jain2016 wrote:Given that N=a^3b^4c^5 where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?
A) a^3b^4c^5
B) a^5b^4c^3
C) a^2b^3c^5
D) a^7b^6c^5
E) a^27b^26c^25
The correct answer is not among the answer choices.
The correct answer is actually zero. When we multiply N by zero, the product is zero, and zero is a perfect square, a perfect cube and a perfect 5th power.
The next smallest number is
[a^(-3)][b^(-4)][c^(-5)]
Notice that (a^3)(b^4)(c^5) x
[a^(-3)][b^(-4)][c^(-5)] = 1, and 1 is a perfect square, a perfect cube and a perfect 5th power.
There are actually several other (smaller) numbers that work, but let's find the smallest number
among the answer choices.
First we need to know what perfect squares, cubes and 5th powers look like.
For example, 7^10 is a perfect square, since we can rewrite 7^10 as (7^5)^2
7^10 is also a perfect 5th power, since we can rewrite 7^10 as (7^2)^5
However, 7^10 is NOT a perfect cube since we CANNOT rewrite 7^10 as (something)^3
Another example:
7^12 is a perfect square, since we can rewrite 7^12 as (7^6)^2
7^12 is also a perfect cube, since we can rewrite 7^12 as (7^4)^3
However, 7^12 is NOT a perfect 5th power since we CANNOT rewrite 7^12 as (something)^5
So, if 7^k is to be a perfect square, a perfect cube and a perfect 5th power, k must be divisible by 2, 3 and 5
In other words, k must be divisible by 30.
So, among the answer choices, only
E will yield a product that's a a perfect square, a perfect cube and a perfect 5th power.
(a^3)(b^4)(c^5) x
[(a^27)(b^26)(c^25)] = (a^30)(b^30)(c^30)
(a^30)(b^30)(c^30) is is a perfect square, a perfect cube and a perfect 5th power.
Cheers,
Brent
