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The last three digits of a number is equivalent to the number reduced $\mod 1000$ . To compute this, we can seperately consider $\mod 125$ and $\mod 8$ , and combine them at the end.

First notice that $2^{100} \equiv 1 \pmod{125}$ by Euler's Totient Theorem. Therefore, for all integers $100|k$ , $2^k \equiv 1 \pmod{125}$ .

Now notice for all integers $n \ge 4$ , $n! \ge 24$ , so $(n!)!$ , when multiplied out, includes $2,5,10$ , meaning $100|(n!)!$ . Therefore, $\displaystyle\sum_{i=4}^{1000} 2^{(i!)!} \equiv \displaystyle\sum_{i=4}^{1000}1 \equiv 997 \equiv -3 \pmod{125}$

Our sum can therefore be written, in $\mod 125$ as $2^{(1!)!} + 2^{(2!)!} + 2^{(3!)!} + (-3) = 2 + 4 + 2^{720} - 3 = 3 + 2^{720}$

This can be reduced further using our formerly mentioned identity $2^{100} \equiv 1 \pmod{125}$ , giving $3 + 2^{720} = 3 + 2^{20}$ . Now computing this is easy because $2^{10} = 1024 \equiv 24$ , so $3 + 2^{20} \equiv 3 + 24^2 \equiv 579 \equiv 79$

Therefore, our expression is $79 \pmod{125}$

It is evident that all terms in our sum save $2^{(1!)!}$ and $2^{(2!)!}$ are divisible by $8$ , so reducing $\mod 8$ gives $2 + 4 + 0 \equiv 6 \pmod{8}$

We have now deduced $N \equiv 79 \pmod{125}$ and $N \equiv 6 \pmod{8}$ .

The values that are $79 \pmod{125}$ can be expressed as $79 + 125y$ . Taking this $\mod 8$ gives $-1 + 5y$ , which we want to be $6 \pmod{8}$ . Trying values shows that $y = 3$ works, so $N \equiv 79 + 3\cdot 125 = 454 \pmod{1000}$ .

Therefore, our answer is $\boxed{454}$ .

We want the number $N\mod{1000}$ , and $1000=8\cdot 125$ . For $n\geq 3$ , $(n!)!\geq (3!)!=720\geq 3$ so $2^{(n!)!}$ is multiple of $8$ for $n\geq 3$ , and $2^{(1!)!}+2^{(2!)!}=2^1+2^2=6$ , so $N\equiv 6 \mod{8}$ .

Now, as $(125,2)=1$ , we can use Euler to deduce(knowing that $\phi (5^3)=5^3-5^2=100$ ) that $2^{100}\equiv 1\mod{125}$ . We know that $n!\equiv 0\mod{100}$ for $n\geq 10$ , and $n!\geq 10$ for $n\geq 4$ . So for $n\geq 4$ , $2^{(n!)!}\equiv 2^{100\cdot k}\equiv (2^{100})^{k}\equiv 1\mod{125}$ . Together with $2^{(3!)!}\equiv 2^{720} \equiv 2^{20}\equiv (2^{10})^2\equiv (1024)^2\equiv 24^2\equiv 576 \mod{1000}$ and $2^{(1!)!}+2^{(2!)!}=6$ , we got $N\equiv 6+576+997\cdot 1\equiv 1579\equiv 79\mod{125}$ .

So $N\equiv 6 \mod{8}$ and $N\equiv 79\mod{125}$ . So $N=125k+79\equiv 5k+7\equiv 6\mod{8}$ , then $5k\equiv -1$ then $k\equiv 3\mod{8}$ . So $N\equiv 125\cdot 3 +79\equiv 454\mod{1000}$ .

Thus, the last three digits are $454$ .

We begin by computing N modulo 8 and N modulo 125.

We know, $N \equiv 6 (mod 8)$ and $2^{100} \equiv 1(mod 125)$ (from Euler's theorem).

Note that, (4!)! onwards all the powers are divisible by 100.

So, $N \equiv 2+4+2^{720}+997 (mod 125)$ $\implies N \equiv 3+2^{20} (mod 125)$ .

Again, $2^{8} \equiv 6 (mod 125)$ $\implies 2^{20} \equiv 36*16 (mod 125)$ $\implies 2^{20} \equiv 76 (mod 125)$

This means $N \equiv 79 (mod 125)$ .

