Difference between revisions of "2021 WSMO Accuracy Round Problems/Problem 7"
Line 12: | Line 12: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Thus, | Thus, | ||
− | <math></math>\begin{align*} | + | <math></math> |
+ | \begin{align*} | ||
2^n+3^n\equiv&\text{ }5\pmod{10}\tag{for <math>n\equiv1,3\pmod{4}</math>}\\ | 2^n+3^n\equiv&\text{ }5\pmod{10}\tag{for <math>n\equiv1,3\pmod{4}</math>}\\ | ||
2^n+3^n\equiv&\text{ }3\pmod{10}\tag{for <math>n\equiv2\pmod{4}</math>}\\ | 2^n+3^n\equiv&\text{ }3\pmod{10}\tag{for <math>n\equiv2\pmod{4}</math>}\\ | ||
Line 18: | Line 19: | ||
\end{align*}<cmath> | \end{align*}<cmath> | ||
This means that | This means that | ||
− | </cmath>\begin{align*} | + | </cmath> |
+ | \begin{align*} | ||
\sum_{n=1}^{100}\left(\sum_{i=1}^{n}r_i\right)&=\sum_{n=4a,1\leq a\leq25}(20a)+\sum_{n=4a+1,0\leq a\leq24}(20a+5)+\sum_{n=4a+2,0\leq a\leq24}(20a+8)+\sum_{n=4a+3,0\leq a\leq24}(20a+13)\\ | \sum_{n=1}^{100}\left(\sum_{i=1}^{n}r_i\right)&=\sum_{n=4a,1\leq a\leq25}(20a)+\sum_{n=4a+1,0\leq a\leq24}(20a+5)+\sum_{n=4a+2,0\leq a\leq24}(20a+8)+\sum_{n=4a+3,0\leq a\leq24}(20a+13)\\ | ||
&=\frac{25\cdot26}{2}\cdot20+\left(\frac{24\cdot25}{2}\cdot20+25\cdot5\right)+\left(\frac{24\cdot25}{2}\cdot20+25\cdot8\right)+\left(\frac{24\cdot25}{2}\cdot20+25\cdot13\right)\\ | &=\frac{25\cdot26}{2}\cdot20+\left(\frac{24\cdot25}{2}\cdot20+25\cdot5\right)+\left(\frac{24\cdot25}{2}\cdot20+25\cdot8\right)+\left(\frac{24\cdot25}{2}\cdot20+25\cdot13\right)\\ |
Revision as of 11:21, 11 July 2022
Problem
Find the value of where
is the remainder when
is divided by 10.
Solution 1
From Fermat's Little Theorem, we find that is periodic in cycles of 4. This means that
for all
Now, we will compute the first 4 values of
Thus,
$$ (Error compiling LaTeX. Unknown error_msg)
\begin{align*}
2^n+3^n\equiv&\text{ }5\pmod{10}\tag{for
}\\
2^n+3^n\equiv&\text{ }3\pmod{10}\tag{for}\\
2^n+3^n\equiv&\text{ }7\pmod{10}\tag{for }\\
\end{align*}
\begin{align*}
\sum_{n=1}^{100}\left(\sum_{i=1}^{n}r_i\right)&=\sum_{n=4a,1\leq a\leq25}(20a)+\sum_{n=4a+1,0\leq a\leq24}(20a+5)+\sum_{n=4a+2,0\leq a\leq24}(20a+8)+\sum_{n=4a+3,0\leq a\leq24}(20a+13)\\
&=\frac{25\cdot26}{2}\cdot20+\left(\frac{24\cdot25}{2}\cdot20+25\cdot5\right)+\left(\frac{24\cdot25}{2}\cdot20+25\cdot8\right)+\left(\frac{24\cdot25}{2}\cdot20+25\cdot13\right)\\ &=325\cdot20+300\cdot20\cdot3+25\cdot(5+8+13)\\ &=6500+6000\cdot3+25\cdot26\\ &=6500+18000+650\\ &=\boxed{25150}\\
\end{align*}$$ (Error compiling LaTeX. Unknown error_msg) ~pinkpig
Solution 2
We will begin by examining and
:
and
From this, we can note that:
We can simplify our sum as follows:
Note that :
~BigKahuna227