Delayed summation

Note that this is not formal. At all. And very tedious.

• $\dfrac11+\dfrac12+\cdots+\dfrac1{100}$
• $=\dfrac{2\times3\times\cdots\times100 + 1\times3\times4\times\cdots\times100 + 1\times2\times\cdots\times99}{100!}$
• in order to know the number of $5$ s in the prime factorizations of the denominator and the numerator, we look at them separately.
• For the denominator, I will skip this part. It is $\frac{100}5+\frac{100}{25}=24$ " $5$ "s.
• So, we have to see whether the numerator also has $24$ " $5$ "s, i.e. if the numerator is divisible by $5^{24}$ .
• Then, we will check whether there are extra " $5$ "s in the numerator.

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• The numerator is more complicated, and has 100 terms.
• Each term can be expressed as $\frac{100!}n$ .
• Note that each term in the numerator is divisible by $5^{22}$ (only). This includes $\frac{100!}{25}$ , $\frac{100!}{50}$ , $\frac{100!}{75}$ , and $\frac{100!}{100}$ .
• Some of the terms are also divisible by $5^{23}$ , because only one " $5$ " is taken away from them. This includes $\frac{100!}5$ , $\frac{100!}{10}$ , $\frac{100!}{15}$ , $\frac{100!}{20}$ , $\frac{100!}{30}$ , ...
• Most of the terms are divisible by $5^{24}$ , because no " $5$ " is taken away from them. This includes $\frac{100!}1$ , $\frac{100!}2$ , $\frac{100!}3$ , $\frac{100!}4$ , $\frac{100!}6$ , ...

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• Note that:
• $\dfrac{100!}n + \dfrac{100!}{100-n}$
• = $\dfrac{100!}{n\times(100-n)} (n+(100-n))$
• = $\dfrac{100\times100!}{n\times(100-n)}$
• which is divisible by $5^{24}$ if and only if $n(100-n)$ is not divisible by $125$ .
• This applies to $\frac{100!}{25}$ , $\frac{100!}{50}$ , $\frac{100!}{75}$ , $\frac{100!}{100}$ only.
• However:
• $\dfrac{100!}{25} + \dfrac{100!}{50} + \dfrac{100!}{75} + \dfrac{100!}{100}$
• $= \dfrac{100!}{25\times1\times2\times3\times4} (2\times3\times4 + 1\times3\times4 + 1\times2\times4 + 1\times2\times3)$
• $= \dfrac{100!}{25\times1\times2\times3\times4} (50)$
• $= \dfrac{100!}{12}$
• which is divisible by $5^{24}$ now after our manipulation.
• Therefore, the numerator is divisible by $5^{24}$ .

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Now, we will check whether it is divisible by $5^{25}$ .

• Note that:
• $\dfrac{100!}n + \dfrac{100!}{100-n}$
• = $\dfrac{100!}{n\times(100-n)} (n+(100-n))$
• = $\dfrac{100\times100!}{n\times(100-n)}$
• which is divisible by $5^{25}$ if and only if $n(100-n)$ is not divisible by $5$ .
• This applies to $\frac{100!}{5}$ , $\frac{100!}{10}$ , ..., $\frac{100!}{100}$ only.
• Let us look at those except $\frac{100!}{25}$ , $\frac{100!}{50}$ , $\frac{100!}{75}$ , and $\frac{100!}{100}$ .
• They can be grouped into 4 groups each containing 4 members in the form of $(25n+5)$ , $(25n+10)$ , $(25n+15)$ , $(25n+20)$ .
• Note that:
• $\frac{100!}{25n+5} + \frac{100!}{25n+10} + \frac{100!}{25n+15} + \frac{100!}{25n+20}$
• = $\frac{100!(50n+25)}{(25n+5)(25n+25)} + \frac{100!(50n+25)}{(25n+10)(25n+15)}$
• = $\frac{100!(2n+1)}{(5n+1)(5n+4)} + \frac{100!(2n+1)}{(5n+2)(5n+3)}$
• = $\frac{100!(2n+1)((5n+2)(5n+3)+(5n+1)(5n+4))}{(5n+1)(5n+2)(5n+3)(5n+4)}$
• = $\frac{100!(2n+1)(50n^2+50n+10)}{(5n+1)(5n+2)(5n+3)(5n+4)}$
• is divisible by $5^{25}$ .
• Therefore, we only need to look at $\frac{100!}{25}$ , $\frac{100!}{50}$ , $\frac{100!}{75}$ , and $\frac{100!}{100}$ .
• From above, their sum is $\frac{100!}{12}$ , which is NOT divisible by $5^{25}$ .

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Hence, the highest power of $5$ that divides the numerator is $5^{24}$ .

The highest power of $5$ that divides the denominator is also $5^{24}$ .

Therefore, neither A nor B is divisible by 5.

Rather than a solution, let me post this hint:

Out of the 100 terms, 92 can be grouped into pairs of the form:

$\frac 1{a} +\frac 1{125-a}$ ,

where a isn't a multiple of 25, or

$\frac 1{b} +\frac 1{25-b}$

where b isn't a multiple of 5, and the other eight terms add up to $\frac 1{12} +\frac 5{12}=\frac 12$ .

X=2^6×3⁴x7²×11×13×17×19×23×29×31×37×41×43×47×53×59×61×67×71× 73×79×83×89×97 B|X, Y=X/1+X/2+...X/100,A|Y X/25=2^25=2×4¹²=2(mod 5) X/50=2^24=1(mod 5) X/75=-2^24=-1(mod 5) X/100=2^23=-2(mod 5) X/r=0(mod 5) when r≠25k So Y=1+2-1-2+96×0=0(mod 5) Similiar X/p=0(mod 25) p≠5l Now we can easily check that X/5+X/10+ ...+X/95+X/100=0(mod 25) So for sure B isn't divided by 5,also Y≠0 (mod 125) so A isn't divided by 5