Question

Fun problem.

find the units digit of the following ?

Answer 1

\(\large \begin{align} \color{black}{12^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
+13^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
+14^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
+19^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
}\end{align}\)

Answer 2

is it 4 ?

Answer 3

yes it is 4

Answer 4

Other than throwing away the 10s to make it 2^p+3^p+...+9^p I don't really know how else to do this, other than make it a geometric series...

Answer 5

that was like the speed of light!

Answer 6

Well, it isn't a geometric series I take that back lol, I need to get a new equation lol

Answer 7

everything just cancels out in mod10 except two terms

Notice \(S = 1! + 2!+3!+\cdots+10! \equiv 1!+2!+3!\equiv 9\equiv 1\pmod{4}\)
then we have
\[12^1 + 13^1 +14^1+15^1+16^1 + 17^1+18^1 +19^1 \\\equiv 2^1 +3^1+4^1+5^1+(-4)^1 + (-3)^1+(-2)^1+(-1)^1\\\equiv 5^1 - 1^1 \\\equiv 4 \pmod{10}\]

Answer 8

this is very quick !

Answer 9

whats your method ? xD

Answer 10

lol it was very traditional one , outdated in front of this

Answer 11

please im bit curious to see another way to work this.. :)

Answer 12

ok wait

Answer 13

\(\large \begin{align} \color{black}{12^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
+13^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
+14^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
+19^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
=2^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
+3^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
+4^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
\cdot \hspace{.33em}\\~\\
+9^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
\normalsize \text{applying mod 4 to 2,3,7, and 8 as they have cycle of 4}\hspace{.33em}\\~\\
2^{\frac{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}{4}}\hspace{.33em}\\~\\
+3^{\frac{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}{4}}\hspace{.33em}\\~\\
+7^{\frac{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}{4}}\hspace{.33em}\\~\\
+8^{\frac{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}{4}}\hspace{.33em}\\~\\
=2^{1}\hspace{.33em}\\~\\
+3^{1}\hspace{.33em}\\~\\
+7^{1}\hspace{.33em}\\~\\
+8^{1}\hspace{.33em}\\~\\
=20\hspace{.33em}\\~\\
\normalsize \text{applying odd and even property to 4,9 as they have cycle of 2}\hspace{.33em}\\~\\
=4^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
+9^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
=4^{odd+even +even+\cdot\cdot\cdot\cdot+even}\hspace{.33em}\\~\\
+9^{odd+even +even+\cdot\cdot\cdot\cdot+even}\hspace{.33em}\\~\\
=4^{odd}\hspace{.33em}\\~\\
+9^{odd}\hspace{.33em}\\~\\
=4+9=13\hspace{.33em}\\~\\
\normalsize \text{last digit of 5 and 6 are same}\hspace{.33em}\\~\\
=5^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
+6^{1!+2!+3!+\cdot\cdot\cdot\cdot+10!}\hspace{.33em}\\~\\
=5+6=11 \hspace{.33em}\\~\\
\normalsize \text{so}\hspace{.33em}\\~\\
\dfrac{20+13+11}{10}\hspace{.33em}\\~\\
\huge =4

}\end{align}\)

Answer 14

@ganeshie8

Answer 15

nice nice! that looks much bettwe than solving with congruences xD