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

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.

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.

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?

Hi Bunuel,

Could you please explain the concept needed to solve this problem? This would be of great help.

Thanks.

Well I am not him but just went through that problem.

There is a rule that you can apply with decimals and remainder question that the digits behind the decimal (in this case 0.32) equivalent "Remainder / Divisor"
In this case this would mean "32/100"

The tricky thing about this rule is that you can't determine with a hundert percent certainty what the actual combination of remainder and divisor is.

Think about it -> 8 / 25 = 0.32 as well which would be the same as a remainder of 32 and a divisor of 100.

In this question they only ask about the sum of the two digit remainders so we go from 8 / 25 up to all the way to 100 in steps of 8.

which means 16 , 24 , 32 .... 96

\((96+16)/2 =56\)
\((56*11)=616\)
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Is there a shorter way for finding sum of all the 2 digit factors of 8?

Is there a shorter way for finding sum of all the 2 digit factors of 8?

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

Right, got it. Also read Veritas Prep's approach on this as well and it made it clearer, thanks!

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.

Nice One...
The remainders must be => 16,24,,,,,96=> using the AP sum sequence => 616
hence B
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