How many distinct positive integers d exist such that, when 48 is

Question

How many distinct positive integers  d  exist such that, when 48 is divided by  d , the remainder is  d - 4 ?

A. Two
B. Three
C. Four
D. Five
E. Six

M40-07

Bunuel · Accepted Answer

How many distinct positive integers  d  exist such that, when 48 is divided by  d , the remainder is  d - 4 ?

A. Two
B. Three
C. Four
D. Five
E. Six

Dividing 48 by  d  and obtaining a remainder of  d - 4  can be expressed as  48 = qd + (d - 4) . Note that since the remainder must be non-negative, we have  d - 4 \geq 0 , which implies  d \geq 4 .
 
Rewriting  48 = qd + (d - 4)  leads to  q + 1= \frac{52}{d} . Here, since  q + 1  is an integer,  \frac{52}{d}  must also be an integer, meaning  d  is a factor of 52. The factors of 52 are 1, 2, 4, 13, 26, and 52. Given that  d \geq 4 , the valid values for  d  are 4, 13, 26, and 52.
 
Therefore, there are four possible values for  d .

Answer: C

Nidzo · Answer

48 divided by d yields a remainder of d - 4  can be written mathematically as:  \frac{48}{d} = Q\frac{d-4}{d}  where Q = Quotient.

In other words  (Q*d)+d-4 = 48

Which simplified is:  Qd + d = 52

Now the question simply becomes: what are the factors of 52 which are 4 or greater (as the lowest value that d-4 can be is 0 given that it is a remainder).

Factors of 52:  
 1*52 
 2*26 
 4*13

Factors of 52 4 or greater:  4, 13, 26  and  52 , all of which when dividing 48 leave a remainder of  d-4 .

Answer is therefore  C , four.

To double check:
 4:   \frac{48}{4}= 12  with a remainder of 0.  4-4=0

13:    \frac{48}{13}= 3  with a remainder of 9.  13-4=9

26:    \frac{48}{26}= 1  with a remainder of 22.  26-4=22

52:    \frac{48}{52} = 0  with a remainder of 48.  52-4=48

Answer C

BrushMyQuant · Answer

We need to find How many different positive integers d are there such that 48 divided by d yields a remainder of d - 4?

48 divided by d yields a remainder of d - 4

Theory: Dividend = Divisor*Quotient + Remainder

48 -> Dividend
d -> Divisor
a -> Quotient (Assume)
d - 4 -> Remainder

=> 48 = d*a + d - 4 
=> 48 + 4 = d*(a+1)
=> d =  \frac{52}{a + 1}

For d to be integer 52 should be divisible by a + 1

Factors of 52 are 1, 2, 4, 13, 26, 52
=> a can be 0, 1, 3, 12, 25, 51
=> d = 52, 26, 13, 4, 2, 1

Since remainder is always ≥ 0 => d - 4 ≥ 0
=> d ≥ 4

So, Possible values are 4, 13, 26, 52

So,  Answer will be C 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

LouisThvnl · Answer

I saw the different logics to solve this problem, but I all find them quite complex when you can do it more easily. My way of finding Answer C was :

48/d =d - 4 => 48 d^2 -4d => 12 = d^2 - d

So only answer C was possible

MartyMurray · Answer

How many distinct positive integers  d  exist such that, when 48 is divided by  d , the remainder is  d - 4 ?

Since the remainder when 48 is divided by  d  is  d  - 4, we can see the following:

48 =  kd  +  d  - 4

52 =  kd  +  d

Thus, 52 is a multiple of  d .

So,  d  must be a factor of 52 such that, when 48 is divided by that factor of 52, the remainder is  d  - 4.

The factors of 52 are the following:

1
2
4
13
26
52

When 48 is divided by 52, the remainder is 48 = 52 - 4.

When 48 is divided by 26, the remainder is 22 = 26 - 4.

When 48 is divided by 13, the remainder is 9 = 13 - 4

When 48 is divided by 4, the remainder is 0 = 4 - 4

Since 2 and 1 less than 4, 2 and 1 are too small to be values of  d  such that the remainder is  d  - 4 since a remainder cannot be negative.

A. Two
B. Three
C. Four
D. Five
E. Six

Correct answer:   C

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=88184 --- Harsha GMAT-Club-Forum-b1zvhxrm.png

Adit_ · Answer

This is a tricky counting exercise and as many have already given the solution, one needs to be careful of not missing out 52 which is a number greater than 48 and 4.
Both these numbers intuitively shouldn't be possible but it just so works because with 4 you get remainder 4-4=0 and with 52, the remainder is the numerator itself.

Are there more such similar questions I can practise from?

pappal · Answer

simply 48=d*k+(d-4) where k=0,1,2,.....
d(k+1)=52 OR d=52/(k+1)-------------i
now putting values of k we can find values of d which satisfy it to be a positive integer and distinct and remainder is 4 less than d, when 48 is divided by d
k=0 d=52 for certain 48%52=48 which is 4 less than 52  satisfies  (here % sign denotes just value remainder)
k=1 d=26 and 48%26=22  satisfies 
k=3 d=13 and 48%13=9  satisfies 
k=12 d=4 and 48%4=0  satisfies 
k=25 d=2 but the remainder 48%2=0  not satisfies 
k=51 d=1 but 48%1=48  not satisfies 
 hence 4 values

How many distinct positive integers d exist such that, when 48 is

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How many distinct positive integers d exist such that, when 48 is

Prep Toolkit

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