2018 AMC 10A Problem 14

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 $\boxed{\textbf{(A)}\ 80}$.

When a fraction has a dominant term in both numerator and denominator, factor out the dominant ratio (here $81$) and study the leftover. The sign of the leftover tells you which side of $81$ you are on, and a crude size comparison tells you how far. The answer choices $80$ and $81$ are designed to punish anyone who does not check the sign.

Let $t = (2/3)^{96}$, a tiny positive number. Dividing top and bottom by $3^{96}$ gives $\dfrac{81 + 16t}{1 + t}$. Since $16 < 81$, adding $16t$ upstairs and $t$ downstairs pulls the ratio below $81$, and with $t$ minuscule it stays above $80$.

• Ignoring the 2-power terms and answering 3^4 = 81 exactly, when the fraction is strictly less than 81 because 2^100 = 16 2^96 is much smaller than 81 2^96.
• Trying to compute or estimate the powers numerically, which is hopeless by hand; the comparison only needs exponent rules.