If $a=3^{p}, b=3^{q}, c=3^{r}$, and $d=3^{s}$ and if $p, q, r$, and $s$ are positive integers, determine the smallest value of $p+q+r+s$ such that $a^{2}+b^{3}+c^{5}=d^{7}$.
(A) 17
(B) 31
(C) 106
(D) 247
(E) 353
If $a=3^{p}, b=3^{q}, c=3^{r}$, and $d=3^{s}$ and if $p, q, r$, and $s$ are positive integers, determine the smallest value of $p+q+r+s$ such that $a^{2}+b^{3}+c^{5}=d^{7}$.
(A) 17
(B) 31
(C) 106
(D) 247
(E) 353

## Solution

If we rewrite the given expression by substituting we arrive at the new expression
$3^{2 p}+3^{3 q}+3^{5 r}=3^{7 s}$. (This is derived by replacing $a$ with $3^{p}, b$ with $3^{q}$ and so on.)

On the left side we remove the lowest power of 3 (whatever it is), $3^{2 p}\left(1+3^{3 q-2 p}+3^{5 r-2 p}\right)=3^{7 s}$.

Both factors on the left side must be multiples of 3 but $1+3^{3 q-2 p}+3^{5 r-2 p}$ cannot be a multiple of 3 unless $3^{3 q-2 p}$ and $3^{5 r-2 p}$ are both exactly 1 . This means that $2 p=3 q=5 r$ or that the exponents are themselves multiples of 30 , say $30 \mathrm{~m}$.

We now have, $3^{30 m}+3^{30 m}+3^{30 m}=3^{7 s}$

$$
\text { or, } \quad \begin{aligned}
(3)\left(30^{30 m}\right) & =3^{7 s} \\
3^{30 m+1} & =3^{7 s}
\end{aligned}
$$

We are now looking for the smallest integers, $m$ and $s$, such that $30 m+1=7 s$.

If we try $m=1,2,3,4, \ldots$ we find that $m=3$ and $s=13$. Thus $2 p=90, p=45 ; 3 q=90, q=30$, $5 r=90, r=18$ and $7 s=91, s=13$.

From this, $p+q+r+s=45+30+18+13=106$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-276.jpg?height=233&width=900&top_left_y=136&top_left_x=404)