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# Two numbers when divided by a certain divisor leave

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Current Student
Joined: 29 Jan 2005
Posts: 5218
Two numbers when divided by a certain divisor leave [#permalink]

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01 Nov 2005, 00:43
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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: 1112
Location: CA

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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: 1885

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01 Nov 2005, 05:14
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: 5218

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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: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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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...
01 Nov 2005, 07:56
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