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X and Y are both integers. If X / Y = 59.32, then what is the sum of a

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X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 07 Mar 2016, 10:28
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 07 Mar 2016, 11:33
2
Remainder = 0.32 --> 32/100 --> Can be written as (32/4) / (100/4) = 8/25

So remainders can be 8, 16, 24, 32, ..... 96.

We need the sum of only 2 digit remainders --> 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616

Answer: B
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 08 Mar 2016, 12:37
Is there a shorter way for finding sum of all the 2 digit factors of 8?
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 08 Mar 2016, 12:37
Is there a shorter way for finding sum of all the 2 digit factors of 8?
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 08 Mar 2016, 16:55
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mosurok wrote:
Is there a shorter way for finding sum of all the 2 digit factors of 8?


Yes. Once you realize that the numbers are multiple of 8 and they are in AP, you can apply the formula \((first term + last term)/2\) * Number of terms. --> 56 * 11 = 616.
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 09 Mar 2016, 03:40
Vyshak wrote:
Remainder = 0.32 --> 32/100 --> Can be written as (32/4) / (100/4) = 8/25

So remainders can be 8, 16, 24, 32, ..... 96.

We need the sum of only 2 digit remainders --> 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616

Answer: B


Can you please explain the first step? I did not understand how doing that step could decide what the remainders could be, escpecially the 8/25.

Thank you :)
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 09 Mar 2016, 03:59
nilleskold wrote:
Vyshak wrote:
Remainder = 0.32 --> 32/100 --> Can be written as (32/4) / (100/4) = 8/25

So remainders can be 8, 16, 24, 32, ..... 96.

We need the sum of only 2 digit remainders --> 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616

Answer: B


Can you please explain the first step? I did not understand how doing that step could decide what the remainders could be, escpecially the 8/25.

Thank you :)


We have to reduce 32/100 to its lowest ratio by dividing both numerator and denominator with a common factor. In other words the original ratio (0.32) gets unaffected by dividing both numerator and denominator with a common factor. We cannot reduce it beyond 8/25 as there are no common factors beyond this point.
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 09 Mar 2016, 05:46
Right, got it. Also read Veritas Prep's approach on this as well and it made it clearer, thanks!
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 16 Mar 2016, 04:31
Vyshak wrote:
Remainder = 0.32 --> 32/100 --> Can be written as (32/4) / (100/4) = 8/25

So remainders can be 8, 16, 24, 32, ..... 96.

We need the sum of only 2 digit remainders --> 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616

Answer: B



Remainder is 8/25 but how did you get the values of remainder as 8,16,24,32,....... 96???
Please explain??
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 16 Mar 2016, 04:45
MeghaP wrote:
Vyshak wrote:
Remainder = 0.32 --> 32/100 --> Can be written as (32/4) / (100/4) = 8/25

So remainders can be 8, 16, 24, 32, ..... 96.

We need the sum of only 2 digit remainders --> 16 + 24 + 32 + 40 + 48 + 56 + 64 + 72 + 80 + 88 + 96 = 616

Answer: B



Remainder is 8/25 but how did you get the values of remainder as 8,16,24,32,....... 96???
Please explain??


Once you find out 8/25, you just have to multiply the numerator and denominator by same values.
(8*2)/(25*2) --> Fractional part = 0.32
(8*3)/(25*3) --> Fractional part = 0.32
.
.
.
(8*12)/(25*12) --> Fractional part remains unchanged, but the remainders vary.

Since we are asked only 2 digit remainders we take 16, 24, .... upto 96.
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 16 Mar 2016, 23:17
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a  [#permalink]

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New post 16 Aug 2018, 10:16
Bunuel wrote:
X and Y are both integers. If X/Y = 59.32, then what is the sum of all the possible two digit remainders of X/Y?

A. 560
B. 616
C. 672
D. 900
E. 1024


So what if the question were If X/Y = 59.325, then what is the sum of all the possible two digit and 3 digit remainders of X/Y?

Would it have been 325/1000 = 13/40

Hence 13, 26, 39,..., (13*76) = 988.

Am I getting this correctly? I guess the crux if my query is, why are we considering 100 as the denominator, because of the '2 digit remainder' that's mentioned?
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Re: X and Y are both integers. If X / Y = 59.32, then what is the sum of a &nbs [#permalink] 16 Aug 2018, 10:16
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