3 ms·

The number has 12 zeroes at the end, which confused me for a bit, so I'm leaving this comment in case anyone else is similarly confused and would like to know w

The number has 12 zeroes at the end, which confused me for a bit, so I'm leaving this comment in case anyone else is similarly confused and would like to know why. There are 10 total numbers between 1 and 52 which include 5 as a factor (5 which also include 2 as a factor to make a factor of 10 and 5 more to be multiplied with a bunch of other 2-factor numbers) so intuitively I was thinking there should be 10 zeroes. The two I missed are additional factors of 5, one each in 25 and 50.

gjm11 2y ago
The cute general theorem is: if p is a prime number then the number of times p divides n factorial is floor(n/p) + floor(n/p^2) + floor(n/p^3) + ... And from this (or by other arguably more elegant means) one can get the related cute theorem: the number of times p divides the binomial coefficient (n choose r) is the number of carries that occur when adding r to n-r in base p. In particular, if n = p^k then unless r=0 or r=n there is always at least one carry because n is "longer" than r and n-r, so all the binomial coefficients (p^k choose r) are multiples of p apart from (p^k choose 0) and (p^k choose p^k). (You can use this sort of idea to understand why, if you write out many many rows of Pascal's triangle mod 2, you get a sort of Sierpinski gasket thing.)
petermcneeley 2y ago
2^49×3^23×5^12×7^8×11^4×13^4×17^3×19^2×23^2×29×31×37×41×43×47 There is a 5^12 and more than that for the 2s. So yes 12 zeros it will be as God intended. https://www.wolframalpha.com/input?i=factors+of+80658175170943878571660636856403766975289505440883277824000000000000 https://www.wolframalpha.com/input?i=factors+of+806581751709...
49pctber 2y ago
25 and 50 contribute two 5 factors each. The multiplicity of 2 factors is even greater, as every other even factor contributes at least two, every fourth at least three, etc.

gjm11 2y ago

The cute general theorem is: if p is a prime number then the number of times p divides n factorial is floor(n/p) + floor(n/p^2) + floor(n/p^3) + ... And from this (or by other arguably more elegant means) one can get the related cute theorem: the number of times p divides the binomial coefficient (n choose r) is the number of carries that occur when adding r to n-r in base p. In particular, if n = p^k then unless r=0 or r=n there is always at least one carry because n is "longer" than r and n-r, so all the binomial coefficients (p^k choose r) are multiples of p apart from (p^k choose 0) and (p^k choose p^k). (You can use this sort of idea to understand why, if you write out many many rows of Pascal's triangle mod 2, you get a sort of Sierpinski gasket thing.)

petermcneeley 2y ago

2^49×3^23×5^12×7^8×11^4×13^4×17^3×19^2×23^2×29×31×37×41×43×47 There is a 5^12 and more than that for the 2s. So yes 12 zeros it will be as God intended. https://www.wolframalpha.com/input?i=factors+of+80658175170943878571660636856403766975289505440883277824000000000000 https://www.wolframalpha.com/input?i=factors+of+806581751709...

49pctber 2y ago

25 and 50 contribute two 5 factors each. The multiplicity of 2 factors is even greater, as every other even factor contributes at least two, every fourth at least three, etc.