A basic power problem.

Question

What is unit digit of $14^{15^{16^{17}}}$? (no brackets given)

This question is simple and it is based on cyclicity, but the confusion arise when I taken this approach: We know that $5$ raised to any natural number will give $5$ in the unit place. Which will make the number odd. And the cyclicity of $4$ is $2$, which is $(4,6)$ so by this concept I got the answer $4$.

But my friend first resolved $14^{15}$ which will give $4$ as unit digit. Again, that unit digit $4$ raised to $16$ will give $6$ as unit place and again $6$ raised to $17$ will return $6$ as unit digit.So by this way answer is $6$.

How do we resolve $a^{b^c}$? Do we find $a^b$ and then raise it to $c$? Or first we resolve $b^c$ and raise $a$ to it?

P.S : This is my first post on stack exchange. Please let me know if I had done anything wrong with description.

Thanks in advance.

One thing you've done wrong is not use MathJax (LaTeX). You haven't explained your notation either, at first I thought "14^15^16^17" meant $14^{15^{16^{17}}}$, but reading on I got to think it might be some bitwise operation? — Henrik supports the community, Dec 19 '18 at 11:08
$a^{b^c}$ or a^b^c is conventionally read as $a^{\left(b^c\right)}$ or a^(b^c) because the alternative of $\left(a^b\right)^c$ can easily be rewritten as $a^{bc}$ — Henry, Dec 19 '18 at 11:08
The question is not about (number-theory) or (cyclic-groups) but about (notations). — Did, Dec 19 '18 at 11:40
@KishanThacker Was your original problem about 14^15^16^17, or about $14^{15^{16^{17}}}$? — Arthur, Dec 19 '18 at 14:00
Possible duplicate of order of operations with many level exponents — Mark S., Dec 19 '18 at 14:38

score 4 · Answer 1 · answered Dec 19 '18 at 11:59

4

Simply, $a^{b^c}$ means you solve for $b^c$ and then raise $a$ to that power. Be sure not to confuse $(a^b)^c$ and $a^{b^c}$. For example:

$$2^{3^2} = 2^9 = 512 \color{red}{\neq (2^3)^2 = 2^6 = 64}$$

answered Dec 19 '18 at 11:59

KM101

7,176

score 2 · Answer 2 · answered Dec 19 '18 at 14:28

Who gave you this problem? I would ask them which interpretation they meant.

From the way you typeset it here, I put it in Wolfram Alpha as 14^(15^(16^17)) and it gave me the last few digits as being 821290624, which is consistent with your conclusion.

But then I put it in as ((14^15)^16)^17, and instead of presenting me with the last few digits, it gave me the first few hundred most significant digits. I could have clicked the "more digits" button, but since it's more than $10^{4676}$, I'm not sure if I have the time to keep clicking "more digits."

So then I tried ((14^15)^16)^17 mod 1000, and it said 176, which is consistent with your friend's conclusion.

Lastly, I put it in Wolfram Alpha without the parentheses and it interpreted that the way you typeset it.

Of course Wolfram Alpha is not the final word on these things. But it does suggest that if the person to posit this problem really meant to give "no brackets," they meant your interpretation to be correct.

score 0 · Answer 3 · answered Dec 19 '18 at 11:11

So basically to cut the long story short how do we resolve a^b^c ? Do we find a^b and the the answer of that is raised to c? Or first we resolve b^c and the answer is raised to a?

Most people who have an opinion uses $b^c$ first, then raise $a$ to the result. Calculating $a^b$, then raise that to the power of $c$ is the same as $a^{bc}$, so it makes sense to let a^b^c mean the other one.

I personally consider a^b^c to be ambiguous and ill-defined.

A basic power problem.

3 Answers3