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@Seth Cohen: Hi Seth,
Your analysis about the multiplicity on primes under 250 is key.
One other thing that helped me is I wrote out '500! x 499! x 498! x 497! x ... x 2! x 1!', stared at it, played with different ideas, and eventually saw that I could rewrite it by grouping together pairs of factorials, which I'll detail in the next paragraph.
I was thinking about how to group that expression into squares, and I eventually lucked out and saw I could do this rewrite: 500! x 499! x 498! x 497! x ... x 2! x 1! = 500 x (499!)^2 x 498 x (497!)^2 x ... x 2 x (1!)^2. This opened up the floodgates for me. I was able to find *an* answer for n. I then used the same analysis you proposed and proved it was the *smallest* answer for n. I hope this helps!
×1
Your analysis about the multiplicity on primes under 250 is key.
One other thing that helped me is I wrote out '500! x 499! x 498! x 497! x ... x 2! x 1!', stared at it, played with different ideas, and eventually saw that I could rewrite it by grouping together pairs of factorials, which I'll detail in the next paragraph.
I was thinking about how to group that expression into squares, and I eventually lucked out and saw I could do this rewrite: 500! x 499! x 498! x 497! x ... x 2! x 1! = 500 x (499!)^2 x 498 x (497!)^2 x ... x 2 x (1!)^2. This opened up the floodgates for me. I was able to find *an* answer for n. I then used the same analysis you proposed and proved it was the *smallest* answer for n. I hope this helps!
(anonymous)
on /blog/107
on /blog/107
×1 @(anonymous): Hi Seth, sorry, I forgot to put my name on my post. I hope it was useful!
Ryan
on /blog/107
on /blog/107
@Ryan: Got it! I like your method -- just keep eliminating square numbers until you're left with what you need.
I still wanted to figure out why my original method was wrong. And it finally dawned on me:
My mistake was not realizing that my answer of 241 was just a lower bound. The value of n needed to be AT LEAST 241, because my analysis said that 241 needed to be divided out. But any number >241 would also do the job of dividing out 241. So I needed to think about higher numbers too. ×1
I still wanted to figure out why my original method was wrong. And it finally dawned on me:
My mistake was not realizing that my answer of 241 was just a lower bound. The value of n needed to be AT LEAST 241, because my analysis said that 241 needed to be divided out. But any number >241 would also do the job of dividing out 241. So I needed to think about higher numbers too.
Seth Cohen
on /blog/107
on /blog/107
×1
I thought about prime factors. For the full expression to be a square, all prime factors need to appear an even number of times. So to find n, I can see what prime factors in the numerator appear an odd number of times, and divide them out.
Looking at the prime numbers less than 500, any that are >=250 will appear an even number of times in the numerator. For example, 251 will appear in the 251! term, the 252! term, all the way up to the 500! term. That's 250 appearances, which is an even number.
But what about less than 250? Let's look at 241. That will appear in 241! up to 500! (260 times), but it will also appear in 482! up to 500! (19 times), because 482=241*2. So 241 appears 260+19=279 times, which is an odd number. So 241 needs to be divided out. Likewise with numbers less than 241, like 239. I didn't count the number of appearances of all numbers below 241, but I figured that if n=241, the denominator being 241! will divide out all the numbers that need to be divided out. But that didn't work.
So what am I missing? Any hint would be appreciated! Thanks!
on /blog/107