Is this even a pattern?

Number Theory Level 4

Let $S_{1}$ represent $1^{2+3}$ and $S_{2}$ represent $2^{3+4}$ and this continues to the infinity. Given that $a_{1}$ is the digit sum of $S_{999}$ and $a_{2}$ is the digit sum of $a_{1}$ and so on and so forth, find the value of $a_{99999999}$

Note

This question is based on a popular number theory question.

The answer is 9.

2 solutions

Trevor Arashiro
Jul 23, 2014

First, the divisibility rule for nine states that all multiples of nine will have a digit sum of a multiple of 9. Thus if the digit sum is divisible by nine, then the digit sum of the digit sum will have to be divisible by nine. And so forth.

Now, we are trying to find $999^{1000+1001}$ . Since 999 is divisible by 9, then by the divisibility rule of 9, the digit sum of all multiples of 999 will have to be divisible by nine and the digit sum of the digit sum will have to be divisible by 9 and so forth. Since all $999^n$ where n is a positive integer are multiples of 999, the rule stated in the previous sentence holds true. Thus the final digit sum of $999^{1000+1001}$ will be either 9 or 0, the only single digit numbers divisible by 9. However, since no positive numbers add up to 0, the final digit sum can only be 9. Thus 9 is our answer.

I could post the proof stating that the 99,999,999th digit sum is enough to reach a single digit number, but I'm far too lazy. However, if anyone would like me to, feel free to ask and I'll be more than happy to post it.

Can you post for me?:)

shivamani patil - 6 years ago

Abhilash Misra
Jan 26, 2015

The resultant of 999^(1000+1001) is obviously a multiple of 9. And all multiples of 9 on adding the digits give result which is again a multiple of 9. So constant adding will ultimately result in a single digit no. , which is 9.

Is this even a pattern?

The answer is 9.

2 solutions

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