Solve the equation with integers

Since $x < y < z$ , let $y = x+a$ and $z = x + a + b$ . Then $a$ and $b$ are integers $\ge 1$ and the equation becomes:

$\begin{aligned} 5^x + 2 \times 5^{x+a} + 5^{x+a+b} & = 4500 & \small \blue{\text{Divide both sides by }5^x} \\ 1 + 2 \times 5^a + 5^{a+b} & = 36\times 5^{3-x} & \small \blue{\text{Add }-1 \text{ on both sides}} \\ 2 \times 5^a + 5^{a+b} & = 36\times 5^{3-x} - 1 \end{aligned}$

Since the left-hand side is divisible by $5$ , this means that the right-hand side $36 \times 5^{3-x} - 1$ is also divisible by $5$ . The only possibility is $5^{3-x} = 1$ or $x=3$ . Then the equation becomes:

$\begin{aligned} 2 \times 5^a + 5^{a+b} & = 35 & \small \blue{\text{Divide both sides by }5^a} \\ 2 + 5^b & = 7 \times 5^{1-a} \end{aligned}$

Since $a \ge 1$ , the only solution is $a=1$ . Then $2+5^b = 7$ , $\implies b = 1$ . Therefore $x+y+z = 3x + 2a + b = 9+2+1 = \boxed{12}$ .

$5^x+2.5^y+5^z=4500$

$\Leftrightarrow 5^x(1+2.5^{y-x} + 5^{z-x}) = 4500 = 5^3.(2^2.3^2)$

$\Rightarrow \left\{\begin{matrix} 5^x=5^3 \Leftrightarrow x=3 \\ 1+2.5^{y-x} + 5^{z-x}=2^2.3^2=36 = 1 + 35 \end{matrix}\right.$

$\Rightarrow 2.5^{y-x} + 5^{z-x}=35$

$\Leftrightarrow 5^{y-x}(2 + 5^{z-y})=35 = 5.(2+5)$

$\Rightarrow \left\{\begin{matrix} 5^{y-x}=5^1 \Leftrightarrow y-x=1 \Leftrightarrow y=x+1=3+1=4\\ 2 + 5^{z-y} = 2+5^1 \Rightarrow z-y=1 \Leftrightarrow z=y+1=4+1=5 \end{matrix}\right.$

Therefore, we have $x=3,y=4,z=5$ . Comparing the result to the condition $x<y<z$ , we can see that the result satisfies the condition.

Hence, the equation has 1 solution, that is $(x,y,z)=(3,4,5)$ . Therefore the answer is $x+y+z = 3+4+5= \boxed{12}$

$\textit{Just convert 4500 to the base of 5}$

4500 = 000121

• That mean 4500 = $5^3$ + 2× $5^4$ + $5^5$

• So x+y+z=3+4+5= $\boxed{12}$ $\color{#D61F06}{Easy Peasy}$ !