2022 AIME I Problem 4

Attempt Problem 4 of the 2022 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AIME I solutions, or check the answer key.

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4.

Let w=3+i2w = \frac{\sqrt{3} + \mathrm{i}}{2} and z=−1+i32,z = \frac{-1 + \mathrm{i}\sqrt{3}}{2}, where i=−1.\mathrm{i} = \sqrt{-1}. Find the number of ordered pairs (r,s)(r, s) of positive integers not exceeding 100100 that satisfy the equation i⋅wr=zs.\mathrm{i} \cdot w^r = z^s.

Answer: 834
Concepts:complex numberDe Moivre’s Theoremmodular arithmetic
Difficulty rating: 2300
Small Hint:

Both numbers lie on the unit circle: ww has argument 30∘30^\circ and zz has argument 120∘,120^\circ, while i\mathrm{i} has argument 90∘90^\circ

Big Hint:

Matching arguments gives 90+30r≡120s(mod360),90 + 30r \equiv 120s \pmod{360}, i.e. r+3≡4s(mod12);r + 3 \equiv 4s \pmod{12}; count how many r∈[1,100]r \in [1, 100] hit each residue

Solution:

Both ww and zz have modulus 1:1: in polar form w=cis⁡30∘w = \operatorname{cis} 30^\circ and z=cis⁡120∘,z = \operatorname{cis} 120^\circ, while i=cis⁡90∘.\mathrm{i} = \operatorname{cis} 90^\circ. The equation i⋅wr=zs\mathrm{i} \cdot w^r = z^s is therefore a statement about arguments: 90+30r≡120s(mod360),90 + 30r \equiv 120s \pmod{360}, i.e. r+3≡4s(mod12).r + 3 \equiv 4s \pmod{12}.

For each s,s, this determines r(mod12):r \pmod{12}: the residue 4s−34s - 3 is 1,1, 5,5, or 99 modulo 1212 according as s≡1,s \equiv 1, 2,2, or 0(mod3).0 \pmod 3. Among 1≤r≤1001 \le r \le 100 there are 99 values with r≡1(mod12)r \equiv 1 \pmod{12} and 88 values each with r≡5r \equiv 5 or r≡9(mod12).r \equiv 9 \pmod{12}. Among 1≤s≤1001 \le s \le 100 there are 3434 values with s≡1(mod3)s \equiv 1 \pmod 3 and 3333 values in each of the other two classes.

The count is 34⋅934 \cdot 9 +33⋅8+ 33 \cdot 8 +33⋅8+ 33 \cdot 8 =306+264+264=834.= 306 + 264 + 264 = 834.

Problem 3#3
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