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ISI MStat 2019 PSA Problem 11 | Multiplication Principle

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.

Multiplication Principle - ISI MStat 2019 PSA - 11


How many positive divisors of 2^55^3{11}^4 are perfect squares?

  • 60
  • 18
  • 120
  • 4

Key Concepts


Basic counting principles

Divisors of a number

Check the Answer


Answer: is 18

ISI MStat 2019 PSA Problem 11

A First Course in Probability by Sheldon Ross

Try with Hints


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}.

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}.

Hence by multiplication principle we have in total 3 \times 2 \times 3 =18 such positive divisors.

Similar Problems and Solutions



ISI MSTAT PSA and PSB Problems
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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.

Multiplication Principle - ISI MStat 2019 PSA - 11


How many positive divisors of 2^55^3{11}^4 are perfect squares?

  • 60
  • 18
  • 120
  • 4

Key Concepts


Basic counting principles

Divisors of a number

Check the Answer


Answer: is 18

ISI MStat 2019 PSA Problem 11

A First Course in Probability by Sheldon Ross

Try with Hints


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}.

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}.

Hence by multiplication principle we have in total 3 \times 2 \times 3 =18 such positive divisors.

Similar Problems and Solutions



ISI MSTAT PSA and PSB Problems
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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