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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] New post 07 Apr 2007, 22:45
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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?

A) 12
B) 13
C) 35
D) 37
E) 59

I'm already familiar with the textbook method. I'm trying to discover what's wrong with the below method.

(1) a/d = x+24 ----> dx+24 = a
(2) 2(a)/d = x+11 ----> 2(dx+24)/d = x+11 ----> 2dx+48/d = x+11
(3) dx+11 = 2dx+48 ----> -dx=37, dx=-37

I managed to produce the correct answer. Nonetheless, there is something wrong with this. Can/will anyone help?
[Reveal] Spoiler: OA

Last edited by mau5 on 31 Jul 2013, 21:37, edited 1 time in total.
Edited the Q,Added OA
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 [#permalink] New post 08 Apr 2007, 02:18
I am sorry, but described steps have mistakes in it.
My quick way of solving this would be:

1) a= d+24
2) 2a=kd+11
3) multiply step 1 by 2 and subtruct 1 from 2
4) 0=d(k-2)-37
5) because 37 is prime number, eq would make sense if k=3, hence answer would be 37 (D)
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 [#permalink] New post 08 Apr 2007, 04:56
botirvoy wrote:
I am sorry, but described steps have mistakes in it.
My quick way of solving this would be:

1) a= d+24
2) 2a=kd+11
3) multiply step 1 by 2 and subtruct 1 from 2
4) 0=d(k-2)-37
5) because 37 is prime number, eq would make sense if k=3, hence answer would be 37 (D)

Hi, Thanks. I know. Hence the answer. Will you let me know what mistakes you found.
Also, how does a = d+24 when the prob says "a number when divided by a divisor leaves a remainder of 24"? Is this some sort of shortcut?
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Re: Remainder fun!! [#permalink] New post 08 Apr 2007, 10:54
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ggarr 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) 13
B) 59
C) 35
D) 37
E) 12

I'm already familiar with the textbook method. I'm trying to discover what's wrong with the below method.

(1) a/d = x+24 ----> dx+24 = a
(2) 2(a)/d = x+11 ----> 2(dx+24)/d = x+11 ----> 2dx+48/d = x+11
(3) dx+11 = 2dx+48 ----> -dx=37, dx=-37

I managed to produce the correct answer. Nonetheless, there is something wrong with this. Can/will anyone help?


See since remainder is 24 therefore the divisor D is > 24. Now by given info we have

N = (D x k) + 24
When 2N is divided by D the remainder will be 24 x 2 = 48 but since it is 11 this means that D = 48 - 11 = 37
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 [#permalink] New post 09 Apr 2007, 06:18
make x/y=z+24
and 2x/y=z+11

so it equals 37

D
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 [#permalink] New post 09 Apr 2007, 09:54
andrehaui wrote:
make x/y=z+24

D



strictly speaking, x/y =z+24 is not correct (it implies x=zy+z24, which is not what we want to do), as garr was trying to do as well.

Garr, when I wrote a=d+24, it is of the form a=dk+r, where I really considered when k=1
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 [#permalink] New post 09 Apr 2007, 22:19
how abt we solve it this way :-

Since the x%y = 24 and 2x%y = 11 we can eliminate AE
since the divisor has to be greater than 24.

also it means 48%y = 11. Hence the answer is 37.
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 [#permalink] New post 10 Apr 2007, 01:48
Let the number is N, the divisor = D,

I will make the two equations-
N = xD+24
2N = yD+11
where x and y are integers

Solving them: D(y-2x) = 37
as D is also integer and 37 is a prime number, the D should be 37 to satisfy the above equation.

Hence answer is 'D'
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Re: No properties [#permalink] New post 30 Apr 2009, 07:36
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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] New post 19 Aug 2009, 11:37
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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).

Cheers
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Re: No properties [#permalink] New post 19 Aug 2009, 16:31
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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] New post 31 Aug 2009, 06:44
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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] New post 31 Aug 2009, 14:18
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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] New post 31 Aug 2009, 15:07
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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] New post 02 Oct 2009, 12:36
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r1+r2-r3

*shortcut ...found in some book
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divisibility and remainders [#permalink] New post 12 Jul 2011, 08:42
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. 13
B. 59
C. 35
D. 37
E. 12.

I solve till
a = fx + 24
2a = fx + 11

After this step I'm lost.


Can someone please explain me the solution, I read the solution but was not totally convinced.

Regards,
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Re: divisibility and remainders [#permalink] New post 12 Jul 2011, 08:48
mustu 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. 13
B. 59
C. 35
D. 37
E. 12.

I solve till
a = fx + 24
2a = fx + 11

After this step I'm lost.


Can someone please explain me the solution, I read the solution but was not totally convinced.

Regards,
Mustu


Looking at the options i know A and E are out ...

so it has to be B ,C or D

59/35 = 24 R
118/35 = 13 R...............wrong

37+24 = 61
61/37 = 24 R
122/37 = 11 R............. right

Hence 37 D
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Re: divisibility and remainders [#permalink] New post 12 Jul 2011, 09:08
sudhir18n wrote:

Looking at the options i know A and E are out ...

so it has to be B ,C or D

59/35 = 24 R
118/35 = 13 R...............wrong

37+24 = 61
61/37 = 24 R
122/37 = 11 R............. right

Hence 37 D


Thanks for the answer,
CaN it be solved Only by picking numbers or is there a shortcut?

Regards,
Must
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Re: divisibility and remainders [#permalink] New post 13 Jul 2011, 10:33
it is actually easier to plug 24 with the answers.
A and E are going out bc the answer need to be bigger than 24.

after that we can just try.

24/35 = (24)
48/35 = 1(13) - and now its obvious that the answer is 37.

done.
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Re: divisibility and remainders [#permalink] New post 13 Jul 2011, 12:58
I used N=DQ+24, 2N=DQ+11. Multiplied the first equation by 2 to get 2N on both sides, then set the equations equal to each other. 2DQ+48=DQ+11. Ended up with DQ=-37. Picked D, got it right. Can anyone explain what was wrong with that approach?
Re: divisibility and remainders   [#permalink] 13 Jul 2011, 12:58
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