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

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Director
Joined: 12 Jun 2006
Posts: 519
A number when divided by a divisor leaves a remainder of 24.  [#permalink]

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Updated on: 31 Jul 2013, 21:37
1
20
00:00

Difficulty:

35% (medium)

Question Stats:

77% (01:39) correct 23% (01:51) wrong based on 537 sessions

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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?

Originally posted by ggarr on 07 Apr 2007, 22:45.
Last edited by mau5 on 31 Jul 2013, 21:37, edited 1 time in total.
Manager
Joined: 14 Aug 2009
Posts: 121

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19 Aug 2009, 16:31
8
1
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
_________________

Kudos me if my reply helps!

##### General Discussion
Director
Joined: 29 Aug 2005
Posts: 778

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08 Apr 2007, 02:18
1
3
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)
Director
Joined: 12 Jun 2006
Posts: 519

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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?
Manager
Joined: 28 Feb 2007
Posts: 186
Location: California

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08 Apr 2007, 10:54
5
3
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
Manager
Joined: 14 Mar 2007
Posts: 228

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09 Apr 2007, 06:18
make x/y=z+24
and 2x/y=z+11

so it equals 37

D
Director
Joined: 29 Aug 2005
Posts: 778

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09 Apr 2007, 09:54
1
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
Intern
Joined: 05 Apr 2007
Posts: 8

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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.
Director
Joined: 14 Jan 2007
Posts: 740

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10 Apr 2007, 01:48
1
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.

SVP
Joined: 29 Aug 2007
Posts: 2378

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30 Apr 2009, 07:36
2
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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GT

Manager
Joined: 22 Jul 2009
Posts: 168

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19 Aug 2009, 11:37
4
2
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
Intern
Joined: 06 May 2009
Posts: 16

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31 Aug 2009, 06:44
1
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
Manager
Joined: 10 Aug 2009
Posts: 129

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31 Aug 2009, 14:18
1
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).
Manager
Joined: 25 Aug 2009
Posts: 162

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31 Aug 2009, 15:07
4
$$N = I1*D + 24$$ --equation 1st

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

Subtract 1st from 2nd..

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

=> $$N + 13 = I*D$$ ---- where$$I = I2-I1$$ 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, $$D = 37$$
Current Student
Joined: 26 May 2005
Posts: 508

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12 Jul 2011, 08:48
1
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
Senior Manager
Joined: 08 Nov 2010
Posts: 341

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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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Intern
Joined: 09 Jun 2011
Posts: 21

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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?
Intern
Joined: 18 Jul 2011
Posts: 43

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19 Jul 2011, 12:44
1
A more algebraic solution:

Starting with what's already been discussed, we have
N = D(Q1) + 24
2N = D(Q2) + 11

We multiply the first equation by 2 and get
2N = 2D(Q1) + 48
2N = D(Q2) + 11

Subtracting, we get
0 = D(2Q1 - Q2) + 37
Which implies D(2Q1 - Q2) = -37.

Since 37 is prime, D is either 1 or 37. If D were 1, the remainder would always be zero, so it must be 37. (It helps to keep in mind that Q1 and Q2 are both integers). This is a somewhat complex approach, and using your answers is a more efficient method here.

BenchPrepGURU
Intern
Joined: 03 Feb 2013
Posts: 1
Re: A number when divided by a divisor leaves a remainder of 24.  [#permalink]

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03 Feb 2013, 15:44
u wrote a/d = x+24
which means a= dx +24d and not dx+24 = a

which is wrong because it changes the whole meaning of the formula dividend(a)= divisor(d)*quotient(x) +remainder(24)

the second mistake which u did was u took the same quotient (x) even when d dividend (a) changed to (2a).

M answering this question years later..but what to do..just saw it!
even i did the same mistakes at my first attempt..but corrected it myself
Director
Joined: 03 Aug 2012
Posts: 744
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Re: A number when divided by a divisor leaves a remainder of 24.  [#permalink]

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31 Jul 2013, 11:33
Easy one from my side:

2N = DB + 11 - (1)
N=DA + 24 - (2)

Multiplying Eq (2) by 2 and thereby subtracting both the equations:

2N = DB + 11
2N = 2DA + 48

0=DB - 2DA - 37

=> D (B-2A) = 37

=> D = (37)/(B-2A)

Since 37 is prime and (B-2A) would always be an integer

Hence (D)

Rgds,
TGC !
Re: A number when divided by a divisor leaves a remainder of 24. &nbs [#permalink] 31 Jul 2013, 11:33

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