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

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