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

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?

make x/y=z+24
and 2x/y=z+11

so it equals 37

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

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'

Should be 24x2 - 11 = 37

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

\(N = I1*D + 24\) --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, \(D = 37\)

----ASIDE--------------
There's a nice rule that says, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3
-------------------------

Let the divisor = d
Let the original number be N

Given: A number when divided by a divisor leaves a remainder 24
We're not told what the quotient is. So, let's just say the quotient is k
In other words: When N is divided by d, we get k with remainder 24
Applying the above rule, we can write: N = dk + 24

Also given: When twice the original number is divided by the same divisor the remainder is 11.
Once again, we're not told what the quotient is. So, let's just say the quotient here is j
In other words: When 2N is divided by d, we get j with remainder 11
Applying the above rule, we can write: 2N = dj + 11

We now have two VERY USEFUL equations:
N = dk + 24
2N = dj + 11

Take the top equation and create an EQUIVALENT equation by multiplying both sides by 2 to get:
2N = 2dk + 48
2N = dj + 11

Since both equations are set equal to 2N, we can write: dj + 11 = 2dk + 48
Subtract 11 from both sides: dj = 2dk + 37
Subtract 2dk from both sides: dj - 2dk = 37
Factor: d(j - 2k) = 37

So, d TIMES (j - 2k) = 37
Notice that d and (j - 2k) are INTEGERS.
Also, 37 is a PRIME number, which means it can be factored in only one way: 37 = (1)(37)
This means EITHER d = 1 and (j - 2k) = 37
OR d = 37 and (j - 2k) = 1

Check the answer choices....
Answer: D

RELATED VIDEO

Let's say the divisor be K.

When a number is divided by a divisor K, leaves a remainder of 24
N= a*K + 24
From this statement ,we can conclude that K > 24, Hence Option A,B can be eliminated

When twice the original number is divided by the same divisor, the remainder is 11.

2N = b*K + 11

2N= 2*a*K + 48

Equating both,
b*K + 11 = 2*a*K + 48

(b-2*a)K= 48-11 = 37

Since 37 is a prime number, 1 * 37 is the only way we can write 37 as the product of two numbers.
=> (b-2*a)*K= 1* 37
from the above statement we can conclude that , Value of K = 37.

Option D is the correct answer.

Thanks,
Clifin J Francis.
GMAT SME

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.