Given that x is a positive integer such that 19 divided by x has a rem

Question

Given that x is a positive integer such that 19 divided by x has a remainder of 3, what is the sum of all the possible values of x?

(A) 12

(B) 20

(C) 24

(D) 28

(E) 32

BrentGMATPrepNow · Answer

Remainder property #1: When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D  
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0
Since we're told that 19 divided by x leaves a remainder of 3, we know that  x > 3

Remainder property #2: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

Since we're not told the quotient in the question, let's say the the quotient is q.
So we can rewrite the given information as follows:  19 divided by x  equals q  with remainder 3 
This means we can conclude that 19 = xq + 3
Subtract 3 from both sides of the equation to get: 16 = xq
If 16 = xq, then we know that  x is a factor of 16 , since x and q are positive integers.

We now have two great pieces of information about the value of x:  x > 3  and  x is a factor of 16 
The factors of 16 are: 1, 2, 4, 8, and 16
Since  x > 3 , the two possible values of x are 4, 8 and 16

4 + 8 + 16 = 28

Answer: D

RELATED VIDEO
 https://www.youtube.com/watch?v=XCxgIR042tE

Crytiocanalyst · Answer

The one possibility that immediately props up is 16

and from thereon we can assert than no value can be greater than 16 otherwise the remainder will be negative

factors of 16 except 2 and 1 fits in therefore our answer = 16+8+4 =28

Therefore IMO D

BrushMyQuant · Answer

Given that x is a positive integer such that 19 divided by x has a remainder of 3 and we need to find the sum of all the possible values of x

Theory: Any number when divided by n gives remainder from 0 to n-1

Since we have a remainder of 3 so the divisor should be more than 3
=> divisor ≥ 4

Let's solve the problem using two methods

Method 1: Algebra

19 divided by x gives 3 remainder

Theory: Dividend = Divisor*Quotient + Remainder

19 -> Dividend
x -> Divisor
a -> Quotient (Assume. Also a is an integer)
5 -> Remainders
=> 19 = x*a + 3 = xa + 3
=> ax = 19-3 = 16

We can take values of x greater than 3 which satisfies this and gives a as an integer
x = 4, a = 4 (possible)
x = 8, a = 2 (possible)
x = 16, a = 1 (possible)

Sum of possible values of x = 4 + 8 + 16 = 28

So,  Answer will be D

Method 2: Taking Numbers

Let's take numbers from 4 to 19 and see which one gives 3 remainder

19 divided by 4 gives 3 remainder =>   POSSBILE  
19 divided by 5 gives 4 remainder =>   NOT POSSIBLE  
19 divided by 6 gives 1 remainder =>   NOT POSSIBLE  
19 divided by 7 gives 5 remainder =>   NOT POSSIBLE  
19 divided by 8 gives 3 remainder =>   POSSBILE  
19 divided by 9 gives 1 remainder =>   NOT POSSIBLE  
19 divided by 10 gives 9 remainder =>   NOT POSSIBLE  
19 divided by 11 gives 8 remainder =>   NOT POSSIBLE  
19 divided by 12 gives 7 remainder =>   NOT POSSIBLE  
19 divided by 13 gives 6 remainder =>   NOT POSSIBLE  
19 divided by 14 gives 5 remainder =>   NOT POSSIBLE  
19 divided by 15 gives 4 remainder =>   NOT POSSIBLE  
19 divided by 16 gives 3 remainder =>   POSSBILE  
19 divided by 17 gives 2 remainder =>   NOT POSSIBLE  
19 divided by 18 gives 1 remainder =>   NOT POSSIBLE  
19 divided by 19 gives 0 remainder =>   NOT POSSIBLE

So, 4 + 8 + 16 = 28

So,  Answer will be D 
Hope it helps!

Watch the following video to learn the Basics of Remainders

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

bumpbot · Answer

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Given that x is a positive integer such that 19 divided by x has a rem

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Given that x is a positive integer such that 19 divided by x has a rem

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