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# A number when divided by a divisor leaves a remainder of 24.

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A number when divided by a divisor leaves a remainder of 24. [#permalink]  17 Apr 2009, 09:22
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A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?

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Re: No properties [#permalink]  17 Apr 2009, 10:58
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I think all those answers are wrong..OA?
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Re: No properties [#permalink]  17 Apr 2009, 11:23
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yes, even i think so.

Divisor must be greater than 24 (remainder of first division)
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Re: No properties [#permalink]  30 Apr 2009, 08:36
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tenaman10 wrote:
A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
A. 6
B. 7
C. 5
D. 8
E. 18

Should be 24x2 - 11 = 37
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Re: No properties [#permalink]  08 May 2009, 01:09
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Maybe 37 or 1, is the number an Interger?
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Re: No properties [#permalink]  17 Aug 2009, 09:09
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GMAT TIGER wrote:
tenaman10 wrote:
A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
A. 6
B. 7
C. 5
D. 8
E. 18

Should be 24x2 - 11 = 37

Hi

How did you arrive at 24*2-11?

could you please explain
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Re: No properties [#permalink]  18 Aug 2009, 02:49
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I am lost..
Quote:
GMAT TIGER

Should be 24x2 - 11 = 37

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Re: No properties [#permalink]  19 Aug 2009, 12:37
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Remainder 1 = R1 = 24
Remainder 2 = R2 = 11
R2 = 2R1 - "excess remainder" -> 11 = 48 - "excess remainder" -> "excess remainder" = 37

We know that:
. The excess remainder is a multiple of the divisor.
. The divisor is greater than the greatest remainder (which is 24).

The only multiple of 37 greater than 24 is 37. Therefore, the divisor is 37.

Lets illustrate this by picking numbers: a=24
remainder equation: a/d=k+r/d
-a. 24/37=0+24/37 ->remainder is 24
-ax2. 48/37=1+11/37 ->remainder is 11

Another way to see it is through algebra:
-a. a/d=k+24/d -> a=dk+24
-ax2. 2a/d=q+11/d -> a=(dq+11)/2
-> 2dk+48=dq+11
Keep in mind that dq = 2dk + "excess remainder"
-> 2dk + 48 = 2dk + "excess remainder" + 11
-> "excess remainder" = 37

Sorry if this is not clear enough. Can't think of another way to explain it.

You may want to refer to the Man NP guide for looking into arithmetic with remainders (pg128).

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Re: No properties [#permalink]  19 Aug 2009, 17:31
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the right way to do:

n=m*x+24, and
2n=p*x+11

therefore, 2mx+48=px+11
x=\frac{37}{p-2m}

x is an integer
therefore x=37 or x=1
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Re: No properties [#permalink]  31 Aug 2009, 07:44
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here is my two cents worth

n = da + 24 (1) and 2n = db +11 (2)
subtract (1) form (2) you get
n = d(b-a) - 13 (3)
subtract (1) from (3), you get
0 = d(b-a-a) - 37
it is a prime so lowest d is 37
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Re: No properties [#permalink]  31 Aug 2009, 15:18
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flyingbunny wrote:
the right way to do:

n=m*x+24, and
2n=p*x+11

therefore, 2mx+48=px+11
x=\frac{37}{p-2m}

x is an integer
therefore x=37 or x=1

The only possible value is 37. How can it be 1? 1 leaves no remainder (0 as remainder).
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Re: No properties [#permalink]  31 Aug 2009, 16:07
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N = I1*D + 24 --equation 1st

2N = I2*D + 11--equation 2nd

Subtract 1st from 2nd..

=> N = (I2 - I1)*D - 13

=> N + 13 = I*D ---- whereI = I2-I1 wil also be an integer..

using 1st equation..

=> I1*D + 24 +13 = I*D

=> \frac{I1*D}{D} + \frac{37}{D} = I

=> I1 + \frac{37}{D} = I

=> \frac{37}{D} should be an integer which is only possible when D = 1 or 37

But 1 never leaves any reaminder..So, D = 37
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Re: No properties [#permalink]  31 Aug 2009, 23:03
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i agree with flyingbunny ...Ans is 37
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Re: No properties [#permalink]  02 Oct 2009, 13:36
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r1+r2-r3

*shortcut ...found in some book
Re: No properties   [#permalink] 02 Oct 2009, 13:36
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