A number theory problem by Puneet Pinku

Number Theory Level 3

If $2^x+2^y+2^z = 2336$ , where $x$ , $y$ , and $z$ are positive integers , then find $x+y+z$ .

2 solutions

Brian Charlesworth
Aug 9, 2016

Without loss of generality let $x \le y \le z$ . The given equation can be rewritten as

$2^{x}(1 + 2^{y - x} + 2^{z - x}) = 2336 = 2^{5}*73$ .

By the Fundamental Theorem of Arithmetic, (FTA), we can conclude that $x = 5$ and that

$1 + 2^{y - x} + 2^{z - x} = 73 \Longrightarrow 2^{y - x}(1 + 2^{z - y}) = 72 = 2^{3}*3^{2}$ .

Again by the FTA we can conclude that $y - x = 3 \Longrightarrow y = 8$ and that

$1 + 2^{z - y} = 9 \Longrightarrow 2^{z - y} = 8 = 2^{3} \Longrightarrow z - y = 3 \Longrightarrow z = 11$ .

Thus $x + y + z = 5 + 8 + 11 = \boxed{24}$ .

The exact solution I was expecting for. How do you manage to get the idea after you see the question. ??

Puneet Pinku - 4 years, 10 months ago

Vishwash Kumar Γξω
Oct 13, 2020

Use binary system.

Looking at the binary representation of 2336 works only if the number of set bits is equal to the number of unknowns.

Krutarth Patel - 6 months ago