Question

The number ${90^9}$ has 1900 different positive integral divisors. How many of these are squares of integers?

Answer 1

Hint: In this question we have to find the squares of integers of different positive integral divisors, first we have to prime factorise the number 90 and then apply the power to all the factors and then we will get the numbers which are perfect squares and then multiplying the combinations we will get the required number.

Complete step by step solution:
Given that the number ${90^9}$ has 1900 different positive integral divisors,
Now factorise 90 into prime its prime factors we get,
$ \Rightarrow 90 = 1 \times 2 \times 3 \times 3 \times 5$,
Now factorising ${90^9}$into its prime factors we get,
 $ \Rightarrow {90^9} = {\left( {2 \times 3 \times 3 \times 5} \right)^9}$
Now simplifying we get,
$ \Rightarrow {90^9} = {\left( {2 \times {3^2} \times 5} \right)^9}$,
Now further simplifying we get,
$ \Rightarrow {90^9} = {2^9} \times {3^{18}} \times {5^9}$,
Squares of integers can be generated in a variety of ways from this factorization:
$ \Rightarrow {2^9} \times {3^{18}} \times {5^9}$,
We can see that ${5^0}$, for example, is a square of an integer and a divisor of ${90^9}$; likewise, ${5^2}$, ${5^4}$,${5^6}$ and ${5^8}$ all meet these conditions as well. Therefore, we have 5 possible ways to configure a divisor of ${90^9}$ that is a square of an integer, using 5’s alone.
We can see that ${3^0}$, for example, is a square of an integer and a divisor of ${90^9}$; likewise, ${3^2}$, ${3^4}$,${3^6}$ and ${3^8}$, ${3^{10}}$, ${3^{12}}$, ${3^{14}}$, ${3^{16}}$, and ${3^{18}}$ all meet these conditions as well. Therefore, we have 10 possible ways to configure a divisor of ${90^9}$ that is a square of an integer, using 3’s alone.
We can see that ${2^0}$, for example, is a square of an integer and a divisor of ${90^9}$; likewise, ${2^2}$, ${2^4}$,${2^6}$ and ${2^8}$ all meet these conditions as well. Therefore, we have 5 possible ways to configure a divisor of ${90^9}$ that is a square of an integer, using 2’s alone.
Furthermore, any combination of these prime divisors who have even powers also satisfies the conditions. For instance, ${\left( {{2^2} \cdot {5^2}} \right)^2}$ is a square of an integer, as is ${\left( {{3^8} \cdot {2^4}} \right)^2}$ and both being made up of divisors of ${90^9}$, are also divisors of ${90^9}$.
Thus the desired number of squares of integers that are divisors of ${90^9}$ is given by $5 \cdot 10 \cdot 5$, which is the multiplication of the possible choices for each prime factor, now simplifying we get,
The number of squares of integers are ${90^9} = 5 \cdot 10 \cdot 5 = 250$.

$\therefore $ The desired number of squares of integers that are divisors of ${90^9}$ are 250.

Note:
To find how many positive integral divisors a number has, there are two methods to choose from. In the first method, you just count the divisors mentally. We know that each number has positive integral divisors in pairs, except for perfect square numbers. If you know your number is not a perfect square, then you can count in pairs.