# Help

• Apr 10th 2009, 02:54 AM
kismyjacq
Help
The numbers 168 and 324, written as the products of their prime factors, are 168=2(cube)×3×7, 324=2(sq)×3(4)

Find the smallest positive integer value of n for which 168n is a multiple of 324.

(Sorry, I can't put type the square or cube on top of the numbers. )
• Apr 10th 2009, 03:06 AM
mr fantastic
Quote:

Originally Posted by kismyjacq
The numbers 168 and 324, written as the products of their prime factors, are 168=2(cube)×3×7, 324=2(sq)×3(4)

Find the smallest positive integer value of n for which 168n is a multiple of 324.

$168 = 2^2 \times 3 \times 2 \times 7$.
$324 = 2^2 \times 3 \times 3^3$.
Therefore $168 \times 3^3 = 324 \times 2 \times 7$.
Since $3^3$ has no common factor with $2 \times 7$ it follows that $n = 3^3 = 27$.