# Pre-RMO 2014/2

The first term of a sequence is $2014$. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the $2014^{\text{th}}$ term of the sequence?

This note is part of the set Pre-RMO 2014 Note by Pranshu Gaba
6 years ago

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Let's define this sequence as $a_1, a_2, a_3, ..., a_{2014}, ...$ where $a_1 = 2014$. The sum of the cubes of the digits of 2014 is 73. $a_2 = 73$. The sum of the cubes of the digits of 73 is 370. $a_3 = 370$. The sum of the cubes of the digits of 370 is 370 again. From this, we yield $a_k = a_{k+1}$ for $k \geq 3$. Therefore, $a_{2014} = a_3 = 370$. Therefore, the answer is 370.

- 6 years ago

That is an Armstrong number. :-)

- 5 years ago

nice

- 6 years ago

An elegant solution.

- 5 years, 9 months ago

$1^{st}$ term $=2014$

$2^{nd}$ term $=73$

$3^{rd}$ term $=370$

$4^{th}$ term $=370$

. . .

Same goes on and $2014^{th}$ term $=\boxed{370}$

- 5 years, 2 months ago

First term is 2014,as per the given question the second term should be 2^3+1^3+4^3=73.Similarly third term will be 7^3 + 3^3 = 370.Now the rest of the terms as we go further comes out to be 370. So the ans is 370.

- 6 years ago

370

- 6 years ago

Here I term is 2014. Second term is (2^3+0^3+1^3+4^3)=73. Third term is (7^4+3^3)=370. Now it is clear that ii and iii digits have only two natural no. And now if we sum the cube of digits then it will remain 370. Hence after iii term all the terms of this series will be 370. Hence answer is 370.

- 6 years ago

370

- 6 years ago

Here 2014=2^3+1^3+4^3=73 Second term =7^3+3^3=370 Third term=3^3+7^3=370 Therefore k>=3. Then a=370

- 6 years ago

370

- 6 years ago

370

- 6 years ago

370

- 6 years ago

370

- 6 years ago

370

- 5 years, 12 months ago

370

- 5 years, 11 months ago

370 :).

- 5 years, 10 months ago

370

- 5 years, 2 months ago

the answer is 370 because all the terms after the third term are 370

- 4 years, 3 months ago

370

- 2 years, 2 months ago