$\large \color{#3D99F6}9^{\color{#20A900}9^{\color{#D61F06}9}}$

What are the last two digits of the number above?

The answer is 89.

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Well done!

The last two digits of the powers of 9 are cyclical with an order of 10. This can easily be shown with a little number crunching (if there is a more elegant way to show this that I am unaware of please share!).

They are (in order and starting with a power of 0): {01, 09, 81, 29, 61, 49, 41, 69, 21, 89}

After this they start to repeat.

Therefore, if k is a specific power of 9, then k mod 10 will tell us the last two digits of $9^{k}$ .

Let $9^{9} = k$

The last 2 digits of $k$ are "89" since 9 mod 10 is 9, so $k$ mod 10 is also 9. We know this because the last digit of a number tells us what it's value mod 10 is.

This means $9^{k}$ has "89" as it's last two digits as well, so the answer is 89.

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In general, the value of $a^b$ mod c repeats every $\phi(c)$ powers.

Samuel Li
- 6 years, 2 months ago

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But that's not in lowest terms. Most have a periodicity of $n$ such that $n|\phi(c)$

Trevor Arashiro
- 6 years, 2 months ago

after we get the order ..or cycle .........we now have 9^(9^9).......the power equal 81...mod 10 gives 1...thus 9 should be the answer !!!!...................well please tell me what's wrong with this logic!!!

Shehanaaz Sk
- 6 years, 2 months ago

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You are computing (9^9)^9, but that is a very different number than 9^(9^9), which is what the problem specified was to be calculated (implied parentheses from precedence rules).

David Moore
- 6 years, 2 months ago

I am also confused with that fact, according to exponents rules (a^b)^c=a^(bc), which in this case is 81, so the answer should be 09, but it isn't so there must be something wrong with this kind of thinking

Javier Naya Hernández
- 6 years, 2 months ago

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This is because $9^{9^{9}} \neq (9^{9})^{9}$

Kevin Klida
- 6 years, 2 months ago

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@Kevin Klida – Thanks, for clarifying :)

Javier Naya Hernández
- 6 years, 2 months ago

$9^{9^{9}} \neq 9^{81}$

$9^{9} = 387420489$ , therefore $9^{9^{9}} = 9^{387420489}$

Kevin Klida
- 6 years, 2 months ago

We use Euler's theorem to solve this problem. The answer we need is simply the residue of the given number modulo $100$ .

Denote by $\phi(n)$ the Euler's totient function. Now,

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

Now, we have $\gcd(9,100)=1$ , so we're free to use Euler's Theorem to get,

$9^{9^9}\equiv 9^{9^9\bmod~40}\equiv 9^{\left((81)^4\cdot 9\right)\bmod~40}\equiv 9^9\equiv 29^3\pmod{100}$

Now, one can use one of two ways:

a) Simply calculate that value since it's fairly easy to compute $29^3$ and find the residue modulo $100$ . (boring)

b) Easily compute the residues modulo $4$ and $25$ and then combine the results using Chinese Remainder Theorem. (interesting)

Either way, we'll get $29^3\bmod 100=89$ and hence it follows that,

$\large 9^{9^9}\equiv 89\pmod{100}$

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Wow , you solved that sum good :3 :D :P

Arnab Mondal
- 6 years, 2 months ago

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Thanks! :3

Prasun Biswas
- 6 years, 2 months ago

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We can use binomial theorem too.

${ 9 }^{ { 9 }^{ 9 } }\quad =\quad { (10-1) }^{ { 9 }^{ 9 } }$ ${ (10-1) }^{ { 9 }^{ 9 } }\quad =\quad { (-1+{ 9 }^{ 9 }\times 10+\frac { { 9 }^{ 9 }\times { (9 }^{ 9 }-1) }{ 2 } \times 100+...) }$

considering only the ten's and unit digit,

${ (-1+{ 9 }^{ 9 }\times 10+...) }$

$={ (-1+({ 10-1) }^{ 9 }\times 10+...) }$

$={ (-1+({ -1+9\times 10) }\times 10+...) }$

$={ (-1-10+900+...) }$

$=89+800+...$

Thus answer is $\boxed{ 89 }$