Towering Product

Let $S=(2^{2^{0}}+1)(2^{2^{1}}+1)...(2^{2^{1024}}+1)$ , As $(2^{2^{0}}-1)=1$ , $S=(2^{2^{0}}-1)(2^{2^{0}}+1)(2^{2^{1}}+1)...(2^{2^{1024}}+1)$ $=(2^{2^{1}}-1)(2^{2^{1}}+1)(2^{2^{2}}+1)...(2^{2^{1024}}+1)$ ... $=(2^{2^{1024}}-1)(2^{2^{1024}}+1)$ $=2^{2^{1025}}-1$ . To evaluate $^{2^{1025}}-1$ ,note that $2^{2^{2}} ≡ 16 (mod100)$ , $2^{2^{3}} ≡ 16*16 ≡ 56 (mod100)$ , $2^{2^{4}} ≡ 56*56 ≡ 36 (mod100)$ , $2^{2^{5}} ≡ 36*36 ≡ 96 (mod100)$ , $2^{2^{6}} ≡ 96*96 ≡ 16 (mod100)$ , This gives us $2^{2^{4k+1}} ≡ 96 (mod100)$ for postive integers $k$ . Then $2^{2^{1025}}-1 ≡ 2^{2^{4(256)}+1}-1 ≡ 96-1 ≡ 95 (mod100)$

First, let $P$ be the product. Multiply $P$ by $(2^{2^{0}} - 1) = 1$ , and using the difference of two squares, we get

$P = (2^{2^{0}} - 1)(2^{2^{0}} + 1)(2^{2^{1}} + 1)...(2^{2^{1024}} + 1)$

$= (2^{2^{1}} - 1)(2^{2^{1}} + 1)(2^{2^{2}} + 1)...(2^{2^{1024}} + 1)$

...

$(2^{2^{1024}} - 1)(2^{2^{1024}} + 1)$

= $2^{2^{1025}} - 1$ .

Next, we proceed by determining the last two digits of $P$ .

First, note that $2^{40} \equiv (2^{10})^4 \equiv (24)^{4} = (576)^{2} \equiv (76)^{2} \equiv 76 \pmod{100}$ . Also, by mathematical induction, we can easily prove that $76^{k} \equiv 76 \pmod{100}$ . Now, we determine $2^{1025} \pmod{40}$ .

$2^{1025} \equiv 0 \pmod{8}$ and $2^{1022} \equiv 4 \pmod{5}$ . So, $2^{1025} \equiv 32 \pmod{40}$ .

Thus, $2^{2^{1025}} \equiv 2^{32} \cdot 76 \equiv (2^{16})^2 \cdot 76 \equiv (65536)^2 \cdot 76 \equiv 36^{2} \cdot 76 \equiv 96 \cdot 76 \equiv 96 \pmod{100}$ .

Thus, the last two digits of $P$ is $96 - 1 = 95$ .

The product, when expanded, contains two raised to all 1024-digit binary numbers. In other words the product can be written as the sum $\sum_{i=0}^{2^{1025}-1} 2^i$ or $2^{2^{1025}}-1 = 4^{2^{1024}}-1.$ We need this huge number modulo 100. Since 4^10=4 mod 100, this is 4^6-1 mod 100 = 95.

We shall use the standard notation for fermat numbers $F_n = 2^{2^n} + 1$ . Then $F_n - 2 = (F_{n - 1} - 1)^2 - 1 \\ F_n - 2 = F_{n - 1}(F_{n - 1} - 2) \\ F_{n} - 2 = F_{n - 1}F_{n - 2}(F_{n - 2} - 2) \\ \vdots$ Hence by mathematical induction $F_n - 2 = \prod_{i = 0}^{n - 1} F_i$ Thus $\prod_{i = 0}^{1024}F_i = F_{1025} - 2 = 2^{2^{1025}} - 1$ Now obviously $2^{2^{1025}} \equiv 0 \pmod{4}$ Also $\phi(25) = 20$ , hence $2^{2^{1025}} \equiv 2^{2^{1025} \mod 20} \pmod{25}$ Now it's easy to prove by induction that $2^{4a + 1} \equiv 12 \pmod{20}$ for $a \in \mathbb{N}$ . Therefore $2^{1025} \equiv 12 \pmod{20}$ and consequently $2^{2^{1025}} \equiv 2^{12} \equiv 21 \pmod{25}$ This means that there exists $x \in \mathbb{N}$ $2^{2^{1025}} = 25x + 21$ , remember that $2^{2^{1025}} \equiv 0 \pmod{4}$ , so $25x + 21 \equiv 0 \pmod{4} \Leftrightarrow x \equiv 3 \pmod{4}$ Hence there exists $y \in \mathbb{N}$ such that $x = 4y + 3$ , hence $2^{2^{1025}} = 25(4y + 3) + 21 = 100y + 96$ That is to say $2^{2^{1025}} - 1 \equiv \boxed{95} \pmod{100}$

Lemma : $(2^{2^0}+1)(2^{2^1}+1)...(2^{2^n}+1)=2^{2^{n+1}}-1$ for all $n \ge 0$ .

Proof :

We would prove this by induction.

Base case: $n=0$ If $n=0$ , we have $(2^{2^0}+1)=3=2^2-1=2^{2^1}-1$ .

Suppose we have $(2^{2^0}+1)(2^{2^1}+1)...(2^{2^n}+1)=2^{2^{n+1}}-1$ .

