The smallest positive integer value of n for which 168 n is a multiple of 324
What you are trying to solve is finding the minimum "n" for a minimum "b" in the following equation, where both n and b are integers:
2^3x3x7xn=2^2x3^4xb
expanding the exponents:
2x2x2x3x7xn=2x2x3x3x3x3xb
cancelling common factors:
2x7xn=3x3x3xb
so: 14n=27b
There are no factors in common between the 14 and 27. In order to obtain integer results, n must be a multiple of 27 (otherwise b would be a non-integer).
The smallest multiple of 27 is 1x27 = 27.
So n=27
((Credits for the answer: friendlyhelp04))
It's been a decade so my answer is probably useless, but I'll say it anyway
(First, we need the smallest possible value which a multiple of both 168 and 324 ie "smallest positive integer...for which 168 n is a multiple of 324" )
168=2³ * 3 * 7
324=2² * 3⁴
(circle the 2s and 3s --which are common)
LCM= 2³ * 3⁴ * 7 (no. w/ bigger power and other 7 is taken)
= 8 * 81 *7
= 4536
(now we have a no. which is a multiple of 168 and 324 "168 n is a multiple of 324" ) (4536 is the value of 168n, through that, we find the value of just n)
168n=4536
n=4536/168
n=27
Hence, the suitable value of n is 27! :)
It's been 11 years, and I am still stuck.
168 n/324
= 2^3×3×7 n/2^2×3^4
Therefore,
n=3^3
=27
168 n/324
= 2^3×3×7 n/2^2×3^4
Therefore,
n=3^3
=27
27/14=1.92(approximately)
168n= 2x2x2x3x7xn
324= 2x2x9x9
so we need 9x9 , which was not contained in 168
n = 81
check: 168n becomes 13608
and 13608/324 = 42