Pi Han's superfactorial

We have that, $SF(2013)=1!2!\cdots 2013!$ A factor of $2013=3\cdot 11\cdot 61$ will occur the minimum number of times $3$ , $11$ , or $61$ occurs. Clearly, 61 will occur the least, so we must count the factors of 61.

We can group the product: $(1!2!\cdots 60!)(61!62!\cdots 121!)\cdots(1952!1953!\cdots 2012!)(2013!)$ The first term has no factors of 61, so we ignore it. The last term has 33 factors of 61. All other terms have 61 terms, each of which has the same number of factors of 61. The total number of 61's is $61(1+2+\cdots+32)+33=61\dfrac{32\cdot 33}{2}+33=32\boxed{241}$

$2013 = 3 \times 11 \times 61$

Clearly, the least power in the superfactorial out of $3,11$ and $61$ would be of $61$ and the same would be $N$ .

Now, $61$ would first come in $61!$ , and till $(61 \times 2- 1)!$ (i.e. next $61$ terms) , the power of $61$ in each individual factorial would be $1$ , then , from $(61 \times 2)!$ to $(61 \times 3 -1)!$ the power of $61$ in each individual factorial would be $2$ , $\dots$ and finally , from $61 \times 32)!$ to $2012!$ each term would have a power of $32$ in it, and power of $61$ in $(2013)!$ would be $33$ .

Hence, we conclude that $N = 61(1 + 2 + 3 + \dots 32) + 33 = 32241$ .

Hence last three digits of $N$ are $\fbox{241}$

$\displaystyle SF(2013) = \prod_{i = 1}^{2013}i!$ $\displaystyle =\left(\prod_{i = 61 \times 0}^{61 \times 1 - 1} i!\right)\left(\prod_{i=61 \times 1}^{61 \times 2 - 1} i!\right)\left(\prod_{i=61 \times 2}^{61 \times 3 - 1} i! \right) \ldots \left(\prod_{i=61 \times 32}^{61 \times 33 - 1} i!\right)\left(\prod_{i=61 \times 33}^{61 \times 33} i!\right)$

Notice that $\displaystyle 2013 = 3 \times 11 \times 61$ and $\displaystyle 61! \pmod{2013} \equiv 0.$ Hence we can see the pattern as below.

$\displaystyle \prod_{i = 61 \times 0}^{61 \times 1 - 1} i! \pmod { 2013^{61 \times 0}} \equiv 0$

$\displaystyle \prod_{i = 61 \times 1}^{61 \times 2 - 1} i! \pmod {2013^{61 \times 1}} \equiv 0$

$\displaystyle \prod_{i = 61 \times 2}^{61 \times 3 - 1} i! \pmod{2013^{61 \times 2}} \equiv 0$

$\vdots$

$\displaystyle \prod_{i = 61 \times 32}^{61 \times 33 - 1} i! \pmod {2013^{61 \times 32}} \equiv 0$

$\displaystyle \prod_{i = 61 \times 33}^{61 \times 33} i! \pmod {2013^{33}} \equiv 0$

Hence, the maximum value of $N$ is

$N = 61(0 + 1 + 2 + 3 \ldots + 32) + 33 = \frac{32 \times 33}{2} + 33 = 32241$

Hence, the last three digits of $N$ is $\boxed{241}.$

Note that the prime factorization of $2013=3 \cdot 11 \cdot 61$ . If we have $61$ or any of it's multiples in a particular factorial, we will undoubtedly have enough of $3's$ and $11's$ to assist in the gleaning of $2013$ . Hence, the sum becomes equivalent to finding the highest power of $61$ that divide's the expression. The number of powers of $61k$ , that appear in $SF(2013)$ , can be written as $2013-61k+1=2014-61k$ .The previous statement implies the number of appearances of $61's$ multiples. Also, we have $1 \leq k \leq 33$ .

Hence, the max power of $61$ is the sum of all the powers of multiples of $61$ in the given range.

So, our answer is $N= \displaystyle\sum\limits_{k=1}^{33} 2014-61k$ $=2014\cdot 33 - 61\cdot \frac{33 \cdot 34}{2}$

$N \equiv 33{14-(61 \cdot 17)} \pmod{1000}$ $N \equiv 33[14 -(37)] \pmod{1000}$ $N \equiv 33 \cdot -23 \equiv -759 \equiv 241 \pmod{1000}$

Since $61$ is the biggest prime factor of $2013$ , we can only care the number of the multiple of $61$ in $SF(2013)$ .

