AMC10B 2004!

Probability Level 3

Let 1 , 4 , … 1,4,\ldots and 9 , 16 , … 9,16,\ldots be two arithmetic progressions . The set S S is the union of the first 2004 terms of each sequence. How many distinct numbers are in S S ?


The answer is 3722.

Novril Razenda by Novril Razenda , 21, Indonesia

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1 solution

A = { 1 + 3 k ; 0 ≤ k ≤ 2003 } , B = { 9 + 7 k ; 0 ≤ k ≤ 2003 } ⇒ S = A ∪ B . A = \{ 1 + 3k \text{ ; } 0 \leq k \leq 2003\}, \quad B = \{ 9 + 7k \text{ ; } 0 \leq k \leq 2003\} \Rightarrow S = A \cup B. Now, because of 3 and 7 are coprime, and the last term belonging to A A is 1 + 2003 ⋅ 3 = 6010 1 + 2003\cdot 3 = 6010 A ∩ B = { 16 + 21 k ; 0 ≤ k ≤ 285 } ⇒ A \cap B = \{16 + 21k \text{ ; } 0 \leq k \leq 285\}\Rightarrow ∣ S ∣ = ∣ A ∪ B ∣ = ∣ A ∣ + ∣ B ∣ − ∣ A ∩ B ∣ = 2004 + 2004 − 286 = 3722 |S| = |A \cup B| = |A| + |B| - |A \cap B| = 2004 + 2004 - 286 = 3722

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