Bunuel wrote:
If a and b are positive integers such that a/b = 82.024, which of the following can be the value of b?
(A) 100
(B) 150
(C) 200
(D) 250
(E) 550
If \(a = bQ + r\), then \(\frac{a}{b} = Q + \frac{r}{b}\)
(a = dividend, b = divisor, Q = quotient, r = remainder)
Change the remainder from decimal to fraction: .024 =\(\frac{24}{1000} =\frac{r}{b}\)
In order for \(a\) and \(b\) to be integers, \((\frac{r}{b} * b)\) must result in an integer.
The decimal .084 = \(\frac{84}{1000}\) * (some # \(b\)) must have a result with no decimals. You cannot add a remainder with decimals to the product a = bQ: \(a\) would not be an integer, which the prompt says it is.
We know**:
1) \(r\) is an integer;
2) \(b\) is an integer;
3) \((\frac{r}{b} * b)\) is an integer, so \(b\) must be a multiple of the denominator in \((\frac{r}{b})\) to stay consistent with: \(a = bQ + r\)
4) No answer choice is b = 1,000 (which would make life easy)
If \(\frac{r}{b} = \frac{24}{1000}\), reduce in succession:
\(\frac{r}{b} =\frac{24}{1000} = \frac{12}{500} =\frac{6}{250} = \frac{3}{125}\)
Any answer choice that equals any one of those denominators, such that \(r\) becomes an integer, could equal \(b\).
Answer D, b = 250, as the only multiple of 125, makes \(r\) an integer.
OR, from the fraction reduced to simpler expressions in succession above, one expression is \(\frac{6}{250}\).
\(\frac{6}{250} *\\
250 = 6 = r\)
From here the dividend \(a\) can be "rebuilt."
b = 250, r = 6, Q is 82. \(a\) = (250)(82) + 6. \(a\) = 20,506. \(r\) is now an integer: \(b\) = 250 clears the fraction \(\frac{6}{250}\).
Answer D
**
It is a little easier to see these concepts, especially the relationship between decimal values and integer values, with a simple example.
\(\frac{12}{5} = 2 +\\
R2\)
\(12 = (5)(2) + 2\)
\(a = bQ + r\)
\(\frac{a}{b} = Q + \frac{r}{b}\)
\(\frac{12}{5} = 2 + \frac{2}{5}\)
\(\frac{12}{5} = 2 + (.4)\)
\(\frac{12}{5} = 2.4\)
\(\frac{r}{b} =\frac{4}{10} = \frac{2}{5}\)
The denominator tells us that \(b\) here is a multiple of 5 -- \((\frac{2}{5})\) * (multiple of 5) = integer.