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@(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
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!
on /blog/107