So Many Exponents!

$9^{2n}$ will always end in a $1$ , while $9^{2n+1}$ will always end in a $9$ (for example, $9^4 = 656\underline{1}$ and $9^5 = 5904\underline{9}$ . As this pattern works for the first digit of $9^n$ , the problem asks for the first two digits. Here is the pattern as shown in the first few values in $9^n$ :

$9^1 = \underline{09}$

$9^2 = \underline{81}$

$9^3 = 7\underline{29}$

$9^4 = 65\underline{61}$

$9^5 = 590\underline{49}$

$9^6 = 5314\underline{41}$

$9^7 = 47829\underline{69}$

$9^8 = 430467\underline{21}$

$9^9 = 3874204\underline{89}$

$9^{10} = 34867844\underline{01}$

$9^{11} = 313810596\underline{09}$

As you can see, the pattern starts over again at 9^11, therefore our pattern is of length $10$ . This means that all we have to do is find the first digit of $8^{7^{6^{5}}}$ and we'll have our answer (another way to phrase this is that we need to find some $n$ such that $8^{7^{6^{5}}} ≡ n$ (mod $10$ ) To do that we need to first find the pattern for the first digit of $8^n$ . $8^1 = \underline{8}$ , $8^2 = 6\underline{4}$ , $8^3 = 51\underline{2}$ , $8^4 = 409\underline{6}$ , and $8^5 = 3276\underline{8}$ , making our pattern is of length $4$ . This means we need to find a number $n$ where $7^{6^{5}} ≡ n$ (mod $4$ ). Finding the remainder when dividing by $4$ is really the same as finding the first digit of something; the first digit is really the remainder of something when you divide by 10. $7^1 = 7 ≡ 3$ (mod $4$ ), $7^2 = 49 ≡ 1$ (mod $4$ ), and $7^3 = 343 ≡ 3$ (mod $4$ ). Therefore our pattern is similar to finding the first digit of $9^n$ . If $n$ is odd, then $7^n ≡ 3$ (mod $4$ ), and if $n$ is even then $7^n ≡ 1$ (mod $4$ ). In this case $n = 6^5$ , which is clearly an even number. From here we just work backwards through our problem:

$n ≡ 0$ (mod $2$ ), therefore $7^n ≡ 1$ (mod $4$ )

$n ≡ 1$ (mod $4$ ), therefore $8^n ≡ 8$ (mod $10$ )

So first digit of $8^{7^{6^{5}}}$ is $8$ , and finally the 8th term in our pattern for $9^n$ is $\boxed{21}$ , our answer.

I did this without any number theory whatsoever, just playing around with numbers.

First observation: The last two digits of $9^n$ is repeating like this: 09-81-29-61-49-41-69-21-89-01

There are 10 possibilities, all we need is the last digit of $8^{7^{6^5}}$ to know which one to choose.

The last digit of $8^n$ is repeating like this:

8-4-2-6

Now we need the reminder of $7^{6^5}$ divided by 4 to choose one of these.

100 is divisible by 4, so all we need the last two digits.

The last two digit of $7^n$ is repeating like this:

07-49-43-01

Observe that $6^5 = 2^5 \times 3^5$ is divisible by 4. Therefor $7^{6^5}$ ends with 01. From this we know that $8^{7^{6^5}}$ ends with 8, and finally $9^{8^{7^{6^5}}}$ ends with 21.

By Euler's theorm

$9^{8^{7^{6^{5}}}} \equiv 9^{8^{7^{6^{5}}} \mod \phi(100)} \pmod{100}$

Note that $\phi(100)=100 \left(1-\frac{1}{2} \right) \left(1-\frac{1}{5} \right)=40$

Now we find the remainder of $8^{7^{6^{5}}}$ devided by $40$

Since, $\gcd(8^{7^{6^{5}}}, 40)\ne 1$ , we break $40$ into $5 \times 8$

Trivially,

$8^{7^{6^{5}}} \equiv 0 \pmod{8}$

and, again, by Euler's theorm

$8^{7^{6^{5}}} \equiv 8^{7^{6^{5}} \mod 4} \pmod{5}$

since $\phi(5)=4$

Once again, by Euler's theorm,

$\begin{aligned}7^{6^{5}} &\equiv 7^{6^{5} \mod 2} \pmod{4}\\ &\equiv 7^0 \\&\equiv 1 \end{aligned}$

Substituding this, we have

$8^{7^{6^{5}}} \equiv 8 \equiv 3 \pmod{5}$

By finding the first positive integer $x$ such that

$x\equiv 0 \pmod{8} \\ x\equiv 3 \pmod{5}$

we get

$8^{7^{6^{5}}} \equiv 8 \pmod{40}$

hence,

$\begin{aligned}9^{8^{7^{6^{5}}}} &\equiv 9^8 \pmod{100}\\&\equiv 6561^2 \\&\equiv 61^2 \\&\equiv 3721\\&\equiv\boxed{ 21}\end{aligned}$

Easy Method... We will frame it as (9^{2})^{8}^{7}^{6} .Remember we will also divide the multiplication of exponents by 2 as we have squared the 9.Now we will find the last digit of 8^{7}^{6}^{5}. To find this we would find the remainder of 7^{6}^{5} when divided of 4(cyclicity of 8). This is calculated as (8-1)^{6}^{5} divide by 4.Remainder is 1(very easy....) Now 8^{1}= 8. Now you remember we have to divide the multiplication by 2. So 8÷2= 4. Now we are in a very good situation. We just have to find the last 2 digits of 81^{4} which is 21...

this problem it's an modular arithmetic problem, we can easily take

$6^{5}=7776$

$7^{n} \mod 100 = n \mod 4$ so the last two digits (that is all we need would be 01 )

Observing the cycle in $8^{n}$ we can induce that it is congruent to $8^{1}$ and finnaly conclude doing the same process in $9^n$ we can observe that it would had the same last digits as $9^{8}$ that are 6721

The pattern repeats after the eleventh power of 9. Following to that, the pattern repeats after the 1672th power, because 1672 is equal to 11 multiplied with 152. The last two Digits of the searched number are the same of the 8th power of 9, 21.

