Sum of last digits of a sum

Question

Let $d_n$ be the last digit of $S_n$ where $S_n = \left(1 + 2 + 3 + .... + n\right)$. Find the remainder when $\sum_{i=1}^{2017}d_i$ is divided by 1000.

My attempt : I found the following $$S_1 =1 , d_1 =1 \\S_2 =3 , d_2 =3 \\S_3 =6 , d_3 =6 \\S_4 =10 , d_4 =0 \\S_5 =15 , d_5 =5 \\S_6 =21 , d_6 =1 \\S_7 =28 , d_7 =8 \\S_8 =36 , d_8 =6 \\S_9 =45 , d_9 =5 \\S_{10} =55 , d_{10} =5 $$ I was expecting to get a sequence, but did not find any. Please help me to solve this question.

Answer 1

The $S_n$ are the triangular numbers and their last digit is given by

$$\frac{n(n+1)}2\bmod10,$$ which is

$$\left(\frac n2\bmod10\right)((n+1)\bmod10)\bmod10$$ for even $n$ and $$\left(\frac {n+1}2\bmod10\right)(n\bmod10)\bmod10$$ for odd $n$. Hence the sequence has period $20$, with a sum of $70$.

Then the sum

$$\sum_{i=1}^{2017}\frac{n(n+1)}2\bmod10$$

has $100$ complete periods and $17$ remaining terms, $7069$ in total.

Answer 2

Ok So note the following things for $n=20m$

$$d_{n+1}+d_{n+2}...+d_{n+4}=10 \\ d_{n+5}+d_{n+6}...+d_{n+15}=50 \\ d_{n+16}+d_{n+16}...+d_{n+20}=10$$

Can you use this to calculate the result?

1. Note that $n=20m\implies \mod(S_n, 10)=0$. Now $$S_{n+1}=S_n+20m+1 \\ S_{n+1}=1\mod(10) \\ S_{n+2}=3\mod(10) \\ S_{n+3}=6\mod(10) \\ S_{n+4}=0\mod(10)$$
2. Note that $n=20m+15\implies\mod(S_n,10)=0$, so similar to above argument we can find the sequence form $S_{n+5}$ to $S_{n+15}$.
3. Same argument work for $S_{n+16}$ to $S_{n+20}$.

So to answer the question $\sum_1^{2000}d_i=70\times\frac{2000}{20}=7000$, $\sum_{2001}^{2015}d_i=60$, $d_{2016}=6, d_{2017}=3$

Hence $T=\sum_{2001}^{2017}d_i=7069$, $T\%1000=69$

Answer 3

continuing $$0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,1・・・$$

You can find something.

Answer 4

In general, if $f(x)\in\mathbb{Q}[x]$ is an integer-valued polynomial of degree $d$ and $m$ is a positive integer, then $$f\big(n+d!\,m\big)\equiv f(n)\pmod{m}$$ for every $n\in\mathbb{Z}$. Hence, $d!\,m$ is a (not necessarily minimal) period of $\big(f(n)\big)_{n\in\mathbb{Z}}$ modulo $m$. To prove this, it is well known that $$f(x)=\sum_{k=0}^d\,t_k\,\binom{x}{k}$$ for some $t_0,t_1,t_2,\ldots,t_k \in\mathbb{Z}$. Thus, $F(x):=d!\,f(x)$ is a polynomial with integral coefficients. It is also commonly known that, for $g(x)\in\mathbb{Z}[x]$, $g(n+N)\equiv g(n)\pmod{N}$ for any $n\in\mathbb{Z}$ and $N\in\mathbb{Z}_{>0}$. That is, $$F(n+d!\,m)\equiv F(n)\pmod{d!\,m}\,.$$ Dividing both sides by $d!$, noting that $d!\mid F(n)$ and $d!\mid F(n+d!\,m)$, we obtain the desired result.

Now, since $S_n=\dfrac{n(n+1)}{2}$ is an integer-valued polynomial of degree $2$, we see that $S_n$ in modulo $10$ is periodic with period $2!\cdot 10=20$. The rest is just as other users have concluded.