Budi's factorial sum

Notice that $i\cdot\;i!\;=\big(i+1-1\big)i!\;=\big(i+1\big)!-i!$ which telescopes.

So we have $\sum_{i=1}^{101} \big(i+1)!-i!\;= 102!-1!$

Now, $102!\;=102\cdot101\cdot100!$ which is divisible by $101$ .

So $102!-1!\;\equiv\;a\big(mod101\big)$

$\Rightarrow\;-1!\;\equiv\;a\big(mod101\big)$

$\Rightarrow\;a\;=\boxed{100}$ which is the remainder we seek.

$\sum_{i=1}^{101}{i!i}$

= $\sum_{i=1}^{101}{i!((i+1)-1)}$

= $\sum_{i=1}^{101}{ (i+1)i!-i! }$

= $\sum_{i=1}^{101}{(i+1)!-i!}$

= $102! - 101! + 101! - 100! +100! - 99! ...... +2! - 1!$

= $102! - 1$

$102 \times 101 \times 100 \times ... \times 2 \times 1 \equiv 0 \pmod{101}$

$102! \equiv 0 \equiv 101 \pmod{101}$

$102! - 1 \equiv 101 -1 \pmod{101}$

$102! - 1 \equiv 100 \pmod{101}$

Hence, the answer is 100. #

We can arrange the given expression as

$\displaystyle \sum_{i=1}^{101} i!(i+1-1)$

$=\displaystyle \sum_{i=1}^{101} (i+1)!-i!$

$=-\displaystyle \sum_{i=1}^{101} i!-(i+1)!$

Then we can observe an exciting sequence,

$=-(1!-2!+2!-3!+...+101!-102!)$

$=102!-1!$

It is obvious that $102!$ is divisible by $101$ , then the remainder will be

$-1 \mod 101$

$=100 \mod 101$

Hence, the answer is $100$ .

---

The idea of solving this problem :

In this typical problem with sigma notation, a good way to solve it is by Telescoping Method . We can try to express it as

$\displaystyle \sum_{k=1}^n f(k)-f(k+1)$

and we can get a surprising

$f(k)-f(n+1)$

I hope some of the users can learn something from me! :)

i*i!=(i+1)!-i!
summing above to 101 terms will give sum as 102!-1 102! is divisible by 101
-1 mod 101 is 100
Thus remainder is 100

Note that $i! \cdot i = i! \cdot \bigl( i+1 - 1\bigr) = i! \cdot (i+1) - i!$ Once this is observed we see that the sum telescopes and we get the value of the summation as $102! - 1!$ from which we conclude that the remainder is 100.

Rewrite the given summation as:

$\displaystyle \sum_{i=1}^{101} i! \cdot i=\sum_{i=1}^{101} i!\cdot(i+1-1)=\sum_{i=1}^{101} (i+1)!-i!$

$\displaystyle =2!-1!+3!-2!+4!-3!+.....+102!-101!$

$\displaystyle = 102!-1$

$\displaystyle =102!-101+100$

The first two terms are divisible by 101, hence 100 is the remainder.

We can prove by induction that 1! 1+2! 2+...+n!*n=(n+1)!-1, so the desired sum is 102!-1. Taking this mod 101, the factorial term is 0, leaving us with -1 which is congruent to 100 mod 101, so the answer is 100

The main relation is that $n\cdot n!=(n+1)!-n! \;\;\; (\star)$ which is easily proved by distributive law and definition of factorial: $\begin{array}{lll}n\cdot n!&=&(n+1-1)\cdot n!\\&=&(n+1)\cdot n!-1\cdot n!\\&=&(n+1)!-n!\end{array}$ Now, $\displaystyle\begin{array}{lll}S:=\sum_{i=1}^{101}i\cdot i!&=&\sum_{i=1}^{101}(i+1)!-i! \\ &=& \sum_{i=1}^{101}(i+1)!-\sum_{i=1}^{101}i! \\ &=& \sum_{i=2}^{102}i!-\sum_{i=1}^{101}i! \\ &=& \sum_{i=1}^{101}i! +102!-1!-\sum_{i=1}^{101}i! \\ &=& 102!-1\end{array}$ Since $101\mid 102!$ , we have $S\equiv 102!-1\equiv-1\equiv100\pmod{101}$