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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.

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

- 60
- 18
- 120
- 4

Basic counting principles

Divisors of a number

But try the problem first...

Answer: is 18

Source

Suggested Reading

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.

Content

[hide]

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?

- 60
- 18
- 120
- 4

Basic counting principles

Divisors of a number

But try the problem first...

Answer: is 18

Source

Suggested Reading

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.

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