2018 AMC 10A Problem 14

Try a small version first to see what happens. Replace the exponents $100$ and $96$ by $4$ and $0$:
$$\frac{3^4 + 2^4}{3^0 + 2^0} = \frac{81 + 16}{1 + 1} = 48.5.$$
Here the $2$-powers matter a lot. But for large exponents the $3$-powers are much, much bigger than the $2$-powers, so the fraction should be close to $\dfrac{3^{100}}{3^{96}} = 3^4 = 81$. The question is whether it is a bit above $81$ or a bit below.

1. Compare with $81$ copies of the bottom. $81$ times the bottom is $81 \cdot 3^{96} + 81 \cdot 2^{96} = 3^{100} + 81 \cdot 2^{96}$.
2. Compare with the top. The top is $3^{100} + 2^{100} = 3^{100} + 16 \cdot 2^{96}$. It has only $16$ copies of $2^{96}$ instead of $81$. So the top is smaller than $81$ times the bottom, and the fraction is less than $81$.
3. How much less? The top falls short by $81 \cdot 2^{96} - 16 \cdot 2^{96} = 65 \cdot 2^{96}$. So
$$\text{fraction} = 81 - \frac{65 \cdot 2^{96}}{3^{96} + 2^{96}}.$$
4. Is that shortfall less than $1$? We need $65 \cdot 2^{96}$ to be smaller than $3^{96} + 2^{96}$, that is, $64 \cdot 2^{96} < 3^{96}$. Now $64 \cdot 2^{96} = 2^{102}$. Write both sides with $48$ equal factors: $2^{102} = 64 \cdot 4^{48}$ and $3^{96} = 9^{48}$. Each $9$ is more than twice the matching $4$, so $9^{48}$ is more than $2^{48}$ times $4^{48}$, far more than $64$ times. So yes, the shortfall is less than $1$.
5. Conclude. The fraction is between $80$ and $81$, so the greatest integer not above it is $80$.

The answer is $\boxed{\textbf{(A)}\ 80}$.

When one kind of term is much bigger than the other, the answer is close to what the big terms alone give. Then the small terms decide the direction: here the top gets a smaller "extra" ($16$ copies) than it would need ($81$ copies) to reach $81$, so the value falls just short. Choices $80$ and $81$ side by side are a warning to check that direction carefully.

Divide the top and the bottom by $3^{96}$ and call $t = \left(\tfrac23\right)^{96}$, a tiny positive number. The fraction becomes
$$\frac{81 + 16t}{1 + t}.$$
If $t$ were $0$ this would be exactly $81$. But $81$ times the bottom is $81 + 81t$, which is more than the top $81 + 16t$, so the value is just below $81$. Because $t$ is so tiny, it is still above $80$.

The $3$-terms dominate, and $3^{100} = 81 \cdot 3^{96}$, so the fraction should be close to $81$. Make that precise by writing the numerator as $81$ copies of the denominator plus a correction:
$$3^{100} + 2^{100} = 81\left(3^{96} + 2^{96}\right) - 81 \cdot 2^{96} + 16 \cdot 2^{96} = 81\left(3^{96} + 2^{96}\right) - 65 \cdot 2^{96}.$$

Dividing by the denominator,
$$\frac{3^{100}+2^{100}}{3^{96}+2^{96}} = 81 - \frac{65 \cdot 2^{96}}{3^{96} + 2^{96}}.$$

The subtracted term is positive, so the value is strictly less than $81$. It is also less than $1$: that needs $65 \cdot 2^{96} < 3^{96} + 2^{96}$, i.e. $64 \cdot 2^{96} = 2^{102} < 3^{96}$, which is true since $3^{96} = 9^{48}$ while $2^{102} = 4^{48} \cdot 2^6$ and $(9/4)^{48}$ dwarfs $64$.

So the value lies strictly between $80$ and $81$, and its floor is $80$.

The answer is $\boxed{\textbf{(A)}\ 80}$.