Also, $GCD(8,125)=1$ , and $8(47)+125(-3)=1$ (from Euclid's Division algorithm).

So, applying Chinese remainder theorem we get, $N \equiv 79*8*47-6*125*3 (mod 1000)$

Thus, $N \equiv 454 (mod 1000)$ .

We attempt to compute this modulo 8 and 125, because we can use CRT to compute this mod 1000. It should be obvious that this is 6 mod 8, because the last 997 terms disappear mod 8. When k >= 4, then k! >= 24. Thus 100 | k, which means that $2^{(k!)!} \equiv 1 \pmod{125}$ . Thus $N \equiv 997 + 2 + 2^2 + 2^720 \equiv 79 \pmod{125}$ . Thus the last three digits of N are 454.

Finding the last three digits is equivalent to determining the remainder modulo 1000. By the Chinese remainder theorem, we can consider modulo 8 and modulo 125 separately.

Modulo 8, we easily see that $2^{(1!)!} \equiv 2$ , $2^{(2!)!} \equiv 4$ and $2^{(k!)!} \equiv 0$ for all $k > 2$ . Therefore, $N \equiv 6 \ (\text{mod } 8)$ .

Modulo 125, by Euler's theorem, we have that $2^{100} \equiv 1$ , since $\phi(125) = 100$ . Therefore, $2^{(k!)!} \equiv 1$ for all $k > 3$ , as the exponent is a multiple of 100. For the smaller cases, we have $2^{(1!)!} \equiv 2$ , $2^{(2!)!} \equiv 4$ and $2^{(3!)!} = 2^{720} \equiv 2^{20} \equiv 76$ . Hence

$N \equiv 2+4+76+997\cdot1 \equiv 1079 \equiv 79 \ (\text{mod } 125)$ .

Combining the two results modulo 8 and 125 (with the Chinese remainder theorem guaranteeing that there is a unique solution), we find that we must have $N \equiv 454 \ (\text{mod } 1000)$ .

1000=125 8 125,8 are coprimes. N=2+4+M(8)... N=6mod8 0r N-6=M(8) since (2,125)=1 so by euler's theorem 2^100=1mod125 in the expression for N from 4th term we see that powers are multiples of100 so they all are 1mod125. N=(2+4+2^720+997) mod125....again 2^720=2^700 2^20=2^20mod125 2^20=76mod125 (2^20=2^14*2^6 and 2^14=(125+3)^2) so N=2+4+76+997mod125=79mod125 or N-79=M(125) our goal is to express N-k=0mod8 and N-k=0mod125... so we find the least x and y satisfying 125x+79=6+8y..we get x=3 and y=56.... we get N-454=0mod8 and 0mod125 this means that N-454=0mod1000(125,8 are coprimes) N=454mod1000 so 454 are the last 3 digits

2^((1!)!) = 2^1 = 2 2^((2!)!) = 2^2 = 4 2^((3!)!) = 2^6! = 2^720 2^((4!)!) = 2^24! = 2^(a multiple of 400)

Now - the key here becomes the cycle of powers of 2 mod 1000. Since φ(1000) = (5^3 - 5^2) * (2^3 - 2^2) = 100 * 4 = 400, the longest possible cycle for powers of 2 mod 1000 is 400.

So the answer will be the last three digits of the first three terms:

2+4+ (2^720 mod 1000)

We can quickly find that: 2^10 = 1024 ≡ 24 mod 1000 2^20 ≡ 24^2 ≡ 576 mod 1000 ... If you continue like this, you'll find that 2^n repeats mod 1000 for every 100 values of n, and 2^100 ≡ 376 mod 1000. So:

2+4+ 2^720 + 997*376 ≡ 6 + 2^20 ≡ 6 + 576 + 872 mod 1000

≡ 454 mod 1000

And the last three digits are 454.

