🚫 Parity & Modular Elimination

Elimination: Sum of 50 odd numbers. Even number of odd terms = even result. Eliminate B, D.

Expected Value: 3 choices remaining, EV = 1.33 points.

Answer: A

Sum of first $n$ odd numbers is $n^2$. Here $n = 50$, so $50^2 = 2500$.

Elimination: $2^3 = 8 \equiv 1 \pmod{7}$, so $2^{2020} = 2^{3 \cdot 673 + 1} \equiv 2 \pmod{7}$.

Expected Value: 1 choice remaining, EV = 6 points.

Answer: B

$2^3 = 8 \equiv 1 \pmod{7}$, so $2^{2020} = 2^{3 \cdot 673 + 1} \equiv 2 \pmod{7}$.

Elimination: Test $n = 1$: remainder 2. Test $n = 2$: remainder 0. Different remainders, so cannot be determined.

Expected Value: 1 choice remaining, EV = 6 points.

Answer: E

$n = 1$ gives remainder 2, $n = 2$ gives remainder 0. Different values produce different remainders.

Elimination: Units digit of $7^n$ cycles as $7, 9, 3, 1$ for $n = 1, 2, 3, 4$. Since $2020 \equiv 0 \pmod{4}$, units digit is $1$.

Expected Value: 1 choice remaining, EV = 6 points.

Answer: A

Units digit cycles every 4 powers: $7, 9, 3, 1$. Since $2020 \equiv 0 \pmod{4}$, units digit is $1$.

Elimination: Use formula: $\frac{n(n+1)(2n+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = 385$.

Expected Value: 1 choice remaining, EV = 6 points.

Answer: A

Using sum of squares formula: $\frac{n(n+1)(2n+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = 385$.

Elimination: $3^2 = 9 \equiv 1 \pmod{8}$, so $3^{100} = 3^{2 \cdot 50} = (3^2)^{50} \equiv 1^{50} \equiv 1 \pmod{8}$.

Expected Value: 1 choice remaining, EV = 6 points.

Answer: A

$3^2 = 9 \equiv 1 \pmod{8}$, so $3^{100} = (3^2)^{50} \equiv 1^{50} \equiv 1 \pmod{8}$.

Elimination: Sum of 50 even numbers. All terms even, so result even. Eliminate odd choices if any.

Expected Value: All choices even, need calculation.

Answer: B

Sum of first 50 even numbers: $2(1 + 2 + 3 + \cdots + 50) = 2 \cdot \frac{50 \cdot 51}{2} = 2550$.

Elimination: $5 \equiv -1 \pmod{6}$, so $5^{2021} \equiv (-1)^{2021} \equiv -1 \equiv 5 \pmod{6}$.

Expected Value: 1 choice remaining, EV = 6 points.

Answer: E

$5 \equiv -1 \pmod{6}$, so $5^{2021} \equiv (-1)^{2021} \equiv -1 \equiv 5 \pmod{6}$.

Elimination: Product of 5 odd numbers is odd. Eliminate even choices B, D.

Expected Value: 3 choices remaining, EV = 1.33 points.

Answer: A

$1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 = 945$.

Elimination: $2^4 \equiv 1 \pmod{5}$, so $2^{100} \equiv 2^0 \equiv 1 \pmod{5}$. $3^4 \equiv 1 \pmod{5}$, so $3^{100} \equiv 3^0 \equiv 1 \pmod{5}$. Sum is $1 + 1 = 2$.

Expected Value: 1 choice remaining, EV = 6 points.

Answer: C

$2^4 \equiv 1 \pmod{5}$ and $3^4 \equiv 1 \pmod{5}$, so $2^{100} + 3^{100} \equiv 1 + 1 = 2 \pmod{5}$.