2011 AIME II reflection notes from H

2011 AIME II — Reflection Notes

Correct: 8, 9, 10, 11, 14, 15

We did #10 during our lesson already.
#15 – made a calculation error but corrected it.

Need Review / Unclear:

#12 – did not understand PIE solution (Principle of Inclusion–Exclusion).
#13 – understood the solution, but it was slightly confusing for me.

2023 Mathcounts school sprint more challenging questions -- Thanks to Ani for trying these problems.

#24 – Probability (Unfair Coin)

Gloria has an unfair coin. Let \( p \) be the probability of heads, with \( \tfrac{1}{2} < p < 1 \). When the coin is flipped three times, the probability of getting exactly two heads is twice the probability of getting exactly two tails. Find the probability of tails when the coin is flipped.

Show Solution

Let \( q = 1 - p \).

\[ P(\text{2 heads}) = 3p^2q, \quad P(\text{2 tails}) = 3q^2p. \] Given \( 3p^2q = 2(3q^2p) \Rightarrow p^2q = 2q^2p. \)
Divide by \( pq \) (nonzero) → \( p = 2q \).

Then \(q = 1-2q \)
Hence \( q = = \boxed{\tfrac{1}{3}}. \)

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#26 – Expected Value (Sum of Cubes)

Johnny rolls a fair six-sided die six times and sums the cubes of all results. What is the expected value of this total?

Show Solution

Let \( X_i \) be the result of the \( i^\text{th} \) roll. We want \( E[T] = E[X_1^3 + X_2^3 + \cdots + X_6^3] \).

By linearity of expectation: \[ E[T] = 6 \, E[X^3]. \]

\[ \displaystyle E[X^3] = \frac{1^3+2^3+3^3+4^3+5^3+6^3}{6} \]
\[ = \frac{(1+2+3+4+5+6)^2}{6} = \frac{21^2}{6} = \frac{441}{6} = 73.5. \] Therefore E[T] = 6 times 73.5 = 441 -- the answer.

(This uses the identity \( \sum_{k=1}^{n} k^3 = \big(\sum_{k=1}^{n} k\big)^2 \).)

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#30 – Powers and Estimation

Find the integer \( n \) such that \[ n^7 = 44{,}231{,}334{,}895{,}529. \]

Show Solution

Since \(80^7 \approx 2.1\times10^{13}\) and \(90^7 \approx 4.78\times10^{13}\), \(n\) is between 80 and 90. The last digit of \(n^7\) matches the last digit of \(n\), which is 9. Modulo-9 check also gives \(n \equiv 8 \pmod9\), so the only candidate is \(n=89\).

Indeed, \(89^7 = 44{,}231{,}334{,}895{,}529\). Thus \(n = \boxed{89}.\)

2021 spring AMC 12 B Reflection Notes from H

✅ Correct: 1–21

❌ Wrong / Unanswered:
#22 – Not sure of optimal strategy
#23 – I got stuck with cases
#24–25 – Didn’t attempt

2021 spring AMC 12 A Reflection Notes from H

2021 Spring AMC 12A — Reflection Notes

1–16 right, 15 time sink.
18, 19, 23 also right — review #19 solution.

Incorrect / Unanswered:
17, 20, 21, 22

#24, #25 → did not even look at (during the test)

At our lesson, we talked about solutions on all wrong, left blank except #25.

2016 AMC 12 B Reflection Notes from H

Wrong

#20

• no idea how to approach

#23

• Right approach
• 2 step away from right answer
• right answer → understood

#24

• review using video

#25

• tried to solve
• I liked linear recurrence / characteristic polynomial approach