There's something special about Ramanujan!
The first term of a sequence is 1729. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 1729-th term of the sequence?
The answer is 370.
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2 solutions
@Chew-Seong Cheong sir I've observed that such recurring solution is there for all numbers. Can u please give a theoretical proof of it?
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Good question. I have checked for natural number a 1 , it appears that there is a closed set of solution { 1 , 1 5 3 , 3 7 0 , 3 7 1 } . But when a 1 = 4 , the solution is cyclical 1 3 3 , 5 5 , 2 5 0 .
I can explain that for large n the a n has to repeat. Because the largest a 1 = 9 9 9 9 and then a 2 = 2 9 1 6 . Therefore, a n > 2 9 1 6 for n > 2 . This means that a n must repeat for n < 2 9 1 7 . But this does not explain why it must be a unique a n .
Nice solution, sir. Did the same way. Upvoted!
- Term 1 = 1729
- Term 2 = 1+343+8+729 = 1081
- Term 3 = 1+512+1 = 514
- Term 4 = 125+1+64 =190
- Term 5 = 1+729 = 730
- Term 6 = 343+27=370
- Term 7 = 343+27=370
The sum continues upto infinite terms but the value remains 370.
Therefore, term 1729 = 370
Let the n t h term be a n , then we have:
\(\begin{array} {} a_1 = 1729 & = 1729 \\ a_2=1+343+8+729 & =1081 \\ a_3=1+0+512+1 & =514 \\ a_4=0+125+1+64 & =190 \\ a_5=0+1+729+0 & = 730 \\ a_6 = 0 + 343 + 27 + 0 & =370 \\ a_7=0+27+343+0 & =370 \\ a_8 = 0 + 27 + 343 + 0 & =370 \\ a_9 = 0 + 27 + 343 + 0 & =370 \end{array} \)
⇒ a n = 3 7 0 for n ≥ 6 , therefore, a 1 7 2 9 = 3 7 0 .