What are the trailing number of the zeroes in the given integer

Question

Problem Statement:- The number of zeroes at the end of the integer $$100!-101!+\ldots-109!+110!$$

I am having a bit of a trouble in thinking how do I proceed. A little push in the right direction would be appreciated.

And if you are posting a full solution do use the spoiler tag, as sometimes I cant stop myself from seeing the whole solution and lose the chance of thinking through it by myself with a push in the right direction from you guys.

Also, I dont know whether I am using the right tag feel free to correct it if its wrong.

Answer 1

I think that it is easiest to factor the expression as

$100!(1-101+101\cdot 102-\cdots +101\cdot 102\cdots 110)$.

Then, it is a standard exercise to compute the number of trailing zeros in

$100!$

The remainder can be computed modulo $10$, $100$, etc, to see how many trailing zeros it has. For instance, modulo $10$, we are looking at whether

$0!-1!+2!-3!+4!-5!+6!-7!+8!-9!+10!\stackrel{?}{\equiv} 0\pmod{10}$

to see if this term contributes any trailing zeros. Since every term with both a factor of $2$ and $5$ is zero modulo $10$, we get that this simplifies to

$0!-1!+2!-3!+4!\equiv 1-1+2-6+24\equiv 20\equiv 0\pmod{10}$

Now, computing this modulo $100$, we consider

$0!-1!+2!-3!+4!-5!+6!-7!+8!-9!+10!\stackrel{?}{\equiv} 0\pmod{100}$

A short calculation shows that this simplifies to

$1-1+2-6+24-20+20-40+20-80+0\equiv 20\not\equiv 0\pmod{100}$

Therefore, the number of trailing zeros is

The number of trailing zeros of $100!$ plus $1$ since the last nonzero digits >! of $100!$ and $(1-101+101\cdot 102-\cdots +101\cdot 102\cdots 110)$ are not $5$s (so no additional zeros can be created in the product).

Answer 2

The number of zero digits at the end of $n!$ can be found by looking at the factors, and decomposing them into $2$ and $5$ factors. You need one $2$ and one $5$ to produce a trailing zero.

How to combine this with that alternating sum, I have no idea yet.

BTW: I explained it to my friend Ruby and she tells me the sum is $157393819470814604687108965690332604480484877$\ $508029687469884051113407737751285106008107839$\ $400103709226880772747397138959112221377791569$\ $61431310006359162880000000000000000000000000$

Answer 3

First of all,

$100!$ has 24 trailing zeroes for the number of factors $5$ in $100!$ is $24$, and there are more factors $2$ than $5$.

Then,

$101!$ also has $24$ trailing zeroes, and so do $102!,103!,104!$, but $105!,106!,107!,108!,109!$ have an extra factor $5$ and thus end in $25$ zeroes. $110!$ ends in $26$ zeroes.

With this,

We know that $-105!+106!-107!+108!-109!+110!$ ends in at least $25$ zeroes.

So

if $100!-101!+102!-103!+104!$ ends in just $24$ zeroes, we know that their sum does. But does it end in more than $24$ zeroes?

We now can

look at the digit before all the zeroes start (the 25th digit from the right). Let's name that digit $d$ for $100!$. Now what is that digit for the other four factorials?

It is

Since going from $100!$ to $101!$ is only a factor $101$, we can find that digit easily looking $\mod 10$. We see $101d\equiv d\mod 10$ for $101!$, and $102\cdot 101\cdot d\equiv 2d\mod 10$ for $102!$, etcetera; Now we can find the alternating sum,

To arrive at

$d-d+2d-6d+4d=0$. Thus, the $25th$ digit from the right is also a $0$. We do the same, but look at the last two digits before the trailing zeroes. Let's call those two digits $d_1$ for $100!$

now,

we need to look $\mod 100$, and obtain the sum $d_1-d_1+2d_1-6d_2+24d_1=20d_1$. Now we need to know, does $20d_1$ end in two zeroes?

Well,

for $20d_1$ to end in two zeroes, $d_1$ needs to be a multiple of $5$.

But

every factor $5$ is in the trailing zeroes, and as such, $d_1$ cannot be a multiple of $5$.

Therefore,

the 26th digit from the right of $100!-101!+102!-103!+104!$ is not $0$, Now what is the 26th digit from the right of $-105!+106!-107!+108!-109!+110!$?

We can use

the same method; looking $\mod 10$, setting the 26th digit from the right of $105!$ to $d_2$, and noting that $-d_2+6d_2-2d_2+6d_2-4d_2+0d_2=5d_2$. Does $5d_2$ end in a $0$?

Because

the last nonzero digit of any factorial greater than $1!$ is even, since there are not enough factors $5$ to cancel out all the factors $2$. Thus, the last non-zero digit of $105!$ must be even; thus, $5d_2$ ends in a $0$.

We conclude the final answer

the number of trailing zeroes must be $25$, since $100!-101!+102!-103!+104!$ has $25$ trailing zeroes and $-105!+106!-107!+108!-109!+110!$ has at least $26$.

Answer 4

$100!$ has $[\frac{100}{5}]+[\frac{100}{5^2}]=24$ trailing $0$'s. Now, $100!-101!+\cdots -109!+110!$ $=m\cdot 100!$ for some positive integer $m$ such that

$m\equiv 0!-1!+2!-3!+\cdots +10!$ (mod $100$)

$\equiv -\{(3!-2!)+(5!-4!)+\cdots +(9!-8!)\}+10!$ (mod $100$)

$\equiv -(2\cdot 2!+4\cdot 4!+\cdots +8\cdot 8!)+10!$ (mod $100$)

$\equiv -(4+96+20+60)$ (mod $100$)

$\equiv -180$ (mod $100$)

$\equiv 20$ (mod $100$)

Thus, we get one more 0 from the factor $m$. Therefore, required number of trailing $0$'s is $25$.