Two numbers when divided by a certain divisor leave

Oldest

Best Reply

Author

Message

TAGS:

Current Student

Joined: 29 Jan 2005

Posts: 5289

Followers: 17

Kudos [?]: 90 [0], given: 0

Two numbers when divided by a certain divisor leave [#permalink]

01 Nov 2005, 00:43

00:00

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?

Joined: 22 Aug 2005

Posts: 1133

Location: CA

Followers: 1

Kudos [?]: 9 [0], given: 0

[#permalink]

01 Nov 2005, 01:04

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: 1913

Followers: 6

Kudos [?]: 55 [0], given: 0

Re: PS Number Properties [#permalink]

01 Nov 2005, 05:14

GMATT73 wrote:

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: 5289

Followers: 17

Kudos [?]: 90 [0], given: 0

[#permalink]

01 Nov 2005, 07:19

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: 3439

Location: New York City

Schools: Wharton'11 HBS'12

Followers: 11

Kudos [?]: 134 [0], given: 2

[#permalink]

01 Nov 2005, 07:56

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...

[#permalink] 01 Nov 2005, 07:56

		Similar topics	Author	Replies	Last post
Similar Topics:		Two numbers when divided by a certain divisor leave	Praetorian	1	29 Sep 2003, 21:06
		Two numbers A, and B when divided by the same divisor leave	sunniboy007	5	07 Mar 2004, 14:19
		A number when divided by a divisor leaves a remainder of 24.	ggarr	10	07 Apr 2007, 23:45
	16	A number when divided by a divisor leaves a remainder of 24.	tenaman10	13	17 Apr 2009, 09:22
	10	Two numbers when divided by a divisor leave reminders of 248	cleetus	8	23 Nov 2011, 09:14

Two numbers when divided by a certain divisor leave

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.

GMAT Courses

Admissions Consulting

Two numbers when divided by a certain divisor leave

Two numbers when divided by a certain divisor leave

Main navigation

GMAT Resources

Partners

MBA Resources