x divided by d, the remainder is 24:

x = md + 24 (where m is a non-negative integer)

2x divided by d, the remainder is 11:

2x = nd + 11 (where n is a non-negative integer)

nd + 11 = 2*(md + 24)
nd + 11 = 2md + 48
nd - 2md = 37
(n-2m)*d = 37

37 is prime number. Since d cannot be 1 (because any number divided 1 will not leave any remainder) and (n-2m) must be an integer, therefore d must be 37.

Choose D.

-BM-

sanju09 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

Small concept again....

If R is remainder when A is divided by D, the the remainder when KxA is divided by D will be KxR provided KxR<D. In case if KxR>D, it is further divided by D and the remainder is given.

Coming to our question here.....

Since A leaves a remainder 24, 2A must leave a remainder 48. But since the remainder is given as 11, it is implied that 48 > our divisor and hence 48 is further divided by our divisor and the remainder is reduced to 11. So the divisor must be 37.

A small note: The divisor obtained here will be the maximum possible divisor. The factors of this divisor could also be the answer depending upon the remainders given.

In this case, since we got the maximum divisor as a prime number, this is the only possible value.

If you consider a case A is multiplied by 3 or 4, we can get a situation where we will have more than one possible divisor (of course not always)

And if the number is multiplied by 3 or 4, what should we do then?

And if the number is multiplied by 3 or 4, what should we do then?

mjjking wrote:And if the number is multiplied by 3 or 4, what should we do then?

Ok let us look at an example.

A number N leaves a remainder 14 when it is divided by D. When thrice the number N is divided by D, the remainder is 2. Find the divisor.

Since N leaves a remainder 14, 3N must leave a remainder 3x14 = 42. But since it is given that the remainder is 2, it is implied that 42>D and hence it is further divided by D and is reduced to 2.

So the maximum possible value of D = 42-2 = 40. Hence the greatest possible divisor is 40. Now all the factors of 40 which are greater than 14 also can be the divisors. In this case we have 20 which is satisfying these conditions.

So D could be 40 or 20. In such case you will have the options like "Cannot be determined" or "more than one value" .

In case if the maximum possible divisor is a prime number then D will have unique solution

sanju09 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 did this problem in this manner:

quotient=q
divisor =d
N=dividend

1)qd + 24 = N
2)qd + 11 = 2N

Multiply the first equation by 2 and subtract the second equation(simultaneous equation)

1) 2qd + 48 = 2N
2) -qd - 11 = -2N
________________
qd + 37 = 0
qd = -37

estimate that the answer is 37.(D)

thanks suresh, now it's clear!

sudi760mba wrote:

sanju09 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 did this problem in this manner:

quotient=q
divisor =d
N=dividend

1)qd + 24 = N
2)qd + 11 = 2N

Multiply the first equation by 2 and subtract the second equation(simultaneous equation)

1) 2qd + 48 = 2N
2) -qd - 11 = -2N
________________
qd + 37 = 0
qd = -37

estimate that the answer is 37.(D)

You did this problem in this manner:

quotient=q
divisor =d
N=dividend

1)qd + 24 = N
2)qd + 11 = 2N

Will the quotients be same in both cases?