1992 OIM Problems/Problem 1

Problem

For each positive integer $n$ , let $a_n$ be the last digit of the number. $1+2+3+\cdots +n$ . Calculate $a_1 + a_2 + a_3 + \cdots + a_{1992}$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

$a_n=\frac{n(n+1)}{2}\text{ mod } 10$

Let $k$ and $p$ be integers with $k \ge 0$ , and $1 \le p \le 20$

$a_{20k+p}=\frac{(20k+p)(20k+p+1)}{2}\text{ mod } 10$

$a_{20k+p}=\frac{20k^2+20k(p+1)+20kp+p(p+1)}{2}\text{ mod } 10$

$a_{20k+p}=\left(10(k^2+k(p+1)+kp)+ \frac{p(p+1)}{2} \right)\text{ mod } 10$

Since $10(k^2+k(p+1)+kp)\text{ mod } 10=0$ , then

$a_{20k+p}=\frac{p(p+1)}{2} \text{ mod } 10 = a_p$

Let $S_n=a_1 + a_2 + a_3 + \cdots + a_n$

Since $a_{20k+p}=a_p$

then, $S_{20k+p}=k\sum_{i=1}^{20}a_i+\sum_{i=1}^{p}a_i$

Now we calculate $a_1$ through $a_{20}$ :

$\begin{cases} a_{1}=\frac{(1)(2)}{2}\text{ mod }10=1\text{ mod }10=1\\ a_{2}=\frac{(2)(3)}{2}\text{ mod }10=3\text{ mod }10=3\\ a_{3}=\frac{(3)(4)}{2}\text{ mod }10=6\text{ mod }10=6\\ a_{4}=\frac{(4)(5)}{2}\text{ mod }10=10\text{ mod }10=0\\ a_{5}=\frac{(5)(6)}{2}\text{ mod }10=15\text{ mod }10=5\\ a_{6}=\frac{(6)(7)}{2}\text{ mod }10=21\text{ mod }10=1\\ a_{7}=\frac{(7)(8)}{2}\text{ mod }10=28\text{ mod }10=8\\ a_{8}=\frac{(8)(9)}{2}\text{ mod }10=36\text{ mod }10=6\\ a_{9}=\frac{(9)(10)}{2}\text{ mod }10=45\text{ mod }10=5\\ a_{10}=\frac{(10)(11)}{2}\text{ mod }10=55\text{ mod }10=5\\ a_{11}=\frac{(11)(12)}{2}\text{ mod }10=66\text{ mod }10=6\\ a_{12}=\frac{(12)(13)}{2}\text{ mod }10=78\text{ mod }10=8\\ a_{13}=\frac{(13)(14)}{2}\text{ mod }10=91\text{ mod }10=1\\ a_{14}=\frac{(14)(15)}{2}\text{ mod }10=105\text{ mod }10=5\\ a_{15}=\frac{(15)(16)}{2}\text{ mod }10=120\text{ mod }10=0\\ a_{16}=\frac{(16)(17)}{2}\text{ mod }10=136\text{ mod }10=6\\ a_{17}=\frac{(17)(18)}{2}\text{ mod }10=153\text{ mod }10=3\\ a_{18}=\frac{(18)(19)}{2}\text{ mod }10=171\text{ mod }10=1\\ a_{19}=\frac{(19)(20)}{2}\text{ mod }10=190\text{ mod }10=0\\ a_{20}=\frac{(20)(21)}{2}\text{ mod }10=210\text{ mod }10=0 \end{cases}$

Using the table above we calculate: $\sum_{i=1}^{20}a_i=70$ ,

Hence, $S_{20k+p}=70k+\sum_{i=1}^{p}a_i$

$S_{1992}=S_{(20)(99)+12}=(70)(99)+\sum_{i=1}^{12}a_i=6930+\sum_{i=1}^{12}a_i$

Using the table above we calculate: $\sum_{i=1}^{12}a_i=54$

Therefore, $S_{1992}=6930+54=6984$

Hence, $a_1 + a_2 + a_3 + \cdots + a_{1992}=6984$

Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. Although I did get the correct answer, I was only given 8 out of 10 points because I didn't prove that the pattern repeats every 20 numbers. I simply listed the calculations for the first 23 to 26 $a_n$ 's and realized that the pattern repeated. Without proof I proceeded to calculate 20 times 99 plus the summation from $a_1$ through $a_{12}$ and despite getting the correct numerical answer, I was not awarded the full points for not sufficient proof. At the competition showing that the pattern repeats every 20 by induction was not sufficient proof. I submitted such proof on this page.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1992 OIM Problems/Problem 1

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