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Two numbers A, and B when divided by the same divisor leave

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Senior Manager
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Two numbers A, and B when divided by the same divisor leave [#permalink]

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07 Mar 2004, 14:19
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Two numbers A, and B when divided by the same divisor leave remainders 11 and 21 respectively. When the sum of the numbers is divided by the same divisor, the remaninder is 4. Find the divisor.

32
24
36
28
23
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Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

Senior Manager
Joined: 05 Feb 2004
Posts: 290
Location: USA
Followers: 1

Kudos [?]: 4 [0], given: 0

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07 Mar 2004, 14:36
D!!.........28 ....jus woke up from a nap and njoying ur questions buddy.....keep posting sunni!!
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07 Mar 2004, 15:03
a and b be the numbers and x be divisor.

a = nx + 11
b = mx + 21
a+b = (n+m)x + 32

if x = 28 then 32 divided by 28 will have the remainder = 4
Senior Manager
Joined: 05 Feb 2004
Posts: 290
Location: USA
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Kudos [?]: 4 [0], given: 0

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07 Mar 2004, 15:06
On 2nd thoughts, this result can be generalized:

For Two numbers A, and B when divided by the same divisor leave remainders X and Y respectively and When the sum of the numbers is divided by the same divisor, the remaninder is Z, then the divisor is X+Y-Z!!!
Senior Manager
Joined: 30 Aug 2003
Posts: 329
Location: BACARDIVILLE
Followers: 1

Kudos [?]: 14 [0], given: 0

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07 Mar 2004, 15:10
cbrf3, exactly. I was waiting to see if anyone would spot that.
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Pls include reasoning along with all answer posts.
****GMAT Loco****
Este examen me conduce jodiendo loco

SVP
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 72 [0], given: 0

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07 Mar 2004, 17:39
I am not sure if you can apply this formula

a = 10 b = 13 and divisor x = 6
10/6 remainder 4
13/6 remainder 1
total remainder = 5

(10+13)/6 remainder 5

Here 5-5 = 0 So this formula is not valid.

However there is a requiredment for this formula to be satisfied.

Let the two numbers be a,b and divisor c. Remainders be x and y respectively. let z be the remainder for (a+b) divided by c

a = mc + x
b = nc + y
(a+b) = (m+n)c+(x+y) = kc + z
so divisor c = [ (x+y)-z ] / [ k - (m+n) ]

Only if k - (m+n) = 1 then the divisor = (x+y)-z

In case of 28 let a = 28+11 = 39 b = 28+21 = 49
m = 1 n = 1
a+b = 88 so k = 3
Hence 3-(1+1) = 1 which satisfies the above equation
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