms =11(in our case), n1=First term of series and nl=last term of series \(=\frac{11*(16+96)}{2}\) \(=\frac{11*112}{2}\) =11*56 =616
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Re: When positive integer x is divided by positive integer y, the result i
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18 Oct 2018, 17:30
Bunuel wrote: When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2digit remainders for x/y?
A. 560 B. 616 C. 672 D. 728 E. 784 Since x/y = 59.32 = 59 32/100 = 59 8/25, we see that the possible remainders are of the form 8k where k is a positive integer. Therefore, the first twodigit remainder is 8(2) = 16 and the last twodigit remainder is 8(12) = 96. We can now use the formula sum = average x quantity to find the sum of all possible twodigit remainders. We see that average = (16 + 96)/2 = 56 and quantity = (96  16)/8 + 1 = 11. Therefore, the sum is: 56 x 11 = 616 Answer: B
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Re: When positive integer x is divided by positive integer y, the result i
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02 Nov 2018, 01:41
chetan2u wrote: ans B 616... remainders = .32=32/100=8/25=16/50 and so on.. so two digit remainders are 16+24+32+....+96.. =8(2+3+4....+12)=616 chetan2uHello... there is a technique which uses the last digits to find the remainders. Where can i read the nuts and bolts of this? thanks in advance




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