Four different numbers $a, b, c$, and $d$ are chosen from the list $-1,-2,-3,-4$, and -5 . The largest possible value for the expression $a^{b}+c^{d}$ is
(A) $\frac{5}{4}$
(B) $\frac{7}{8}$
(C) $\frac{31}{32}$
(D) $\frac{10}{9}$
(E) $\frac{26}{25}$
Consider $a^{b}$ and choose $a$ and $b$ to be two different numbers from $-1,-2,-3,-4$, and -5 .

What is the largest possible value for $a^{b}$ ?

Since $b$ will be negative, we write $a^{b}=\frac{1}{a^{-b}}$ and here $-b>0$.

If $b$ is odd, then since $a$ is negative, $a^{b}$ will be negative.

If $b$ is even, then $a^{b}$ will be positive.

So to make $a^{b}$ as big as possible, we make $b$ even (ie. -2 or -4 ).

Also, in order to make $a^{b}=\frac{1}{a^{-b}}$ as big as possible, we want to make $a^{-b}$ as small as possible, so $a$ should be as small as possible in absolute value.

Therefore, the largest possible value of $a^{b}$ will be when $a=-1$ and $b$ is either -2 or -4 , giving in either case 1 (ie. $(-1)^{-2}=(-1)^{-4}=1$ ).

What is the second largest possible value for $a^{b}$ ?

Again, we need $b$ to be even to make $a^{b}$ positive, and here we can assume that $a \neq-1$.

To make $a^{b}$ as large as possible, then using similar logic, we choose $b=-2$ and $a=-3$, giving $a^{b}=\frac{1}{(-3)^{2}}=\frac{1}{9}$.

Therefore, the two largest possible values for $a^{b}$ are 1 and $\frac{1}{9}$.

Thus, looking at $a^{b}+c^{d}$, since -1 can only be chosen for one of these four numbers, then the largest possible value for this expression is the sum of the largest two possible values for $a^{b}$, ie. $1+\frac{1}{9}=\frac{10}{9}$, which is obtained by calculating $(-1)^{-4}+(-3)^{-2}$.

ANSWER: (D)