We need to take mod 1000 so we can take mod 125 and mod 8 and then use the chinese remainder theorem. Also, $\phi(125) = 100$ so we see that when $n \ge 4$ $2^{(n!)!} \equiv 2^0 \equiv 1\pmod{125}$ by Euler's Theorem (note that when $n \ge 4$ , we have $n! \ge 24$ which will give $5^2$ and $2^2$ as factors). Also, we know that $(n!)! > 3$ for all $n > 4$ so $2^{(n!)!} \equiv 0\pmod{8}$ Thus we get $2^{(n!)!} \equiv 376 \pmod{1000}$ by the Chinese Remainder Theorem. Since there are $997$ terms such that $n > 4$ so our answer must be $2^1 + 2^2 + 2^{720} + 997\cdot 376.$ We can use Euler's Theorem again on $2^{720}$ to get $2^{720} \equiv 2^{20} \equiv (2^8)^2\cdot 2^4 \equiv 6^2\cdot 2^4 \equiv 576 \pmod {125}$ Also $2^{720} \equiv 576\pmod8\implies 2^{720} \equiv 576 \pmod{1000}$ . Now we have the answer. $2 + 4 + 576+ 997\cdot 376 \equiv 454 \pmod{1000}$

$2^{1!)!}$ = $2^{1}$ = 2 2^(2!)! = 2^2 =4 2^(3!)! = 2^720 2^(4!)!=2^24!=2^(a multiple of 400)

We are taking 1000 since we want last three digits of the no. 2^100=376 mod 1000 2^200 = 376 mod 1000 2^300= 376 mod 1000 2^400 = 376 mod 1000 ................ 1000 as factors 5^3,2^3,1 The last three digits will repeat after (5^3-5^2)*(2^3-2^2)= 400

Hence after 2^(4!)! the last three digits will always be 376 since each and every no. after 24! is a multiple of 400.

Therefor the answer:- 2+4+(last three digits of 2^720)+ (last three digits of 997*376) 997 since leaving the first three terms (1000-3) 997 terms are remaining

Since 720 is divisible of 24 therefore last three digits of 2^24 is 576 (2^24=576 mod 1000) therefor last three digits of 720 is also 576 997*376=374872 last three digits 872 2+4+576+872 1454 answer=454

This problem's answer is given by $\left(\sum_{i=1}^{1000} 2^{\left(i!\right)!}\right) \mod{1000}$ . Now notice that $2^{100n} \equiv 376 \mod{1000}$ for all $n>0$ . Also, since $\left(n!\right)! \equiv 0 \mod{100}$ for all $n>0$ , the answer would be the last three digits of $376\left(1000-4+1\right) + \sum_{i=1}^3 2^{\left(i!\right)!}$ . This can easily be computed to be $454$ .

We begin by simplifying the exponent of some of the first few terms. The first four terms will then be: 2 + 2^2 + 2^6! + 2^24! 2^n*100 will always have 376 as its last three digits. Since, 10! has zero as its two last digits, any k! in which k is greater than 10 will yield a 376 as last three digits when it is raised to two. Thus, the terms starting from 2^(4!)! will have 376 as its last three digits.

the only problem now is the last three digits of 2^6! 2^6! = 2^720, we can write this in a different way, that is, 2^6! = (2^700)(2^10)(2^10) 2^700 will have a last three digits of 376. and 2^10 has a 024 as its last three digits. 376 24 24 has a last three digits of 576. Now, replacing all the terms with the last three digits we get, we will have 2 + 4 + 576 + 997(376) by getting 997(376) mod 1000, we will obtain 872. We can now replace 997(376) by 872 from our equation. 2 + 4 + 576 + 872, adding them all and getting its mod 1000, we will have 454.

We will utilize the Euler's formula: $a^{\varphi(m)} \equiv 1 \bmod m$ when $\gcd(a,m) = 1$ .

First, $\varphi(125) = 125 \cdot \dfrac{4}{5} = 100$ , by the formula of Euler's totient function . So $2^{100} \equiv 1 \pmod{125}$ , and so $2^{100n} \equiv (2^{100})^n \equiv 1^n \equiv 1 \pmod{125}$ for all positive integer $n$ . Also, $2^{n} \equiv 0 \pmod{8}$ for all $n \ge 3$ ; in particular, $2^{100n} \equiv 0 \pmod{8}$ for all positive integer $n$ . So, by the Chinese remainder theorem we get $2^{100n} \equiv 376 \pmod{1000}$ for all positive integer $n$ .

Now, when $n \ge 4$ , $n! \ge 24$ , so $(n!)!$ contains the factors $10, 20$ and so $(n!)!$ is divisible by $100$ . So for all $n \ge 4$ , $2^{(n!)!} = 2^{100k}$ for some $k$ . We invoke the result we proved in the second paragraph, so $2^{(n!)!} \equiv 376 \pmod{1000}$ for all $n \ge 4$ .

