0. A number theory problem by Sompong Chuisurichy

We seek $a, b, c \in \mathbb{Z}_{\gt}$ such that $2^a + 2^b + 2^c = 33554466$ . Without loss of generality, assume $a \le b \le c$ . After factoring out $2^a$ and observing that $33554466 = 2 \cdot 3^3 \cdot 621379$ , this equation becomes $(*) \qquad 2^a (1 + 2^{b-a} + 2^{c-a}) = 2 \cdot 3^3 \cdot 621379 \; .$ There are two possibilities: $a \lt b$ or $a = b$ .

$\underline{\boldsymbol{ a = b :}}$ When $a = b$ , then $(*)$ reduces to: $2^{a+1}(1 + 2^{c-a-1}) = 2 \cdot 3^3 \cdot 621379 \; .$ By assumption, $c\ge a$ . If $c=a$ , then this means $2^a \cdot 3 = 2 \cdot 3^3 \cdot 621379$ , which is not possible for any $a \in \mathbb{Z}_{\gt}$ . If $c=a+1$ , then $2^{a+2} = 2 \cdot 3^3 \cdot 621379$ , which again is not possible. If $c > a+1$ , then $1 + 2^{c-a-1}$ is odd, so $2^{a+1} = 2 \Rightarrow a = 0$ , which is contrary to the given positivity of $a$ . Therefore, $a \ne b$ .

$\underline {\boldsymbol{ a < b :}}$ When $a < b$ , then the factor $1 + 2^{b-a} + 2^{c-a}$ in $(*)$ is odd for any $a, b, c \in \mathbb{Z}_{\gt}$ when $a<b \le c \Rightarrow 2^a = 2 \Rightarrow \boldsymbol{a = 1}.$ Using this value of $a$ in $(*)$ and simplifying and then factoring out $2^{b-1}$ gives:

$(**) \qquad 2^{b-1} (1 + 2^{c-b}) = 3^3 \cdot 621379 - 1 = 16777232 = 2^4 \cdot 1048577 \; .$

There are two possibilities: $b < c$ or $b = c$ . If $b = c$ , then $(**)$ reduces to $2^b = 2^4 \cdot 1048577$ , which is not possible for any $b \in \mathbb{Z}_{\gt}$ . Therefore, $b < c$ . Then the factor $1 + 2^{c-b}$ is odd for any $b, c \in \mathbb{Z}_{\gt}$ when $b<c \Rightarrow 2^{b-1} = 2^4 \Rightarrow \boldsymbol{b = 5}$ . Using this value of $b$ in $(**)$ yields

$1 + 2^{c-5} = 1048577 \Rightarrow 2^{c-5} = 1048577-1 = 1048576 = 2^{20} \Rightarrow \boldsymbol{c=25} \; .$

Therefore, $a + b + c = 1 + 5 + 25 = \boxed{31}$ .

By construction, this solution is unique up to permutations in the ordering of $a, b,$ and $c$ .

It isn't reduced generalization to assume that: $a \le b \le c$ So: $2^a+2^b+2^c=2^a(1+2^{b-a}+2^{c-a})=33554466$ $\Rightarrow a=1$ $2^{b-a}+2^{c-a}=2^{b-a}(1+2^{c-b})$ $\Rightarrow b-a=4, c-b=20$ $b=5, c=25, a+b+c=\boxed{31}$