I used the Chinese remainder theorem and I considered the number mod 5 and mod 20 . Our number is congruent to 1 mod 20 and mod 5, so the last 2 digits are 21

Consecutive powers of 9 are $9,81,29,61,49,41,69,21,89,1,... \pmod{100}$ , repeating every 10th power. So we need to know what is $8^{7^{6^{5}}} \pmod{10}$ .

Consecutive powers of 8 have end digit $8,4,2,6,8,...$ repeating every 4th power, so it matters what $7^{6^{5}} \pmod{4}$ is.

Consecutive powers of 7 $\pmod{4}$ are $3,1,3,1$ alternating, so 7 to an even power is always 1 (mod 4), and 6^5 is certainly even.

Now working our way back:

$7^{6^{5}}=1 \pmod{4}$

$8^{7^{6^{5}}}=8 \pmod{10}$

so $9^{8^{7^{6^{5}}}} =21 \pmod{100}$

$9^{k}\equiv 1\pmod{10}$ for even k, and $9^{k}\equiv 9\pmod{10}$ for odd k.

Now observing the pattern in the last 2 digits of powers of 9: $\begin{array}{|c|c|} \hline 9^{1} & 09 \\ \hline 9^{2} & 81 \\ \hline 9^{3} & 29 \\ \hline 9^{4} & 61 \\ \hline 9^{5} & 49 \\ \hline 9^{6} & 41 \\ \hline \end{array}$

Starting from $9^{2}$ and looking at only even powers the digit in the 10s column decreases by 2 after each power increase, it will do this until it reaches $0$ and then starts the cycle again, there are 5 things in the cycle.

Now to check if $8^{7^{6^5}}$ is odd or even. Well $5\equiv 1\pmod{2} \implies 7^{6^5}\equiv 1\pmod{2} \\ 8^{7^{6^5}}\equiv 0\pmod{2}$ So the power is even meaning the last digit is a $1$ and we are looking at the even power cycle.

$8^{7^{6^5}}\equiv 4\pmod{5}$ So counting down to the 4th entry in our even powers pattern means the last two digits of $9^{8^{7^{6^5}}}\equiv 21\pmod{100}$

Last two digits are $\boxed{21}$

I went with Euler's little theorem. So I had :

• 6^5 = 7776
• In finding the last two digits of the whole number, I just needed the last digit of 7^7776.
• Luckily, 7776 = 4x1944, so for the last digit, 7^7776 (mod10) = (7^4)^1944(mod10) = 1
• I then found myself with 9^8^1. Calculating it gave me the last two digits.

Given the pattern of last two digits in the powers of 9: $\begin{aligned} 9^1 &= 09 \\ 9^2 &= 81 \\ 9^3 &= ...29 \\ 9^4 &= ...61 \\ 9^5 &= ...49 \\ 9^6 &= ...41 \\ 9^7 &= ...69 \\ 9^8 &= ...21 \\ 9^9 &= ...89 \\ 9^{10} &= ...01 \\ ... \end{aligned}$

and last digit in the powers of 8: 8-4-2-6-8-..., it is easy to see that an odd power of 8 will end up in either 8 or 2. Any positive integral power of 7 is odd. And so, the answer corresponds to either $9^2$ or $9^8$ . There, you can use 2 attempts at maximum to get the write answer!

Last two digits means remainder with 100. And 9 = -91 mod 100.

Also 8^{7}^{6}^{5} is EVEN, so 9^{EVEN} = 91^{EVEN} mod 100

Also noteworthy is that ten's digit of (A1)^{BCD} is obtained by unit digit of A\times D. As (A1)^{BCD} = (10A + 1)^{BCD} = BCD\times 10A + 1.

Here unit digit of 8^{7}^{6}^{5} = 8, and unit digit of 9\times 8 = 2.

Hence the required last two digits are \boxed{21}.

Here 6^5 can be find out..Now 7 has period 4(7,9,3,1)..and we knew that 6^5 is divisible by 4..so last digit of 7^6^5 will be 1..Now 8 also has period 4(8,4,2,6)..So check the divisibility of 7^6^5 by 4, But it is a long process..So If the last digit of 7^6^5 is 1 we are left with 2 options for the last digit of 8^7^6^5=x(Here i suppose the value as 'x' for simplicity) which are 8 or 2..as remainder can be 1 or 3, when it is divided by 4..Now 8^7^6^5~1 mod(4), and so last digit will be 8, as per (8,4,2,6,) period of 8... Now for (10-1)^x the last 2 digits can be find out using Binomial theorem as follows... -(Xc1) 10+1=-(.....8 ) 10+1= (10-8)*10+1=21(Here the last 2 terms of Binomial Expansion is to be considered, as other terms will have more than one 0's at the end)...(Here the property that last digit of 9^2n equals to the [10-(last digit of 11^2n )] is used)..and here 'x' is even integer..

Simple; Take ${ 9 }^{ { 8 }^{ { 7 }^{ { 6 }^{ 5 } } } }(mod,100)=21$