Question

For integer n>1 the digit at units place in the number
Summation r! +2^2^n from r=0 to 20

Answer 1

@michele_laino

Answer 2

I'm sorry, I don't know this answer :(

Answer 3

I'm not good with discrete mathematics

Answer 4

Alryt ...no problem...

Answer 5

I think that @ganeshie8 will solve that problem

Answer 6

\[5! \mod 10=5 \cdot 2 \cdot 4 \cdot 3 \cdot 1 \mod 10=10(4 \cdot 3 \cdot 1) \mod 10=0(4 \cdot 3 \cdot 1) \\ 0 \\ \text{ so } k! \mod 10=0 \text{ for integer } k \ge 5\]

so if you write out some of the terms you know you have 21 of those 2^(2^n) terms
and you also know you are going to have +0!+1!+2!+3!+4!+0+0+0+....+0
=1+1+2+6+24=4+6+24=30+24
but 30 mod 10 is 0
and 24 mod 10 is 4
so you have your sum simplified to:
\[(4+21 \cdot 2^{2^n} )\mod 10 \\ \]
but 21 mod 10 is 1
so you are actually left with
\[(4+1 \cdot 2^{2^n}) \mod 10 \\ =(4+2^{2^n}) \mod 10\]

now I'm trying to figure out how to find 2^(2^n) mod 10 in general for n greater than 1
but you could cheat in plug in n=2 and apply the mod 10 part afterwards

Answer 7

actually I do see a pattern with repeated squares method

Answer 8

for \(n\gt 1\), we have \(2^n = 4k\)

Answer 9

so \(2^{2^n} = 2^{4k} = 16^k\)

Answer 10

easy to see that \(16^k \equiv 6^k \equiv 6\pmod{10}\)

Answer 11

Oh great !! Thanks @ganeshie8 and @myininaya