Last visit was: 06 May 2024, 16:08 It is currently 06 May 2024, 16:08

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 93060
Own Kudos [?]: 621714 [125]
Given Kudos: 81767
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 93060
Own Kudos [?]: 621714 [28]
Given Kudos: 81767
Send PM
General Discussion
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11216
Own Kudos [?]: 32312 [4]
Given Kudos: 301
Send PM
User avatar
Intern
Intern
Joined: 26 Dec 2014
Posts: 10
Own Kudos [?]: 6 [1]
Given Kudos: 1
Schools: CBS '19 (A)
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
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
avatar
Manager
Manager
Joined: 27 Oct 2013
Posts: 176
Own Kudos [?]: 225 [2]
Given Kudos: 79
Location: India
Concentration: General Management, Technology
GMAT Date: 03-02-2015
GPA: 3.88
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
1
Kudos
1
Bookmarks
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
avatar
SVP
SVP
Joined: 27 Dec 2012
Status:The Best Or Nothing
Posts: 1562
Own Kudos [?]: 7221 [0]
Given Kudos: 193
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
\(\frac{32}{100} = \frac{16}{50} = \frac{24}{75} =\) .....

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
avatar
Senior Manager
Senior Manager
Joined: 26 Jul 2010
Posts: 294
Own Kudos [?]: 154 [0]
Given Kudos: 41
Location: European union
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
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
VP
VP
Joined: 07 Dec 2014
Posts: 1071
Own Kudos [?]: 1567 [0]
Given Kudos: 27
Send PM
When positive integer x is divided by positive integer y, the result i [#permalink]
.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.
User avatar
Current Student
Joined: 18 Oct 2014
Posts: 680
Own Kudos [?]: 1763 [4]
Given Kudos: 69
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.98
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
2
Kudos
2
Bookmarks
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
Manager
Manager
Joined: 03 Apr 2013
Posts: 222
Own Kudos [?]: 241 [0]
Given Kudos: 872
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
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.
Senior Manager
Senior Manager
Joined: 08 Dec 2015
Posts: 258
Own Kudos [?]: 117 [1]
Given Kudos: 36
GMAT 1: 600 Q44 V27
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
1
Kudos
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)?
Intern
Intern
Joined: 09 Jun 2017
Posts: 21
Own Kudos [?]: 8 [0]
Given Kudos: 34
Location: United States (MA)
Concentration: Healthcare, General Management
GMAT 1: 710 Q43 V44
GPA: 3.31
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
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
Senior Manager
Senior Manager
Joined: 05 Jul 2018
Status:Current student at IIMB
Affiliations: IIM Bangalore
Posts: 384
Own Kudos [?]: 404 [0]
Given Kudos: 326
Location: India
Concentration: General Management, Technology
GMAT 1: 600 Q47 V26
GRE 1: Q162 V149
GPA: 3.6
WE:Information Technology (Consulting)
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
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
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18808
Own Kudos [?]: 22147 [2]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
1
Kudos
1
Bookmarks
Expert Reply
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
Manager
Manager
Joined: 29 May 2017
Posts: 154
Own Kudos [?]: 19 [0]
Given Kudos: 63
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
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
Manager
Manager
Joined: 11 Mar 2012
Posts: 176
Own Kudos [?]: 57 [0]
Given Kudos: 103
Location: India
Concentration: General Management, Operations
GMAT 1: 670 Q50 V31
GPA: 4
WE:Project Management (Real Estate)
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
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.


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
Senior Manager
Senior Manager
Joined: 12 Jan 2019
Posts: 404
Own Kudos [?]: 216 [0]
Given Kudos: 372
Location: India
Concentration: Strategy, Leadership
GMAT 1: 660 Q47 V34
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
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.



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.
GMATWhiz Representative
Joined: 07 May 2019
Posts: 3409
Own Kudos [?]: 1802 [0]
Given Kudos: 68
Location: India
GMAT 1: 740 Q50 V41
GMAT 2: 760 Q51 V40
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
Expert Reply
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.
Intern
Intern
Joined: 05 Jun 2018
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 8
Send PM
When positive integer x is divided by positive integer y, the result i [#permalink]
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
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 32790
Own Kudos [?]: 827 [0]
Given Kudos: 0
Send PM
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: When positive integer x is divided by positive integer y, the result i [#permalink]
Moderators:
Math Expert
93060 posts
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne