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Calculate the number of perfect squares which are factors of 1!2!3!...100!

I tried to produce the prime factorisation of the product and then combine its factors but messed up. For example, starting from the right, we definitely have 99^2, 98^2 and so on but what about the rest of their factors?

Carlos Lopez
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1 Answers1

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Let $p_i$ be the various prime factors of the product, and $q_i=p_i^2$ (we have $p_i=2,3, \dots 97)$

We can write the product as $$\Pi q_i^{r_i}\Pi p_i^{\epsilon_i}$$ with $\epsilon_i\in \{0,1\}$ so that the product of the $p_i^{\epsilon_i}$ is square free.

Then any product of the $q_i$ (including the empty product) is a square by construction, and the number of such products is the product of terms of the form $r_i+1$. You simply need to use what you know of the prime factorisation here to determine the $r_i$.

So what you get is a prime factorisation of form $2^{2r_1+\epsilon_1}\cdot 3^{2r_2+\epsilon_2}\cdot \dots 97^{2r_{25}+\epsilon_{25}}$ where the epsilons are $1$ or $0$. There are $r_1+1$ ways of choosing an even power of $2$, $r_2+1$ ways of choosing an even power of $3$ etc and when you multiply even powers of primes together you get a square.

Now when you create the product $(r_1+1)(r_2+1)\dots (r_{25}+1)$ you are counting the ways of combining the different primes raised to even powers - i.e. the number of possible square factors.

Mark Bennet
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  • I think the biggest perfect square which is a factor of the product is 47^2. – Alex.vollenga Oct 18 '17 at 19:06
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    @Alex.vollenga Well you have $97$ as a factor of $97!$, $98!$, $99!$ and $100!$ so you have $(97^2)^2$ as a factor. At the other end of things there is a huge power of $2$ which divides the product of the factorials. And note that the product of two squares is itself a square. – Mark Bennet Oct 18 '17 at 19:18
  • Ooops you are right, I forgot that it's the product of factorials! Thank you! – Alex.vollenga Oct 18 '17 at 19:20
  • OK, so, all in all, we have 99 squares plus 25 4th powers plus 12 8th powers plus 6 16th powers plus 3 32nd powers? – Alex.vollenga Oct 18 '17 at 19:57
  • @Alex.vollenga You have the exponent of $2$ in $100!$ as $50+25+12+6+3+1=97$, in $99!$ it is $49+24+12+6+3+1=95$, in $98!$ it is also $95$ (because $99$ is odd). The exponent of $3$ is $33+11+3+1=48$ in $100!$ and the same in $99!$ etc. You get $2^{287+}3^{96+}\dots$ so the number of squares here is at least $144\times 49 \times \dots$ (some of these are indeed fourth powers). I've only considered the prime factors $2$ and $3$ and there are others. – Mark Bennet Oct 18 '17 at 20:22
  • Wow... so we also have the powers of 5, 7, 11 and all the primes up to 97. How do we calculate the total number? – Carlos Lopez Oct 18 '17 at 20:30
  • @CarlosLopez See this question and answers https://math.stackexchange.com/questions/141196/highest-power-of-a-prime-p-dividing-n – Mark Bennet Oct 18 '17 at 20:52
  • @MarkBennet: Thank you. So, according to the above link, the highest power of 2 dividing 100! is 97? And we do the same for all primes up to 100 and add them up? – Carlos Lopez Oct 19 '17 at 11:16
  • @CarlosLopez That's right - you need to do it for all the factorials for each prime - once you have an exponent for each prime you can then multiply the terms for each prime together. – Mark Bennet Oct 19 '17 at 11:46
  • @Mark Bennet: So the problem expects you to do this for all 25 primes (from 1 to 100) and for each of the factors? I.e. 1!, 2!, 3! etc? And in your above comment to me, why do we stop at 287? Aren't there any more factors in 97!, 96! etc and up to 2!? (sorry for being a bit naive!!) – Alex.vollenga Oct 19 '17 at 19:11
  • And also just thought of the following: How about the combinations of two prime powers? For example, 2^5^2 or 7^211^2? Isn't this also a perfect square? – Carlos Lopez Oct 19 '17 at 20:29
  • @CarlosLopez I have added a bit at the end of my answer which may help. And I put $287+$ in my exponent to indicate that I hadn't finished the calculation, but had just given an example of how to proceed. – Mark Bennet Oct 19 '17 at 20:50