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

 It is currently 05 May 2015, 22:18

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

Two numbers when divided by a certain divisor leave

Author Message
TAGS:
Current Student
Joined: 29 Jan 2005
Posts: 5244
Followers: 23

Kudos [?]: 179 [0], given: 0

Two numbers when divided by a certain divisor leave [#permalink]  31 Oct 2005, 23:43
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?
VP
Joined: 22 Aug 2005
Posts: 1123
Location: CA
Followers: 1

Kudos [?]: 43 [0], given: 0

599?

x = 431 mod d
y = 379 mod d

therefore,
(x+y) = (431 + 379) mod d

as is given remainder from sum is 211
therefore, d divides (431 + 379) - 211 = 599

now 599 is prime number so d = 599
SVP
Joined: 24 Sep 2005
Posts: 1893
Followers: 11

Kudos [?]: 139 [0], given: 0

Re: PS Number Properties [#permalink]  01 Nov 2005, 04:14
GMATT73 wrote:
Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

Surely duttsit's approach is the shortest one. In case, one doesn't know those mod rules, here is the way:
Let A be the divisor, X and Y be the two numbers. ( A is an integer)
We have: X= Ax + 431
Y= Ay+ 379
X+Y= Az + 211= A(x+y) + 810 ( x, y and z and integers)
---> A ( z-x-y)= 810-211= 599
Since 599 is a prime number as duttsit pointed out, we have two cases:
A= 599, z-y-x=1 ( the vice versa is impossible coz if A=1, there won't be any remainder since every number is divisible by 1)
A= -599, z-x-y=-1

Uhm, I think the negative value is valid...
Current Student
Joined: 29 Jan 2005
Posts: 5244
Followers: 23

Kudos [?]: 179 [0], given: 0

Another problem that looks difficult, but actually is extremely easy.

OA/OE:

Two numbers when divided by a common divisor, if they leave remainders of x and y and when their sum is divided by the same divisor leaves a remainder of z, the divisor is given by x + y - z.

In this case, x and y are 431 and 379 and z = 211.
Hence the divisor is 431 + 379 - 211 = 599

Memorize the rule!
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

I get 599 too...

my method is simple...no need to memorize anything..understand everyting....

let say number one is z=xm+431
lets say number two is y=xK+319

z+y=x(m+K)+810...now focus in 810, this reduces to 211 well lets see 810/599 gives remainder 211.... 599 it is...
Similar topics Replies Last post
Similar
Topics:
11 When a number is divided by a divisor it leaves a remainder 11 05 Aug 2014, 08:36
23 Two numbers when divided by a divisor leave reminders of 248 11 23 Nov 2011, 08:14
A number when divided by a divisor leaves a remainder of 24. 0 16 Oct 2013, 12:23
25 A number when divided by a divisor leaves a remainder of 24. 26 07 Apr 2007, 22:45
When a number is successively divided by two divisors d1 and 2 08 Jan 2006, 11:37
Display posts from previous: Sort by