This is a beautiful problem from ISI MSTAT 2019 PSA problem 11 based on multiplication principles. We provide sequential hints so that you can try.
How many positive divisors of \( 2^55^3{11}^4 \) are perfect squares?
Basic counting principles
Divisors of a number
But try the problem first...
Answer: is 18
ISI MStat 2019 PSA Problem 11
A First Course in Probability by Sheldon Ross
First hint
See in order to get the positive divisors of \( 2^55^3{11}^4 \) that are perfect squares , we need to take only the even powers of the primes {2,5,11}.
Second Hint
Now the maximum powers of 2 is 5 . So there are 3 choices for even powers {0,2,4} .
The maximum powers of 5 is 3 . So there are 2 choices for even powers {0,2}.
Now the maximum powers of 2 is 5 . So there are 3 choices for even powers {0,2,4} .
The maximum powers of 11 is 4 . So there are 3 choices for even powers {0,2,4}.
Final Step
Hence by multiplication principle we have in total \( 3 \times 2 \times 3 =18 \) such positive divisors.
From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:
