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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

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.

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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

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.

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