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When positive integer x is divided by positive integer y, the result i

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When positive integer x is divided by positive integer y, the result i  [#permalink]

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New post 28 Jan 2015, 07:40
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Re: When positive integer x is divided by positive integer y, the result i  [#permalink]

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New post 02 Feb 2015, 03:03
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Bunuel wrote:
When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2-digit 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 "thirty-two hundredths" or 32/100. Since 32/100 reduces to 8/25, you know that in order to have a two-digit 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 two-digit 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  [#permalink]

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New post 28 Jan 2015, 09:59
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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  [#permalink]

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New post 28 Jan 2015, 16:58
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  [#permalink]

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New post 28 Jan 2015, 22:09
1
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  [#permalink]

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New post 04 Feb 2015, 01:50
\(\frac{32}{100} = \frac{16}{50} = \frac{24}{75} =\) .....

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  [#permalink]

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New post 10 Apr 2015, 02:32
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  [#permalink]

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New post Updated on: 21 Aug 2017, 14:53
.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, 15:16.
Last edited by gracie on 21 Aug 2017, 14:53, edited 5 times in total.
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Re: When positive integer x is divided by positive integer y, the result i  [#permalink]

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New post 13 Jun 2016, 17:22
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Bunuel wrote:
When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2-digit 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  [#permalink]

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New post 17 Jun 2016, 06:46
Bunuel wrote:
When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2-digit 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  [#permalink]

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New post 07 Jul 2016, 03:18
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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  [#permalink]

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New post 21 Aug 2017, 13:25
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 double-digit 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  [#permalink]

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New post 16 Oct 2018, 20:31
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  [#permalink]

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New post 18 Oct 2018, 18:30
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1
Bunuel wrote:
When positive integer x is divided by positive integer y, the result is 59.32. What is the sum of all possible 2-digit 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 two-digit remainder is 8(2) = 16 and the last two-digit remainder is 8(12) = 96. We can now use the formula sum = average x quantity to find the sum of all possible two-digit 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  [#permalink]

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New post 02 Nov 2018, 02:41
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


chetan2u
Hello...:)

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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Re: When positive integer x is divided by positive integer y, the result i   [#permalink] 02 Nov 2018, 02:41
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