When a Fraction Makes a Perfect Power

Note that $9910 = 10^4 - 3^4 - 2^3 - 1$ , hence using the factorizations of $a^3 - b^3$ and $a^2 - b^2$ , we have $\begin{aligned} 17^3 + 108^2 + 9910 & = 17^3 + 108^2 + 10^4 - 3^4 - 2^3 - 1 \\ & = (17^3 - 2^3) + (108^2 - 1) + (10^4 - 3^4) \\& = (17 - 2)(17^2 + 17 \cdot 2 + 2^2) + (108 - 1)(108 + 1) + (10^2 - 3^2)(10^2 + 3^2) \\& = (15)(327) + (107)(109) + (91)(109) \\& = 109(15 \cdot 3 + 107 + 91) \\& = 109 \cdot 243 \\& = 3^5 \cdot 109 \end{aligned}$ Thus, $\frac{17^3 + 108^2 + 9910}{109} = 3^5,$ so $a = 3$ and $b = 5$ , making the answer $a + b = 3 + 5 = 8$ .

$109 \cdot a^b = 17^3 + 108^2 + 9910$

• Note that $17^3 \equiv 17 \equiv -1 , 108 \equiv 0, 9910 = 9909 + 1 \equiv 1 \pmod{3}$

$\begin{aligned} 17^3 + 108^2 + 9910 &= (18-1)^3 + (2^2\cdot 3^3)^2 + 9909 +1 = (2\cdot 3^2 - 1)^3 + 2^4\cdot 3^6 + 367 \cdot 3^3 + 1 \\ &= 2^3\cdot 3^6 - 3\cdot 2^2 \cdot 3^4 + 3 \cdot 2 \cdot 3^2 -\cancel{1 }+ 2^4\cdot 3^6 + 367 \cdot 3^3 +\cancel{1} \\ &= (2^3 + 2^4) 3^6 - 2^2 \cdot 3^5 + (2 +367 ) 3^3 \\ &= (24 \times 3) 3^5 - 4 \cdot 3^5 + (41 \times 3^2 ) 3^3\\ &= (72-4+41) 3^5\\ &= 109 \cdot 3^5 \end{aligned}$

$\therefore a = 3, b=5, a+b = 3+5 = \boxed{8}$

$17^3+108^2+9910=4913+11664+9910=26487$

$\frac{26487}{109}=243$

$243/2/2/2/2/2/2/2/2=2^8=0.949\ldots$

$243/3/3/3/3/3=3^5=\boxed{1}$

$3+5=8$

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$\text{*Author's Notes*:}$

$\text{All calculations were done on a calculator except for the last one}$

$\text{All of this is}$ $LaTeX\text{ed}$