Problem 28 · 2019 Math Kangaroo
Stretch
Algebra & Patterns
digit-sumspiral-pattern
The sequence \(a_1, a_2, a_3, \ldots\) starts with \(a_1 = 49\). To find \(a_{n+1}\) for \(n \ge 1\), you add 1 to the digit sum of \(a_n\) and square the result. For example, \(a_2 = (4 + 9 + 1)^{2} = 196\). Find \(a_{2019}\).
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Answer: C — 64
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Hint 1 of 2
Compute terms in order — \(a_1 = 49\), \(a_2 = 196\), \(a_3 = 289\), \(a_4 = 400\), \(a_5 = 25\), … — and watch for a repeating cycle.
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Hint 2 of 2
Once the sequence cycles, locate term 2019 by its position within the cycle.
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Approach: iterate until the sequence settles into a cycle
- \(a_1 = 49\), \(a_2 = (4+9+1)^2 = 196\), \(a_3 = 17^2 = 289\), \(a_4 = 20^2 = 400\), \(a_5 = (4+0+0+1)^2 = 25\), \(a_6 = 8^2 = 64\), \(a_7 = 11^2 = 121\), \(a_8 = (1+2+1+1)^2 = 25\).
- From \(a_5\) on, the values cycle \(25 \to 64 \to 121 \to 25 \to \cdots\) with period 3.
- For \(n \ge 5\) the term depends on \((n-5) \bmod 3\); since \(2019 - 5 = 2014\) and \(2014 \equiv 1 \pmod 3\), we get \(a_{2019} = a_6 = 64\).
- Answer (C) 64.
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