Hello,
Can you please assist with this:
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^27b26c^25
oa: e
I tried to solve this by taking a = 2, b = 3 and c = 5 => N = 2^3 3^4 5^5
a) 2^3 3^4 5^5 x 2^3 3^4 5^5 = 2^6 3^8 5^10 - Not all three
b) 2^5 3^4 5^3 x 2^3 3^4 5^5 = 2^8 3^8 5^8 - Not all three
This way I tested all the way till D and when none of them satisfied all three conditions I selected E. I was wondering if there is a better method for solving this. Thanks for your help.
Best Regards,
Sri
Can you please assist with this:
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^27b26c^25
oa: e
I tried to solve this by taking a = 2, b = 3 and c = 5 => N = 2^3 3^4 5^5
a) 2^3 3^4 5^5 x 2^3 3^4 5^5 = 2^6 3^8 5^10 - Not all three
b) 2^5 3^4 5^3 x 2^3 3^4 5^5 = 2^8 3^8 5^8 - Not all three
This way I tested all the way till D and when none of them satisfied all three conditions I selected E. I was wondering if there is a better method for solving this. Thanks for your help.
Best Regards,
Sri
