Let a, b, c, d and e be integers such that `a=6b=12c,and2b=9d=12e`. Then which of the following pairs contains a number that is not an integer?

Options

A. `(a/(27),b/(e))`
B. `(a/(36),c/(e))`
C. `(a/(12),(bd)/(18))`
D. `(a/(6),c/(d))`

Answer

To solve the problem, we need to analyze the given equations involving the integers \( a, b, c, d, \) and \( e \).

1. **Given Equations**:
- \( a = 6b \)
- \( a = 12c \)
- \( 2b = 9d \)
- \( 2b = 12e \)

2. **Expressing Variables in Terms of \( b \)**:
- From \( a = 6b \), we can express \( a \) as:
\[ a = 6b \]
- From \( a = 12c \), we can express \( c \) in terms of \( b \):
\[ 12c = 6b \implies c = \frac{6b}{12} = \frac{b}{2} \]
- From \( 2b = 9d \), we can express \( d \) in terms of \( b \):
\[ d = \frac{2b}{9} \]
- From \( 2b = 12e \), we can express \( e \) in terms of \( b \):
\[ e = \frac{2b}{12} = \frac{b}{6} \]

3. **Analyzing Integer Conditions**:
- Since \( a, b, c, d, \) and \( e \) are all integers, we need to ensure that \( \frac{2b}{9} \) (for \( d \)) is an integer. This means \( b \) must be a multiple of 9.
- Let \( b = 9k \) for some integer \( k \).

4. **Substituting \( b \) Back**:
- Substitute \( b = 9k \) into the expressions for \( a, c, d, \) and \( e \):
- \( a = 6b = 6(9k) = 54k \)
- \( c = \frac{b}{2} = \frac{9k}{2} \) (This will only be an integer if \( k \) is even)
- \( d = \frac{2b}{9} = \frac{2(9k)}{9} = 2k \)
- \( e = \frac{b}{6} = \frac{9k}{6} = \frac{3k}{2} \) (This will only be an integer if \( k \) is even)

5. **Conclusion**:
- The integers \( c \) and \( e \) will not always be integers unless \( k \) is even. If \( k \) is odd, \( c \) will not be an integer, and if \( k \) is odd, \( e \) will also not be an integer.
- Therefore, the pairs that contain a number that is not an integer are \( c \) and \( e \) when \( k \) is odd.