Re: [math-fun] MoMath's Mindbenders for the Quarantined
Mike Beeler <mikebeeler2@gmail.com> wrote:
Voted most favorite puzzle (24% of votes) was ?Factorials and Squares?: from the product 1! * 2! * 3! * ? * 100!, remove one ?term? K!, to make the result a perfect square. Attributed to Jeremy Kun.
I just brute-force solved it, but the unique solution gives me no insight. I suspect that there's a symmetry argument which I'm missing. I don't know how many decimal digits the product of all the factorials through 100 has, but the last 1124 of them are zeros. And the last 1112 digits of the perfect square are zeros. (Of course the latter must be an even number.) I reinvented A248663 in the process. Note that in the digest option of this list, all non-ASCII characters are turned into question marks.
I solved this almost instantly. My intuition was to make a triangle of the factors, then try to eliminate the ones that occurred an odd number of times. These are 2, 4, ..., 100. But that's just 2^50 * 50!. Since 2^50 is a perfect square, we just need to eliminate 50!. Done. Tom Keith F. Lynch writes:
Mike Beeler <mikebeeler2@gmail.com> wrote:
Voted most favorite puzzle (24% of votes) was ?Factorials and Squares?: from the product 1! * 2! * 3! * ? * 100!, remove one ?term? K!, to make the result a perfect square. Attributed to Jeremy Kun.
I just brute-force solved it, but the unique solution gives me no insight. I suspect that there's a symmetry argument which I'm missing.
I don't know how many decimal digits the product of all the factorials through 100 has, but the last 1124 of them are zeros. And the last 1112 digits of the perfect square are zeros. (Of course the latter must be an even number.)
I reinvented A248663 in the process.
Note that in the digest option of this list, all non-ASCII characters are turned into question marks.
Yup, nice work! Apparently the problem dates back at least to the 2011 Invitational World Youth Mathematics Intercity Competition in Bali; there, the final term was 2012! instead of 100!. On math.stackexchange, Darij Grinberg gives a proof that there is only one answer, N/2, but his proof is only for N = 0 mod 4 and neither N/2 nor N/2 + 1 a perfect square. My computer search was all N through 27725^2; some have 2 solutions; all solutions are within 2 of N/2. NSW numbers (A002315 in OEIS) make a surprising (to me) appearance in categorizing N. I wish I had the expertise to prove the results. Keith, thanks for noting those question marks; sorry; I hate the confusion. — Mike
On Aug 28, 2020, at 12:13 AM, Tom Karzes <karzes@sonic.net> wrote:
I solved this almost instantly. My intuition was to make a triangle of the factors, then try to eliminate the ones that occurred an odd number of times. These are 2, 4, ..., 100. But that's just 2^50 * 50!. Since 2^50 is a perfect square, we just need to eliminate 50!. Done.
Tom
Keith F. Lynch writes:
Mike Beeler <mikebeeler2@gmail.com> wrote:
Voted most favorite puzzle (24% of votes) was ?Factorials and Squares?: from the product 1! * 2! * 3! * ? * 100!, remove one ?term? K!, to make the result a perfect square. Attributed to Jeremy Kun.
I just brute-force solved it, but the unique solution gives me no insight. I suspect that there's a symmetry argument which I'm missing.
I don't know how many decimal digits the product of all the factorials through 100 has, but the last 1124 of them are zeros. And the last 1112 digits of the perfect square are zeros. (Of course the latter must be an even number.)
I reinvented A248663 in the process.
Note that in the digest option of this list, all non-ASCII characters are turned into question marks.
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participants (3)
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Keith F. Lynch -
Mike Beeler -
Tom Karzes