Next, we figure out the remaining three values:

• $2^{(1!)!} = 2^{1!} = 2^1 = 2$
• $2^{(2!)!} = 2^{2!} = 2^2 = 4$
• $2^{(3!)!} = 2^{6!} = 2^{720} = 2^{700} \cdot 2^{20}$ , so $2^{(3!)!} \equiv 2^{700} \cdot 2^{20} \equiv 376 \cdot 2^{20} \pmod{1000}$ . Also, $2^{20} = 1048576$ is computable by a calculator, so $2^{(3!)!} \equiv 376 \cdot 1048576 \equiv 394264576 \equiv 576 \pmod{1000}$ .

Finally, we just sum up everything:

$2^{(1!)!} + 2^{(2!)!} + 2^{(3!)!} + (2^{(4!)!} + \ldots + 2^{(1000!)!})$

$\equiv 2^{(1!)!} + 2^{(2!)!} + 2^{(3!)!} + (376 + \ldots + 376) \pmod{1000}$

$\equiv 2^{(1!)!} + 2^{(2!)!} + 2^{(3!)!} + (376 \cdot 997) \pmod{1000}$

$\equiv 2 + 4 + 576 + (376 \cdot 997) \pmod{1000}$

$\equiv 2 + 4 + 576 + 374872 \pmod{1000}$

$\equiv 375454 \pmod{1000}$

$\equiv \boxed{454} \pmod{1000}$

...which is the last three digits of $N$ .

The last three digits of $N$ is the remainder in the division of $N$ by $1000.$ We first compute the number $N$ modulo $8$ and the number $N$ modulo $125.$ Since $2^3 \equiv 0 \pmod{8}$ and $(n!)! \equiv 0 \pmod{3}$ for all $n\ge 3$ , it follows that $2^{(n!)!} \equiv 0 \pmod{8}$ for all $n\ge 3$ , so that $N \equiv 2^{(1!)!}+2^{(2!)!} \equiv 6 \pmod 8.$ Now by the Fermat-Euler Theorem, we have $2^{\varphi(125)} \equiv 1 \pmod{125}.$ Since $\varphi(125) = 5^3-5^2 = 100$ , this implies $2^{100} \equiv 1 \pmod{125}.$ Since $(n!)! \equiv 0 \pmod{100}$ for all $n\ge 4,$ we have $2^{(n!)!} \equiv 1 \pmod{125}$ for all $n\ge 4,$ so that $N \equiv 2^{(1!)!} + 2^{(2!)!} + 2^{(3!)!} +\underbrace{(1 + \cdots+1)}_{997} \equiv 3 + 2^{720} \pmod{125}.$ By using the fact that $2^{100} \equiv 1 \pmod{125}$ and $2^{10} = 1024 \equiv 24 \equiv -101 \pmod{125},$ we further reduce $2^{720} \equiv 2^{20} \equiv 101^2 \equiv 10000+200 + 1\equiv 201 \equiv 76 \pmod{125}.$ Hence, $N \equiv 3 + 76 \equiv 79 \pmod{125}.$ From $N \equiv 6 \pmod{8},$ we have $N =6 +8s$ for some $s\in\mathbb{Z}.$ Substituting this into the congruence $N\equiv 79 \pmod{125},$ we have $8s \equiv 73 \pmod{125} \implies s \equiv 73(8^{-1}) \equiv 73\cdot 47 \equiv 56 \pmod{125}.$ That is $s=56+ 125t$ for some $t\in\mathbb{Z}.$ So $N = 6 + 8s = 6 +8(56+125t) =454 + 1000t \equiv 454 \pmod{1000} .$ Hence the last three digits of the integer $N$ is $\boxed{454}.$

Working this equation modulo $8$ , it is clear that except the first two terms the rest are all divisible by $8$ . So, we write $N \equiv 2 + 2^{(2!)} + 0+0...+0 \pmod{8}$ . $\Rightarrow N \equiv 6 \pmod{8}$ .

Now we will work this equation modulo $125$ . Note that $\phi(125)=100$ .

So by the Euler's theorem we can state that $2^{100k} \equiv 1 \pmod{125}$ [ $k$ is any non-negative integer.]

