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.
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.
