2021 AMC 10A Problem 10

The factors have the form $(3^k + 2^k)$ with $k = 1, 2, 4, \dots, 64$. Multiplying by $(3 - 2) = 1$ does not change the value but starts a chain of differences of squares:
$$(3-2)(3+2) = 3^2 - 2^2,\quad (3^2-2^2)(3^2+2^2) = 3^4 - 2^4,\quad (3^4-2^4)(3^4+2^4) = 3^8 - 2^8,$$
and so on. Each step doubles the exponent, and after absorbing the last factor $(3^{64} + 2^{64})$ we are left with
$$3^{128} - 2^{128}.$$

The answer is $\boxed{\textbf{(C)}\ 3^{128}-2^{128}}$.

Products of the form $(a+b)(a^2+b^2)(a^4+b^4)\cdots(a^{2^n}+b^{2^n})$ always telescope to $\frac{a^{2^{n+1}} - b^{2^{n+1}}}{a - b}$. When $a - b = 1$, the division is free. The exponents double at each stage, so seven factors reach $2^7 = 128$, not $127$.

Check small cases: $(2+3) = 5 = 3^2 - 2^2$, and $(2+3)(2^2+3^2) = 65 = 81 - 16 = 3^4 - 2^4$. The pattern "difference of powers with the next doubled exponent" is clear, and only (C) has that shape.

• Ending at exponent 127 instead of 128, or picking the plus sign 3^128 + 2^128 after losing track of the pattern.
• Assuming $(2+3)(2^2+3^2)\cdots$ behaves like $5 \cdot 5^2 \cdots$ and choosing (E) $5^{127}$; sums of powers do not multiply that way.