Clearly the term $2^{(4!)!} = 2^{24!}$ and those after it will have powers greater than $100$ .[There are $997$ such terms]

So, we write $N \equiv 2 + 4 + 2^{720} +997(1) \pmod{125}$

Now, using the properties of modular arithmetic we get that $2^{720} \equiv 2^{95} \equiv 76 \pmod{125}$

$\Rightarrow N \equiv 79 \pmod{125}$

So, we have the following system of linear congruence's:

$N \equiv 6 \pmod{8}$

$N \equiv 79 \pmod{125}$

Using the Chinese remainder Theorem we get that $N \equiv 454 \pmod {1000}$

So the last $3$ digits are $\boxed{454}$ .

Let $\mathrm{f}(n) = 2^n \left(\mathrm{mod}\: 1000 \right)$ .

Observe that $\mathrm{f}(100) = (2^{10})^{10} \left(\mathrm{mod}\: 1000 \right)\ = 24^{10} \left(\mathrm{mod}\: 1000 \right) = (2^3 \times 3)^{10} \left(\mathrm{mod}\: 1000 \right)$ . This can be reduced to $(2^{10})^3 \times 3^{10} \left(\mathrm{mod}\: 1000 \right) = (24^3 \times 49) \left(\mathrm{mod}\: 1000 \right) = 376$ .

Also observe that $376^2 = 376 \left(\mathrm{mod}\: 1000 \right)$ . From this, we can deduce that $\mathrm{f}(100k) = 376$ for integer $k$ .

Thus, letting $\mathrm{g}(n) = \mathrm{f}((n!)!)$ we need $N = \sum_{i = 1}^{1000} \mathrm{g}(i)$ . Evidently, for all $i \geq 4$ , $(i!)!$ is some multiple of $100$ and thus $\mathrm{g}(i) = \mathrm{f}((i!)!) = 376$ .

It remains to calculate $\mathrm{g}(1)$ through $\mathrm{g}(3)$ . We can easily compute $\mathrm{g}(1) = 2$ and $\mathrm{g}(2) = 4$ .

We have $\mathrm{g}(3) = 2^{720} \left(\mathrm{mod}\: 1000 \right) = 376 \times 2^{20} \left(\mathrm{mod}\: 1000 \right) = 576$ .

Thus, the last three digits of $N$ are given by $376 \times 997 + 576 + 4 + 2 \:\left(\mathrm{mod}\: 1000 \right)$ . Since we are working modulo $1000$ this can be simplified to $376 \times (-3) + 576 + 4 + 2 = -546 \left(\mathrm{mod}\: 1000 \right)$ , giving us a final answer of $\boxed{454}$ (it is obvious that $N$ is positive).

To find $N \pmod{10^3}$ , it suffices to find $N \pmod{2^3}$ and $N \pmod{5^3}$ .

.

For $k \ge 3$ , we have $k! \ge 6$ , and $(k!)! \ge 6! = 720$ .

Hence, $2^{(k!)!} \equiv 0 \pmod{2^3}$ for all $k \in \overline{3,1000}$ . Therefore:

$N \equiv 2^{(1!)!} + 2^{(2!)!} + 0 + \cdots + 0 \equiv 2^1 + 2^2 = 6 \pmod{8}$ .

.

Since $\phi(5^3) = 4 \cdot 5^2 = 100$ and $\text{gcd}(2,5) = 1$ , we have $2^{100} \equiv 1 \pmod{5^3}$ .

For $k \ge 4$ , we have $k! \ge 24 > 10$ , and thus, $(k!)! \equiv 0 \pmod{100}$ .

Hence, $2^{(k!)!} \equiv 1 \pmod{5^3}$ for all $k \in \overline{4,1000}$ . Therefore:

$N \equiv 2^{(1!)!} + 2^{(2!)!} + 2^{(3!)!} + \underbrace{1 + \cdots + 1}_{997 \ \text{ones}}$ $\equiv 2^1 + 2^2 + 2^{720} + 997$ $\equiv 2^{20} + 3$ $\equiv (2^{10})^2 + 3$ $\equiv 1024^2 + 3$ $\equiv 24^2+3$ $\equiv 579$ $\equiv 79 \pmod{125}$ .

.

Since $N \equiv 6 \equiv 454 \pmod{8}$ and $N \equiv 79 \equiv 454 \pmod{125}$ , by the Chinese Remainder Theorem , we must have $N \equiv 454 \pmod{1000}$ .