1. In $n!$ , there are $[\frac{n}{61}]$ numbers of multiple 61. $[n]$ is the floored value of $n$

Some little facts from $1$ :

a. For $n = [0, 60]$ , each of $n$ gives $0$ number of multiple $61$

b. For $n = [61, 121]$ , each of $n$ gives $1$ number of multiple $61$

c. For $n = [61 \times i, 61 \times (i+2) - 1]$ , each of $n$ gives $i$ number of multiple $61$

Resuming from all of those little facts, in $SF(2013)$ , there are:

$61 \times (1+2+3+4+...+32) + 33 = 32241$ numbers of multiple 61

The last 3 digits of it is $\boxed{241}$

The given sequence is a product of factorials till 2013 and since SF(2013) is divisible by 2013^N, thus we have to factorize the number. Upon doing that, the limiting factor would be the highest power of the largest prime number that divides SF(2013) which in this case would be 61. Now, 61 starts from 61! and is present only from 61! till 121! and from 122! onwards there are two 61's present in the factorial. (we do not have to worry about the higher powers of 61 as 61^2 is 3721 which is beyond the range of SF given here. Now 61 occurs once from 61 and goes till 121! that is 61 numbers and from 122! to 182! there are two 61's present again 61 numbers and from 183! onwards till 243! there are 3 61's.... thus we get a series as follows 1 61 + 2 61 + 3 61 +............ +61 32 <---(2nd last term) and 2013 which is 61* 33 but that occurs only once so there is only term that contains 33 61's adding up all we get 32241, thus the last three digits are 241

Since 61 is the largest factor of 2013, it grows the fastest and we can count how many 61's there are in the superfactorial, since it is a sort of limiting factor in a sense. Listing the product out: 1! * 2! * 3! .... 61! .... (61 * 2)! .... (61 * 3)! ............ (61 * 33)! All the factorials in the interval [61, 61 * 2) have one factor of 61 in them. This means there are: 1 * 61 factors of 61 contained within this interval. All the factorials in the interval [61, 61 * 3) have two factors of 61 in them. This means there are: 2 * 61 factors of 61 contained within this interval. Continuing on in the same way, we get: 1 * 61 + 2 * 61 + 3 * 61 +.....+ 32 * 61= 61* (1+2+3+...+32)=32208 When we get to the last number, however, there is one extra factor of 61 (2013= 33 * 61) So we add 33 to get a final answer of 32241=> 241

We can divide $2013 = (61\times11\times3)$ in 3 prime numbers. Hence, $2013^N=(616^N\times11^N\times3^N)$

Now, We have to calculate number of 61 and it's multiple in SF(2013). 61 contain once from $2013!$ to $61!$ . No. of 61 contain in $SF(2013)={2013-(1\times61)}+1.$

$2\times61$ contain once from $2013!$ to $122!$ . No. of 122 contain in $SF(2013)={2013-(2\times61)}+1.$ Preceeding this way, we are to calculate upto, $33\times 61$

Hence total number of 61 or its multiple is $={(33\times2013)-61(1+2+3+.....+33)+33}$ $={66429-(61\times17\times33)+33}$ $=32208+33=32241$ Therefore, the answer is $\boxed{241}$

The thing worth noticing, which will make the solution so much simpler, is that

$2013 = 3 \times 11 \times 61$

Therefore, we need to find the maximum exponent in SF(2013) for any of the 3 prime factors of 2013. Obviously, of all 3 primes, 61 will have the smallest exponent. Hence, we calculate the value of the exponent of 63 in SF(2013):

Exponent of $61 = 1953 + 1892 + ... + 62 + 1$ , because we count one exponent value of 61 for every time

$61n | n \in Z$

appears in the factorials.

$1953 + 1892 + ... + 62 + 1 = 33 + 61\sum_{x = 0}^{32} {x}$ , which can easily be computed to $32 241$ .

Hence, the last 3 digits are $\boxed{241}$ .

2013 = 3 * 11 * 61 therefore, we just find the greatest value of 'N' in 61^N.

SF(2013) = prod( i! ), from i=1 to 2013

SF(2013) = (1!) * (2!) * (3!) * . . . * (2013!)

SF(2013) = (1) * (1 * 2) * (1 * 2 * 3) * . . . * (1 * 2 * 3 * . . . * 2013)

SF(2013) = (1^2013) * (2^2012) * (3^2011) * . . . * (2013^1)

SF(2013) = prod( i^(2014-i) ), from i=1 to 2013

Thus, the greatest value of 'N' in 61^N is

N = sum( 2014-i ), i=61k, k=1 to 33

N = 32241

Therefore, the last three digits is 241. :))