Then

$((2^{2^0}+1)(2^{2^1}+1)...(2^{2^n}+1))$ $(2^{2^{n+1}}+1)$

$=(2^{2^{n+1}}-1)$ $(2^{2^{n+1}}+1)$

$=2^{2^{n+1} \cdot 2}-1$ $=2^{2^{n+2}}-1$ $=2^{2^{(n+1)+1}}-1$ .

Hence our lemma is proved.

Thus the original expression is equivalent to $2^{2^{1025}}-1$ .

Now we would find the remainder when $2^{2^{1025}}$ is divided by $25$ and when it is divided by $4$ .

We would first find the remainder when $2^{1025}$ is divided by $40$ . Note that $2^2=4 \cong 4 mod 20$ , and $2^4 \cong 16 mod 20$ . Now suppose that $2^{4k+2} \cong 4 mod 20$ . Then $2^{4k+6}=2^{4(k+1)+2} \cong (4)(16) \cong 4 mod 20$ . By mathematical induction, $2^{4k+2} \cong 4 mod 20$ . Hence $2^1022 \cong 4 mod 20$ , and $2^1025 \cong 32 mod 20 \cong 12 mod 20$ .

By Euler's Theorem, since $\phi(25)=20$ , $2^{2^{1025}} \cong 2^12 \cong 4096 \cong 21 mod 25$ .

$2^{2^{1025}}$ is clearly divisible by $4$ .

If a number is congruent to $21 mod 25$ and is divisible by $4$ , then it is congruent to $96 mod 100$ . (We could check by trying $21$ , $46$ , $71$ , and $96$ .

Hence the last $2$ digits of $2^{2^{1025}}$ is $96$ , and the last $2$ digits of $2^{2^{1025}}-1$ is $\boxed{95}$ .

Let A is the expression, we have: A = ( $2^{ 2^{0} }$ - 1 )A = $2^{1025}$ - 1.

The $2^{k}$ (k = 0, 1, 2,..) have remainder mod 25 is 2, 4, {16, 6, 11, 21}

1025 = 1 (mod 4}, so $2^{1025}$ = 21 (mod 25).

Moreover, $2^{1025}$ = 0 (mod 4)

So, $2^{1025}$ = 96 (mod 100)

96 is the answer

temos q isso é igual a 2^(2^1025) - 1 q é congruente a 56^(2^1022) - 1 (mod 100), mas isso é igual a : 55(56^(2^1022-1)+...+1), chamemos (56^(2^1022-1)+...+1) de K, temos q K é congruente a 9 (mod 20), temos q 55K será congruente a 95 (mod 100) o q conclui nossa resposta

We can easily find out through smaller values that $\prod_{i=1}^x (2^{2^x}+1) = (2^{2^{x+1}}-1)$ . Since $2^{1025}$ ends with 32 and $2^{32}$ ends with 96, $\prod_{i=1}^{1024} (2^{2^x}+1) = (2^{2^{1025}}-1)$ ends with 95.

Notice that $2^{2^{0}} + 1 = 3 = 2^2 - 1$ , $(2^{2^{0}} + 1)(2^{2^{1}} + 1) = 15 = 2^4 - 1$ , $(2^{2^{0}} + 1)(2^{2^{1}} + 1)(2^{2^{2}} + 1) = 255 = 2^8 -1$ . Hence, $(2^{2^{0}} + 1)(2^{2^{1}} + 1)(2^{2^{2}} + 1)... (2^{2^{N}} + 1) = 2^{2^{N+1}} - 1$ . Thus, $(2^{2^{0}} + 1)(2^{2^{1}} + 1)(2^{2^{2}} + 1)... (2^{2^{1024}} + 1) = 2^{2^{1025}} - 1$

Note that the last 2 digits of the $2^N$ series repeats for 20 values of N. That is, $2^0 = 01 , 2^1 = 02 , 2^2 = 04 , 2^3 = 08 , 2^4 = 16..., 2^{22} = 04 , 2^{23} = 08 , 2^{24} = 16,...$ etc. In this case, we would note the case of $2^5 = 32$ and $2^{12} = (40)96$

Hence, $2^{1025 (mod 20)} = 2^5 = 12$ implies that $2^{2^{1025}} - 1 = 2^{...12} - 1$ , and thus , $2^{...12(mod20)} - 1 = 2^{12} -1 = 96 - 1 =$ $95$ Thus, the answer is 95

Multiplying above and below by 2^(2^0) - 1 i.e. 1, we reduce the sum into 2^(2^1025) - 1.

Now, 2^1025 = 32mod100

2^(a large number with units and tens digits both zero + 32) = (2^10)^(a large number ending in zero) x 2^32.

The first term ends in 76 as given by number theory. Now, 2^8 = 56mod100 so 2^32 = 56^4mod100 = (50+6)^4mod100 = 4x50x6^3 + 6^4

Computation yields the final answer for 2^2^1025 as 96 (last two digits) and you subtract 1 from it so you get 95.

Note $2^{2^0}+1=2^{2^1}-1$ , and a massive sequential difference of squares will develop. You then reach $2^{2^{1025}}-1$ .Note that $2^2\equiv 2^6\pmod{20}$ and that $2^{2^4}\equiv 2^{2^{22}}\pmod{100}$ , so $2^n$ is $4$ -periodical $\pmod{20}$ and $20$ -periodical $\pmod{100}$ .

Since $1025\equiv 5 \pmod{4}$ (the first power is power is the non-periodical part), our answer is $2^{2^{1025}}-1\equiv 2^{2^5}-1\equiv 2^{12}-1\equiv\ 95\pmod{100}$