Therefore, the last three digits of $N$ are $\boxed{454}$ .

We have to evaluate the sum $N= \sum_{i=1}^{1000} 2^{(i!)!}$ modulo $1000$ , which is equivalent to evaluating it modulo $125$ and modulo $8$ separately. Note that $2^{100} \equiv 1 (mod 125)$ (from Euler's Totient Theorem). Thus, $100$ is the order of $2$ mod $125$ [note that the factors of $100$ are $1$ , $2$ , $4$ , $5$ , $10$ , $20$ , and $50$ , and none of them satisfy the condition]. So if $m$ is a positive integer which is divisible by $100$ , $2^m \equiv 1 (mod 125)$ . Now we shall find the smallest positive integer $p$ such that $(p!)!)$ is a multiple of $100$ . If we can find such $p$ , then we have to evaluate the sum up to $2^{p-1}$ , because for all larger terms $q$ , $100 \mid (q!)!$ , hence $2^{(q!)!} \equiv 1 (mod 125)$ , and the rest of the sum can be evaluated easily. Checking, we find out that $n= 1$ does not work, $n= 2$ does not work, $n= 3$ does not work, but $n= 4$ works! So, $\sum_{i= 4}^{1000} 2^{(i!)!} \equiv \sum_{i= 4}^{1000} 1 \equiv 997 \equiv -3 mod(125)$ . Hence, the total sum is $\equiv 2^{(1!)!} + 2^{(2!)!} + 2^{(3!)!} - 3 \equiv 2 + 4 + 2^{720} - 3 \equiv 3 + 2^{720} mod(125)$ . Again, $2^{100} \equiv 1 (mod 125)$ , so $2^{720} \equiv 2^{20} \equiv (2^{10})^2 \equiv (1024)^2 \equiv 24^2 \equiv 576 mod(125)$ . Hence, the total sum is $\equiv 579 \equiv 79 (mod 125)$ .

Now we have to evaluate this modulo $8$ . Clearly, all terms after $2^{(3!)!}$ are divisible by $8$ , so we have to evaluate only the first $2$ terms, which is $6$ . Hence, $N \equiv 6(mod 8)$ .

From the Chinese Remainder Theorem, $x$ will have only $1$ possible value modulo $1000$ . Checking cases, we find out that $x \equiv 454 (mod 1000)$ .

Since that $(n!)!>3, \forall n>2$ , we have $8|2^{(n!)!}, \forall n>2$ and so $N\equiv 2^{(1!)!}+2^{(2!)!}\equiv 6\mod 8$ .

Now, by the Euler's Theorem, we have $2^{100}\equiv 1\pmod(125)$ since that $mdc(2, 125)=1$ and $\varphi(125)=100$ and it's easy to see that $100|n!, \forall n\ge 10\Rightarrow 100|(n!)!, \forall n\ge 4$ .

Then $2^{(n!)!}\equiv 2^{100k}\equiv 1\pmod{125}, \forall n\ge 4$ and $(*)\quad N\equiv 2^{(1!)!}+2^{(2!)!}+2^{(3!)!}+\underbrace{1+1+\cdots+1}_{997}\pmod {125}$ .

We have $2^{(1!)!}+2^{(2!)!}+2^{(3!)!}\equiv 2+2^{2}+2^{720}\equiv$

$2+2^{2}+2^{20}\equiv 2+4+76\equiv 82 \pmod{125}$ .

So, by the $(*)$ we have $N\equiv 997+82\equiv 79\pmod{125}$ , and how $N\equiv 6\pmod 8$ , solving the two congruences we find $N\equiv 454\pmod{1000}$ .

Thus the last three digits of $N$ are $454$ .

2^100≡376 (mod 1000) ⟺2^100k≡〖376〗^k≡376 (mod 1000) 2^120≡2^100.2^20≡376.576≡576(mod 1000) 2^(1!)!=2≡2 (mod 1000) 2^(2!)!=2^2≡4 (mod 1000) 2^(3!)!=2^120≡576(mod 1000) From h=4 to 1000 we have 2^(k!)!=2^(k.100)≡376 (mod 1000) ⟹∑_(i=1)^1000▒2^(i!)! ≡2+4+576+376×997≡454 mod 1000 So the last three digits of N is 454

We note that if $k \ge 4$ , then $(k!)!$ is divisible by 100. Proof: Note that when $k=4$ then we get $(4!)!=24!$ , which is divisible by 100. Note also that $((k+1)!)!)$ is a multiple of $(k!)!$ , so all of $(k!)!$ where $k \ge 4$ is divisible by 100.

