Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 May 2015, 15:06

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# When a number divided by 35, the remainder is 2;divided by

Author Message
TAGS:
Senior Manager
Joined: 15 Aug 2004
Posts: 329
Followers: 1

Kudos [?]: 5 [0], given: 0

When a number divided by 35, the remainder is 2;divided by [#permalink]  06 Sep 2006, 01:37
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
When a number divided by 35, the remainder is 2;divided by 31, the remainder is 3. The number is:

*** Wanted to know the way to do this apart from plugging in numbers...
Director
Joined: 05 Feb 2006
Posts: 902
Followers: 1

Kudos [?]: 51 [0], given: 0

Construct the two equations and solve for X...

x/35=q+2
x/31=q+3.....

I probably made a mistake somewhere in the process and got the number 78.75....
SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 115 [0], given: 39

35m+2=31k+3

35m-1=31k

k= (35m-1)/31= (4m-1)/31 +m

Since k and m are an integers , 4m-1 must be a multiple of 31

m=8 is the smallest positive value that will do the trick- k would be 9

35(8)+2=282=31(9)+3

I hope this solve ur problem
Senior Manager
Joined: 11 May 2006
Posts: 260
Followers: 1

Kudos [?]: 6 [0], given: 0

SimaQ wrote:
Construct the two equations and solve for X...

x/35=q+2
x/31=q+3.....

I probably made a mistake somewhere in the process and got the number 78.75....

you are assuming q to be same in two cases - which is not necessary..
SVP
Joined: 24 Aug 2006
Posts: 2134
Followers: 3

Kudos [?]: 98 [0], given: 0

yezz wrote:
35m+2=31k+3

35m-1=31k

k= (35m-1)/31= (4m-1)/31 +m

Since k and m are an integers , 4m-1 must be a multiple of 31

m=8 is the smallest positive value that will do the trick- k would be 9

35(8)+2=282=31(9)+3

I hope this solve ur problem

Hi, remainder problems give me headaches. Can you explain how you get

(4m-1)/31 +m

Only thing I can assume is that 35-31 = 4

Thanks
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 123 [0], given: 0

I did this way:

n = 35x+2
n = 31y + 3
35x+2 = 31y+3
x = (31y+1)/35

because x is an integer. 31y+1 must have last digit as 0 or 5.
Hence y must be an integer with unit digit 4 or 9.

When y = 4 then (31*4 + 1)/35 i.e 125/35 is not an integer.
when y = 9 then (31*9+1)/35 i.e 280/35 = 8 is an integer.
Hence n = 31 * 9 + 3 = 282
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 115 [0], given: 39

since

x= 35m+2
and x = 31k+3 where k,m are positive intigers

thus

35m+2 = 31k+3

thus k = 35m+2-3/31 = 35m-1/31

35m-1 could be expressed as 31m+4m-1

and since K is intiger

therfore

31m+(4m-1)/31 is intiger

31m/31 + 4m-1/31 and since 31m is completely divisable by 31

therfore 4m-1 must be divisible by 31

ie least possible value of 4m-1 = 31

therfore 4m =32 ie m = 8
SVP
Joined: 24 Aug 2006
Posts: 2134
Followers: 3

Kudos [?]: 98 [0], given: 0

YES!!!! Now it makes perfect sense!!! thank you yezz & ps_dahiya

So I guess that's a standard technique for remainder problem. "That" being breaking up a variable such as 35m into 31m+4m to simplify into manageable components.

Thanks, very insightful.
SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 115 [0], given: 39

Let me tell you that remainder problems come in a wide range of representations

and there is multiple ways to approach each

so my suggestion is solve as much as u can

take care
Similar topics Replies Last post
Similar
Topics:
What is the remainder when 7^n + 2 is divided by 5 (1) when 2 19 Jun 2008, 03:51
When a number divided by 35, the remainder is 2;divided by 1 19 Aug 2006, 01:41
What is the remainder when k^2 is divided by 8? 1). When k 8 17 Aug 2006, 07:20
remainder when p^2 is divided by 4. 1 17 Jul 2006, 14:26
What is the remainder when k^2 is divided by 8? 1). When k 4 26 May 2006, 23:24
Display posts from previous: Sort by