June 18th, 2019, 03:03 AM  #1 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0  Sum of the digits
Please help my to solve this problem: $N = 1111 ….11$ (a total of 1989 digits). What will be the sum of the digits of $N^2$? My attempt was like: sum of the digits of $N^2=($sum of the digits of $N)^2$ But I was wrong because this will only work for up to 9 digits. I really appreciate your help. 
June 18th, 2019, 08:51 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,554 Thanks: 1403 
I suggest you try squaring smaller numbers such as 11, 111, 1111, etc. until you see the pattern. Having determined the pattern see if you can then prove (probably with induction) that the pattern still applies with 1989 digits. 
June 18th, 2019, 08:56 AM  #3 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
This is exactly what I did and it worked with less than nine 1s but after that I couldn't see any pattern greater than nine 1s

June 18th, 2019, 09:45 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,554 Thanks: 1403 
Toying with this I find that $p(x) = 1 + x + x^2 + \dots + x^n$ $p(x)^2 = 1+2x + 3x^2 + \dots +(n+1)x^n + n x^{n+1} + \dots + 2x^{2n1}+x^{2n}$ The above polynomial evaluated at $x=1$ is the sum you are after. $\left . (p(x))^2 \right_{x=1} = (n+1) + 2\sum \limits_{k=1}^n~k = \\ (n+1) + 2\dfrac{n(n+1)}{2} = (n+1)^2$ I leave you to plug in $n=1989$ 
June 18th, 2019, 10:01 AM  #5 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
I am sorry I don't know what you did. Can you please elaborate it with an example? Let's say N=111, then what would be the sum of the digits of N^2 with your solution?

June 18th, 2019, 10:20 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,554 Thanks: 1403 
3 digits would be $1+x+x^2$ so in this case $n=2$ $(2+1)^2 = 9$ $111^2 = 12321$ whose digits are readily seen to sum to 9. 4 digits includes an $x^3$ term and thus $n=3$ $4^2 = 16$ $1111^2 = 1234321$ whose digits sum to $16$ So the general formula is given $N$ digits the sum of the digits of $(1111\dots 1)^2 = N^2$ 
June 18th, 2019, 10:34 AM  #7 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
When there are $10$ digits then $n=9$ then $(n+1)^2=100$ but $1111111111^2=1234567900987654321$ which the sum of these digits is $82$

June 18th, 2019, 10:57 AM  #8 
Senior Member Joined: Sep 2015 From: USA Posts: 2,554 Thanks: 1403 
Yeah, you're correct. Seems to be working for the polynomials but not for the numbers... I'll get back to you. 
June 19th, 2019, 06:22 PM  #9 
Senior Member Joined: Aug 2008 From: Blacksburg VA USA Posts: 354 Thanks: 7 Math Focus: primes of course 
I suggest listing the first twenty digit sums, and you should spot the pattern. For example, 1 sum is 1 11 sum of (121)=4 111 sum of (12321) = 9 Focus on the change of the digit sum between each term, in sets of nine. If I am correct, the answer falls out directly as As 1989/9=221, 221*( predictable set of digit sum changes) 
June 20th, 2019, 01:46 AM  #10 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
I have made that list before but I couldn't find that "predictable set of digit sum changes". I need to find a general formula that works for any number of "1"s. 

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