Thus by Euler Theorem and since $\phi(125)=100$ , $2^{100} \cong 1 (mod 125)$ . Hence all of $2^{(k!)!}$ where $k \ge 4$ is congruent to $1 (mod 125)$ . Also, it is congruent to $0 (mod 8)$ , so it is congruent to $376 (mod 1000)$ .

Note that $2^{(1!)!}=2$ and $2^{(2!)!}=4$ , and $2^{(3!)!}=2^{720}$ . Note that $2^{20}=1048576 \cong 576 (mod 125)$ and $2^{100} \cong 1 (mod 125)$ . Thus $2^{720} \cong 576 (mod 125)$ . It is also congruent to $0 (mod 8)$ , thus it is congruent to $576 (mod 1000)$ .

Hence the sum that we need to evaluate is congruent to $2+4+576+(376)(997)=375454$ , which is congruent to $454 (mod 1000)$ . Thus, the last $3$ digits of the sum is $\boxed{454}$ .

By Euler's theorem, we know that for two relatively prime numbers $a$ and $b$ , $a^{\phi (b)} \equiv 1$ (mod 125), where $\phi (b)$ is the Euler totient function, which counts the number of positive integers less than $b$ that are relatively prime to $b$ . Thus $2^{\phi (125)} = 2^{100} \equiv 1$ (mod 125).

Using this, we examine $N$ mod 125. For all integers $n \geq 4$ , $(n!)!$ is divisible by 100, and therefore we know that there is always some integer $k$ for which $2^{(n!)!} = 2^{100k} \equiv 1^k \equiv 1$ (mod 125).

$\therefore N \equiv 2^{(1!)!} + 2^{(2!)!} + 2^{(3!)!} + 997$

$\equiv 2+4+2^{720} + 997 \equiv 3+2^{720} \equiv 3+2^{20}$ (mod 125)

Note that $2^7 = 128 \equiv 3$ (mod 125)

$\therefore 3+2^{20} = 3+2^{14} \times 2^6 \equiv 3+3^2 \times 2^6$

$\equiv 579 \equiv 79$ (mod 125)

Thus we know that $N$ mod $11$ must be in the form $125k +79$ for some integer $0 \leq k < 8$ , and thus must be one of the following: $79, 204, 329, 454, 579, 704, 829, 954$

However, we can also examine $N$ mod $8$ . For all integers $n \geq 3$ , $(n!)! > 8$ , thus $2^{(n!)!} \equiv 0$ (mod 8).

$\therefore N \equiv 2^{(1!)!} + 2^{(2!)!} \equiv 6$ (mod 8)

The only number in the earlier list of numbers that satisfies this is 454.

We want to find $N \pmod{1000},$ so we take each term of $N$ in modulo $8$ and in modulo $125.$ Notice that if $k \ge 4,$ then $(k!)! \ge 24!,$ so $(k!)!$ is divisible by $\phi(125) = 100.$ Then by Euler's Theorem, $2^{(k!)!} \equiv 2^0 = 1 \pmod{125}.$ Clearly $2^{(k!)!} \equiv 0 \pmod 8,$ so for $k \ge 4$ we have the system of congruences $2^{(k!)!} \equiv 1 \pmod{125},$ $2^{(k!)!} \equiv 0 \pmod 8.$ By the Chinese Remainder Theorem, $2^{(k!)!} \equiv 376 \pmod{1000}$ is the only solution in modulo $1000.$ Thus, all terms of $N,$ except for the first three, end in $376.$

Now we compute the last three digits of the first three terms.

• The first term is $2^{(1!)!} = 2^1 = 2.$

• The second term is $2^{(2!)!} = 2^2 = 4.$

• The third term is $2^{(3!)!} = 2^{720}.$ Using Euler's Theorem with modulo $125,$ we have $2^{720} \equiv 2^{20} = 128^2 \cdot 64 \equiv 3^2 \cdot 64 = 576 \equiv 76 \pmod{125}.$ Therefore, $2^{720} \equiv 576 \pmod{1000}.$