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

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Two numbers A, and B when divided by the same divisor leave [#permalink] New post 07 Mar 2004, 13:19
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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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 [#permalink] New post 07 Mar 2004, 13:36
D!!.........28 :sleeping: ....jus woke up from a nap and njoying ur questions buddy.....keep posting sunni!!
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 [#permalink] New post 07 Mar 2004, 14:03
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
So 28 is the answer.
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 [#permalink] New post 07 Mar 2004, 14:06
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!!!
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 [#permalink] New post 07 Mar 2004, 14:10
cbrf3, exactly. I was waiting to see if anyone would spot that.
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 [#permalink] New post 07 Mar 2004, 16:39
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
  [#permalink] 07 Mar 2004, 16:39
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