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

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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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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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

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

\(\frac{32}{100} = \frac{16}{50} = \frac{24}{75} =\) .....

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

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

.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

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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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.

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)?

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

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

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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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

The least reminder is 8 for a y of 25.

And the other reminders being 16 (y=50), 24 (y=75), 32 (y=100), 40 (y=125), 48 (y = 150), 56 (y = 175), 64 (y=200), 72 (y=225), 80 (y=250), 88 (y = 275) and 96 (y=300), are of two digits.

The sum of the above reminders = 616.

OA: B
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In the problem above:
Remainder = R
Divisor = 7
Decimal = 0.32 = 32/100 = 8/25
Plugging these values into remainder/divisor = decimal, we get:
R/y = 8/25

The resulting equation implies that the remainder must be a MULTIPLE OF 8.

For any EVENLY SPACED SET:
Count = (biggest - smallest)/(increment) + 1.
Average = (biggest + smallest)/2.
Sum = (count)(average).
The INCREMENT is the difference between successive values.

Here, we must sum the 2-digit multiples of 8 between 16 and 96, inclusive.
Since the integers are multiples of 8, the increment = 8.
Thus:
Count = (96-16)/8 + 1 = 11
Average = (96+16)/2= 56
Sum = (count)(average) = 11*56 = integer with a units digit of 6

The correct answer is B.

When x is divided by y, the result is 59.32. The decimal part of the result, when multiplied with the divisor, represents the remainder when x is divided by y.

Therefore, remainder = 0.32 * y = \(\frac{32 }{ 100}\) * y = \(\frac{8 }{25}\) * y.

Note that remainder is always a non-negative integer, therefore, y should be a multiple of 25.

If y = 25, remainder = 8. Ignore since we are looking for two-digit remainders.

If y = 50, remainder = 16. This is the first two-digit remainder when x is divided by y.

If y = 75, remainder = 24.

We see that the remainders are consecutive multiples of 8, starting with 16 and going on until 96 (since that’s the last 2-digit multiple of 8).
This is an Arithmetic sequence with a common difference of 8.

16 = 8 * 2; 96 = 8 * 12.

Therefore, number of terms, n = (12 – 2) + 1 = 11.

Sum of terms of an Arithmetic sequence = \(\frac{n}{2}\) (First term + Last term)

Therefore, sum of all the 2-digit remainders = \(\frac{11 }{ 2}\) (16 + 96) = \(\frac{11 }{ 2}\) ( 112) = 11 * 56 = 616.

The correct answer option is B.
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Concept : X= QY+ R
Dividing both sides by Y
X/Y= Q+R/Y
Since X/Y= 59.32 Therefore Q+ R/Y= 59+ 32/100

Now we know R/Y is roughly equal to 1/3 (32/100 = ~33/100 = 1/3)
Therefore, 3R=Y
Since we need only 2 digit numbers the maximum value of Y can be 33

To take sumation of all 2 digit remainders we take the sum from 10 to 33
= n/2 [2a+ (n-1)d]
=24/2[2(10)+23]
=616

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