How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$?

Hmmm... So I didn't know what to do here so I tested some cases for a rule.

If a number had the factors $3^2$ and $2^1$, you can make $5$ distinct factors: $2^1$, $3^1$, $3^2$, $2^1 \cdot 3^1$, $2^1 \cdot 3^2$... I don't see a pattern yet.

How does one go about this?

I don't think the answer is $5!$....

  • $\begingroup$ Not sure why you put parantheses there, you mean the product of these numbers, right? $\endgroup$ – k.stm Feb 12 '14 at 8:12
  • $\begingroup$ $2^1$ can be written as $2^1\cdot 3^0$.$3^1$ can be written as $2^0\cdot 3^1$. The factor you forgot is $2^0\cdot3^0$. So if in the product, the power of a prime number is $x$, we will multiply $x+1$ while calculating the number of factors. $\endgroup$ – Shaurya Gupta Feb 12 '14 at 18:55

If $\begin{equation}x = a^p \cdot b^q\cdot c^r+...\end{equation}$
then there are $(p+1)(q+1)(r+1)...$ numbers that divde $x$.
Any number that divides $x$ will be of the form $a^\alpha\cdot b^\beta\cdot c^\gamma$ .
So we have p p+1 options for $\alpha$ because we need to consider $\alpha = 0$ also. Similarly, we have $q + 1$ options for $\beta$ and $q+1$ options for $\gamma$.
Therefore, we multiply these out.

To get the intuition of why this is so, let us take the example of the number $24$.
$24 = 2^3\cdot3^1$
Any number that divides $24$ will be of the form $2^\alpha\cdot3^\beta$.
We have 4 choices for $\alpha:0,1,2,3$ and 2 choices for $\beta:0,1$.
So we have $4\cdot2 = 8$ numbers that divide 24. These can be listed out as:


So, $2^5*3^4*5^3*7^2*11^1$ has $6\cdot5\cdot4\cdot3\cdot2 = \boxed{720}$

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  • $\begingroup$ That should be $x = a^p b^q c^r \cdots$, not with addition. $\endgroup$ – Richard D. James Feb 12 '14 at 18:51
  • $\begingroup$ @SpamIAm THanks for noticing the typo! $\endgroup$ – Shaurya Gupta Feb 12 '14 at 18:53

The number you are talking about is $2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1$ if i understand what you write.
You can write this number as $ 2^5\cdot3^4\cdot2^6\cdot3^2\cdot2^1=2^{12}\cdot3^6$ which is a product of prime numbers.
So,the number of distinct factors is exactly $$(12+1)\cdot(6+1)=91$$.
You should look here for more details and to see how to solve any similar problem.

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Your approach with $3^2$ and $2^1$ is very close. You only forgot to account for the factor $1 = 3^0\cdot2^0$. This brings us to $6$ distinct factors for this smaller example. You can calculate this by multiplying the number of all the different exponents used for each factor. $3$ can occur either $2$, $1$ or $0$ times, while $2$ can only occur $1$ or $0$ times. This brings us to $3\cdot2 = 6$ combinations of these exponents.

Now, the "real" question has another little trick to it because the factors aren't all that nice just yet. $2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1 = 2^6\cdot3^6\cdot4^3$
Much better.
Now, we can have between $0$ and $6$ occurences of $2$, making it $7$ possibilities.
Another $7$ possibilities for occurences of $3$ and finally $4$ possibilities for $4$.
This makes a total of $7\cdot7\cdot4 = 196$ possible distinct factors of $2^6\cdot3^6\cdot4^3 = 2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1$.

Edit Alright, I didn't pay attention either. As Konstantinos Gaitanas stated, $4^3 = 2^6$, making the whole thing $2^{12}\cdot3^6$.
So the solution should indeed be $13\cdot7$ = 91.

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