Find the last digit of a series

[anemone]

What is the last digit of $1+2+\cdots+n$ if the last digit of $1^3+2^3+\cdots+n^3$ is 1?

 

[castor28]

[sp]
We have:
\begin{align*}
S_1 &= 1 + \cdots + n = \dfrac{n(n+1)}{2}\\
S_3 &= 1^3 + \cdots + n^3 = \dfrac{n^2(n+1)^2}{4}
\end{align*}
This shows that $S_3 = S_1^2$. Therefore, if $S_3\equiv1\pmod{10}$, then $S_1\equiv\pm1\pmod{10}$.
It is rather obvious that $S_1\equiv S_3\pmod2$.
We may write $S_1\equiv3n(n+1)\pmod5$, since the multiplicative inverse of $2$ is $3$.
We list the value of $S_1$ for $n\equiv0\dots4\pmod5$:
$$
\begin{array}{c|c|c|c|c|c}
n&0&1&2&3&4\\
\hline
S_1\pmod5&0&1&3&1&0
\end{array}
$$
and we see that we cannot have $S_1\equiv-1\pmod5$; therefore, $S_1\equiv1\pmod{10}$
[/sp]

 

[kaliprasad]

Spoiler: my solution

We see that the $1^3+2^3+\cdots+n^3 = (1+2+\cdots+n)^2$

we know that the LHS has last digit 1(given) so $(1+2+\cdots+n)$ has last digit 1 or 9.

$(1+2+\cdots+n) = \frac{n(n+1)}{2} = 10k + m $ say for some k and m

so $n(n+1) = 20 k + 2m$

or $4n(n+1) + 1 = 80k + 8m + 1$
or $(2n+1)^2 = 80k + 8m + 1$

if m =9 then 8m + 1 ends with 3. so the square ends with 3. As no square ends with 3 so m cannot be 9. but 1 is possible

So last digit is 1 that is for n of the form 5k +1