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- Feb 14, 2012

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

- Thread starter anemone
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- #1

- Feb 14, 2012

- 3,588

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

- Oct 18, 2017

- 242

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, $S_1\equiv1\pmod{10}$

- Mar 31, 2013

- 1,279

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

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