Question

Fool's problem of the day,

Today, it's a cute number theory problem,

Can you find the last digit \( 77{\uparrow\uparrow}9 \)?

[Solved by @KingGeorge]

Notation Reference: \( \large \begin{matrix}
a\uparrow\uparrow b & = {\ ^{b}a} = & \underbrace{a^{a^{{}^{.\,^{.\,^{.\,^a}}}}}} &

\\

& & b\mbox{ copies of }a
\end{matrix} \)

Good luck!

Answer 1

what does the double up arrow signify?

Answer 2

\[\Large 77 \uparrow \uparrow 9=77^{77^{77^{...^{77}}}}\]Where there are 9 "77's" in the stack. Since we're looking mod for the last digit, we need to take mod 10.

Answer 3

knowing FFM I'm sure we need to make use of primes factors somewhere: 77 = 7 * 11

Answer 4

11 raised to any power will always have last digit equal to 1

Answer 5

so we just need to concentrate on the 7 I would guess

Answer 6

@asnaseer: I love primes! But the general solution i have for any \( x{\uparrow\uparrow}a\) doesn't uses prime factorization.

Answer 7

FFM: you have a GENERAL solution for \(x{\uparrow\uparrow}a\) !!! you MUST be from another planet! :D

Answer 8

This is equivalent to asking \[\Large 7^{7^{7^{...^7}}} \mod 10\] with 9 7's. Since \(7^4 \equiv 1 \mod 10\), we need to find \[7 \uparrow\uparrow8 \mod 4\]Then, we know that \(7 \equiv -1 \mod 4\), so we're finding \[\Large -1^{3^{3^{...^3}}} \mod 4\]Since 3 to any power is odd, this is equivalent to -1.

This leaves us with \(7^{-1} \mod 10\) This is easy to calculate, and is equivalent to 3.

Answer 9

@KingGeorge: What about the general problem?

Answer 10

What level of math is this?

Answer 11

In general, I'll have to think about it some more. I've got class now, so I'll return later hopefully with a solution.

btw, I am correct in thinking 3 was the answer right?

Answer 12

Yes you are.

Answer 13

looks like out of my league

Answer 14

i have no experience with modular arithmatic, am i allowed to have mod functions within other functions?

Answer 15

Back now.

While I'm not positive about getting a formula for \(a\uparrow\uparrow b \mod 10\) for all \(a, b\), I have noticed some striking patterns.

First pattern (trivial):\[a \uparrow\uparrow 1\equiv a \mod 10\]
Second pattern: If the \(\gcd(a, 10)=1\) and \(b \geq2\),\[a\uparrow\uparrow b\equiv a^{-1} \mod 10\]
Pattern three: If \(a=5\) or \(a=6\)\[a\uparrow\uparrow b =a \mod 10\]
When a equals the powers of 2 however, it gets a little funky. For 2, \[2 \uparrow\uparrow b=5+(-1)^{b-1} \mod 10 \qquad\quad b\geq2\]So it alternates between 4 and 6. When \(a=4=2^2\), \[4 \uparrow\uparrow b \equiv 5+(-1)^{2(b-1)}\equiv5+1\equiv6 \mod 10 \qquad\quad b\geq2\]Finally, if \(a=8=2^3\), we have that \[8 \uparrow\uparrow b= 5+(-1)^{3(b-1)} \equiv 5-(-1)^{b-1} \mod10\qquad\quad b\geq2\]So it alternates between 4 and 6 in the opposite manner of \(a=2\).

Answer 16

It should also be noted that \(a\uparrow\uparrow b\) can be defined recursively as\[\Large a\uparrow\uparrow b=a^{a\uparrow\uparrow (b-1)}\]