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Difference between revisions of "1992 OIM Problems/Problem 1"
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then, <math>S_{20k+p}=k\sum_{i=1}^{20}a_i+\sum_{i=1}^{p}a_i</math>
 
then, <math>S_{20k+p}=k\sum_{i=1}^{20}a_i+\sum_{i=1}^{p}a_i</math>
  
−
&\begin{cases}
+
<math>\begin{cases}
 
a_1=\frac{(1)(2)}{2}\text{ mod }10=1\text{ mod }10=1\\
 
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_2=\frac{(2)(3)}{2}\text{ mod }10=3\text{ mod }10=3\\
Line 49: Line 49:
 
a_19=\frac{(19)(20)}{2}\text{ mod }10=190\text{ mod }10=0\\
 
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
 
a_20=\frac{(20)(21)}{2}\text{ mod }10=210\text{ mod }10=0
−
\end{cases}$
+
\end{cases}</math>
  
  

Revision as of 23:39, 13 December 2023

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 $1 \le p \le 20$

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

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

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

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

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

$\begin{cases} a_1=\frac{(1)(2)}{2}\text{ mod }10=1\text{ mod }10=1\\ a_2=\frac{(2)(3)}{2}\text{ mod }10=3\text{ mod }10=3\\ a_3=\frac{(3)(4)}{2}\text{ mod }10=6\text{ mod }10=6\\ a_4=\frac{(4)(5)}{2}\text{ mod }10=10\text{ mod }10=0\\ a_5=\frac{(5)(6)}{2}\text{ mod }10=15\text{ mod }10=5\\ a_6=\frac{(6)(7)}{2}\text{ mod }10=21\text{ mod }10=1\\ a_7=\frac{(7)(8)}{2}\text{ mod }10=28\text{ mod }10=8\\ a_8=\frac{(8)(9)}{2}\text{ mod }10=36\text{ mod }10=6\\ a_9=\frac{(9)(10)}{2}\text{ mod }10=45\text{ mod }10=5\\ a_10=\frac{(10)(11)}{2}\text{ mod }10=55\text{ mod }10=5\\ a_11=\frac{(11)(12)}{2}\text{ mod }10=66\text{ mod }10=6\\ a_12=\frac{(12)(13)}{2}\text{ mod }10=78\text{ mod }10=8\\ a_13=\frac{(13)(14)}{2}\text{ mod }10=91\text{ mod }10=1\\ a_14=\frac{(14)(15)}{2}\text{ mod }10=105\text{ mod }10=5\\ a_15=\frac{(15)(16)}{2}\text{ mod }10=120\text{ mod }10=0\\ a_16=\frac{(16)(17)}{2}\text{ mod }10=136\text{ mod }10=6\\ a_17=\frac{(17)(18)}{2}\text{ mod }10=153\text{ mod }10=3\\ a_18=\frac{(18)(19)}{2}\text{ mod }10=171\text{ mod }10=1\\ a_19=\frac{(19)(20)}{2}\text{ mod }10=190\text{ mod }10=0\\ a_20=\frac{(20)(21)}{2}\text{ mod }10=210\text{ mod }10=0 \end{cases}$


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

See also

https://www.oma.org.ar/enunciados/ibe7.htm

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