An algebra problem by Doli Paik
What is the 2018th term of a sequence whose first term is 8161 and every following term is the sum of the cube of the digits of the previous number?
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2 solutions
a 1 a 2 a 3 a 4 a 5 ⋯ a 2 0 1 8 = 8 1 6 1 = 8 3 + 1 3 + 6 3 + 1 3 = 5 1 2 + 1 + 2 1 6 + 1 = 7 3 0 = 7 3 + 3 3 + 0 3 = 3 4 3 + 2 7 + 0 = 3 7 0 = 3 3 + 7 3 + 0 3 = 2 7 + 3 4 3 + 0 = 3 7 0 = 3 7 0 = ⋯ = 3 7 0
8^3+6^3+1^3+1^3=730 7^3+3^3+0^3=370 Similarly the Next term is also 370 as the digits of the Number remains same that is 3,7,0. Thus the answer is 370.