Page 1 of 1

BdMO National 2021 Junior Problem 2

Posted: Mon Apr 12, 2021 11:35 am
by Anindya Biswas
নিচের সমীকরণটার সব সমাধানের যোগফল বের করো।
\[5^{2r+1}+5^2=5^r+5^{r+3}\]

Find the sum of all solutions of the equation, \[5^{2r+1}+5^2=5^r+5^{r+3}\]

Re: BdMO National 2021 Junior Problem 2

Posted: Wed Apr 14, 2021 5:00 pm
by gwimmy(abid)
$\begin{align*} &5^{2r+1}+5^2 = 5^r+5^{r+3} \\
\Longrightarrow & 5 \cdot (5^r)^2 + 25 = 5^r + 125 \cdot 5^r\\
\Longrightarrow & 5 \cdot (5^r)^2 + 25 = 126\cdot 5^r \\
\Longrightarrow & 5 \cdot (5^r)^2 - 126\cdot 5^r + 25 = 0 \\
\Longrightarrow & 5 \cdot (5^r)^2 -125\cdot 5^r -5^r + 25 = 0\\
\Longrightarrow & 5 \cdot (5^r) (5^r - 25) - 1 (5^r -25)=0\\
\Longrightarrow & (5^r - 25)(5^r - 1) =0 \\
\Longrightarrow & 5^r = 25\text{ or }5^r = \frac{1}{5}\\
\Longrightarrow & r = \boxed{2} \text{ or } r = \boxed{-1}
\end{align*}$

Re: BdMO National 2021 Junior Problem 2

Posted: Thu Apr 15, 2021 6:04 am
by Marzuq
My solution
It can be easily observable that an equation where both side are sum of two number with same base but different power is not possible unless the base is $-1, 0, 1, 2$

So we will consider the powers are same.
In the equation
$5^{2r+1} + 5^{2} = 5^{r} + 5^{r+3}$
We will split the equation into two case:
Case 1: $2r+1 = r+3$ and $ 2=r$
This two equation can be true at the same time and here $r = 2$

Case 2: $2r + 1 = r$ and $2 = r+3$
This two equation can also be true at the same time and here $r = -1$

So the solution of $r = (2,-1)$
And their sum $= (2) + (-1) = \boxed{1}$
And if there is any error , then inform me. A solution without using algebra :mrgreen:

Re: BdMO National 2021 Junior Problem 2

Posted: Fri Apr 16, 2021 4:05 pm
by gwimmy(abid)
Marzuq wrote:
Thu Apr 15, 2021 6:04 am
My solution
It can be easily observable that an equation where both side are sum of two number with same base but different power is not possible unless the base is $-1, 0, 1, 2$

So we will consider the powers are same.
In the equation
$5^{2r+1} + 5^{2} = 5^{r} + 5^{r+3}$
We will split the equation into two case:
Case 1: $2r+1 = r+3$ and $ 2=r$
This two equation can be true at the same time and here $r = 2$

Case 2: $2r + 1 = r$ and $2 = r+3$
This two equation can also be true at the same time and here $r = -1$

So the solution of $r = (2,-1)$
And their sum $= (2) + (-1) = \boxed{1}$
And if there is any error , then inform me. A solution without using algebra :mrgreen:
i like the solution!