finding the number of factors $2^{15}\times3^{10}\times5^6$

Question

The number of factors of $2^{15}\times3^{10}\times5^6$ which are either perfect square or perfect cubes(or both)
I don't know how to start this even!
Plz solve this!

Answer 1

A perfect square will have even exponents for each term, and a perfect cube will have multiples of three in the exponents of each term. Using the inclusion-exclusion formula, the number of factors which are squares or cubes will be the number of squares plus the number of cubes minus the number of factors which are both squares and cubes (i.e. have multiples of six in the exponents).

The even exponents of the first term are $\{0,2,\dots,14\}$, so there are 8. For the second and third terms, there are 6 and 4 respectively. So, there are $8\times 6\times 4=192$ squares.

The exponents which are multiples of 3 in the first term are $\{0,3,\dots, 15\}$, which is 6 possible exponents. For the second and third terms, there are 4 and 3 exponents that are multiples of 3 respectively. So there are $6\times 4\times 3=72$ possible cubes.

To complete the question, I will leave it up to you to subtract the number of factors which have a sixth root (i.e. have multiples of 6 in the exponents). We are subtracting the number of factors which are both cubes and squares because of the inclusion-exclusion principle, which says this: Let $A$ be the set of squares and $B$ the set of cubes. Then $|A\cup B|=|A|+|B|-|A\cap B|$.

Answer 2

$2^2=4$, $3^2=9$, and $5^2=25$ are the irreducible square factors, i.e. they have no square factors of their own besides themselves and the trivial factor $1$. You have $$ (2^2)^7\times 2 \times(3^2)^5 \times (5^2)^3 = \underbrace{4^7\times 9^5\times 25^3}_\text{squares} \times \underbrace{ \qquad 2 \qquad}_\text{square-free part}. $$ You can count the square factors the same way you count all the factors just by treating $4$, $9$, and $25$ the same way you treat prime factors when you're counting all of the factors.

Answer 3

Case 1: factors that are perfect squares

\begin{align} A & =\{2^0,2^2,2^4,2^6,2^8,2^{10},2^{12},2^{14}\} \\ B & =\{3^0,3^2,3^4,3^6,3^8,3^{10}\} \\ C & =\{5^0,5^2,6^4,5^6\} \end{align}

Choose 1 from each set. We have $8\times6\times4=192$ choices.

Case 2: factors that are perfect cubes

\begin{align} A& =\{2^0,2^3,2^6,2^9,2^{12},2^{15}\} \\ B & =\{3^0,3^3,3^6,3^9\} \\ C & =\{5^0,5^3,5^6\} \end{align}

Choose 1 from each set. We have $6\times4\times3=72$ choices.

Therefore, we have $72+192-3\times2\times2=252$ factors in total.

Note: subtract $3\times2\times2$ from the total because some sixth powers have been counted twice.

Answer 4

A factor can be described by a tuple of exponents $f \in \{ 0, \dotsc, 15 \} \times \{ 0, \dotsc, 10 \} \times \{ 0, \dotsc, 6 \} $. As $$ (2^{f_1} 3^{f_2} 5^{f_3})^{1/n} = 2^{f_1/n} \, 3^{f_2/n} \, 5^{f_3/n} $$ we need $n$ to divide the exponents $f_i$.

The perfect squares ($n=2$) are combined from $ S = \{ 0, 2, 4, 6, 8, 10, 12, 14 \} \times \{ 0, 2, 4, 6, 8, 10 \} \times \{ 0, 2, 4, 6 \} $, the perfect cubes ($n=3$) from $ C = \{ 0, 3, 6, 9, 12, 15 \} \times \{ 0, 3, 6, 9 \} \times \{ 0, 3, 6 \} $. Then $S \cap C = \{ 0, 6, 12 \} \times \{ 0, 6 \} \times \{ 0, 6 \} $. This gives $$ \lvert S \rvert + \lvert C \rvert - \lvert S \cap C \rvert = 8 \cdot 6 \cdot 4 + 6 \cdot 4 \cdot 3 - 3 \cdot 2 \cdot 2 = 252 $$