Question

Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12e. Then which of the following pairs contains a number that is not an integer?

• A. \(\left(\frac{a}{6}, \frac{c}{d}\right)\)
• B. \(\left(\frac{a}{36}, \frac{c}{e}\right)\)
• C. \(\left(\frac{a}{12}, \frac{bd}{18}\right)\)
• D. \(\left(\frac{a}{27}, \frac{b}{e}\right)\)

Hint:

Call the common value 2b = 9d = 12e as k, and write a, b, c, d, e in terms of k. Then test each fraction.

Answer 1

The Correct Option is A

Step 1: Express everything in k:
Let \(2b = 9d = 12e = k\). Then \(b = \frac{k}{2}\), \(d = \frac{k}{9}\), \(e = \frac{k}{12}\), \(a = 6b = 3k\) and \(c = \frac{a}{12} = \frac{k}{4}\). Since b, d, e are integers, k is a multiple of 2, 9 and 12, so k is a multiple of 36.

Step 2: Option (4):
\(\frac{a}{27} = \frac{3k}{27} = \frac{k}{9} = d\), an integer. \(\frac{b}{e} = \frac{k/2}{k/12} = 6\), an integer.

Step 3: Option (2):
\(\frac{a}{36} = \frac{3k}{36} = \frac{k}{12} = e\), an integer. \(\frac{c}{e} = \frac{k/4}{k/12} = 3\), an integer.

Step 4: Option (3):
\(\frac{a}{12} = c\), an integer. \(\frac{bd}{18} = \frac{(k/2)(k/9)}{18} = \frac{k^2}{324}\); with k = 36m this is \(\frac{1296m^2}{324} = 4m^2\), an integer.

Step 5: Option (1):
\(\frac{a}{6} = b\), an integer, but \(\frac{c}{d} = \frac{k/4}{k/9} = \frac{9}{4}\), which is not an integer. So option (1) contains a non-integer.

Final Answer:
The correct option is (1): \(\left(\frac{a}{6}, \frac{c}{d}\right)\)