Premature Termination

Number Theory Level 1

Let there be a sequence $a_1, a_2, a_3 \dots \dots a_{2014}$ . Let $a_1$ be $2014$ . If the next number in the series is the sum of the cubes of the digits of the previous number, what is the $2014^{\text{th}}$ number in the sequence?

The answer is 370.

2 solutions

Abdur Rehman Zahid
Dec 13, 2014

$\color{#3D99F6}{a_2=2^2+0^3+1^3+4^3=8+0+1+64=73\\a_3=7^3+3^3=343+27=370\\a_4=3^3+7^3+0^3=27+343+0=370}$ .So the sum of cubes of the digits of 370 is 370 so the 2014th term is simply $\boxed{370}$ .

370 is simply unlucky for Malaysian. Remember MH370?

Chew-Seong Cheong - 6 years, 4 months ago

These kinds of numbers are also called narcissistic numbers.You can read more about them on Wikipedia.

Abdur Rehman Zahid - 6 years, 6 months ago

Why does the question say numbers and not "the previous number"?

Adam Pet - 6 years, 3 months ago

Yeah , I agree too .

@Mehul Arora Can you please make the edit ?

A Former Brilliant Member - 6 years, 3 months ago

Yeah... It's edited (by some noble soul...)

Mehul Arora - 6 years, 3 months ago

@Mehul Arora – It is not?

Adam Pet - 6 years, 2 months ago

Venkata Karthik Bandaru
Mar 5, 2015

Just identify the pattern. We can see that the sequence is defined as T(n) = 370 for all n greater than or equal to 3. T(1) = 2014 and T(2) = 73.

Premature Termination

The answer is 370.

2 solutions

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