Question

What are the last 2 digits in following expression:
(unit's place and ten's place digit)
9^(8!)
I know a method to find last digit,but i want the simplest way to do it.
Don't know how to find digit at ten's place.
Any good reference for such questions...?
Another example:9^(9^9)

Answer 1

\[9^{8^{7^{6^{5^{4^{3^{2^{1}}}}}}}}\]
or
\[9^{8!}\]

Answer 2

I'm using:
\[\approx\]
as "congruent with" because I didn't find the symbol.
\[9^{n}\approx9^{r}(10)\]
Where r the residue of n/2.
Then the last digit can be 1 and 9.
Because n! is even for all n higher than 1, we now know 9^(8!) last digit is 1.

Answer 3

The method I know is to search the cycle:
1, 9, 81, 29, 61, 49, 41, 69, 21, 89
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
In this case the cycle has 10 elements.
So we procede to find the residue of n/10
As n! is divisible for 10 to all n equal or higher than 5, we know the residue is 0.
So, the last two digits are 01.

Answer 4

\[9^{n} \approx 9^{r} (100)\]
Is the equation in this case.

Answer 5

\[9^{8!} \equiv (-1)^{8!} \equiv1 \ \ \ \ \text{mod} \ 10 \]

Answer 6

Explain it more so that I can understand it more clearly..

Answer 7

water im searching for a good tutorial about modular arithmetic

Answer 8

Oh wow..

Do it for me..

Answer 9

I want to learn this Modular Arithmetic..

Answer 10

and one more thing\[9^{8^{7^{6^{5^{4^{3^{2^{1}}}}}}}}\neq9^{8!}\]

Answer 11

\[\large 9^{8!} = 9^{8 \times 7 \times 6 \times5 \times 4 \times 3 \times 2 \times 1}\]

Answer 12

\[9^{8!}=(((((((((9)^{8})^{7})^{6})^{5})^{4})^{3})^{2})^{2})^{1} \neq 9^{8^{7^{6^{5^{4^{3^{2^{1}}}}}}}}\]
I understood want he meant, but mukushla's observation is true.

Answer 13

oh and the question is last 2 digit not one last digit

Answer 14

\[9^{8!} \equiv ? \ \ \text{mod} \ 100\]

Answer 15

No it wasn't an error, it was 10 because it was the lengh of the cycle xD, not the mod. (if the mod was 10, the lengh of the cycle would be 2).

Answer 16

\[9^{8!} ≡ 9^{0} (100)\]

Answer 17

\[9^{8!}=(9^{10})^{3\times4\times6\times7\times8}\]
\[9^{10}≡01(100)\]
And any number congruent with 01 mod 100 multiplied by himself will be congruent with 01 mod 100.

Answer 18

yeah

Answer 19

last 2 digit is 01

Answer 20

Proof:
n is congruent with 1 mod m, then
n=s*m+1
(s*m+1)(s*m+1)=s^2*m^2+2s*m+1=m(s^2*m+2s)+1

Answer 21

Is this easier??

Oh wow..

I am just looking around...

Ha ha ha..

Answer 22

oh,yes! sorry
i got what u all did.
i also want for
\[9^{8^{7^{6^{5^{4^{3^{2^{1}}}}}}}}\]

Answer 23

and ofcourse,
a good reference.

Answer 24

i am sorry i didn't know u mean do the same thing for 2 of them....i thought u say they are equal

Answer 25

i said that by mistake.
actually i wanted it for first, but simplified it incorrectly.

Answer 26

Thats a behemot number...
2, 4, 8, 6
3, 9, 7, 1
4, 6
5
6
7, 9, 3, 1
8, 4, 2, 6
9, 1
07, 49, 43, 01
It was very nice that 7 cicle was so short.
6^5/4 rest is 0 so the two last digits of 7^(6^...) is 01
08 64 12 96 68 44 52 16 28 24 92 36 88 04 32 56 48 84 72 76
As 100 is divisible for 20, only the 2 last digits of 7^... matters to 8. So last digit of 8^(expresion) is 08. 9^(thatexpresion) then, is congruent with 9^8 (as we previously saw, 9^(n10+k) was congruent with 9^k. So his last two digits are 21.

Answer 27

I don't even know how many digits the number has xD
2 digits 5 cycle is:
25
2 digits 6 cycle is:
36 16 96 76 56
So it's 56, 56 is divisible by 4 so last digits of 7^(last expresion) are 01.
I decided that it was easier to make the entire process than to explain why I assumed that things in a foreign language.

Answer 28

The basic idea:
We have cycles, when the cycle is of 1 number we don't need anything, if this is of 2 numbers we need the last digits, if it's of 2^n*5^m with:
0=<n=<2
0=<m=<2
We need the last two digits
So, when we start with 5 because so it doesn't matter the previous numbers because 5^(4^...) is obviously higher than 5, so his last 2 digits are 25 not 5.
And we repeat using modular algebra, we calculate the last or last two digits as we need to the next cycle.

Answer 29

PD: I hate being so limited in this language.

Answer 30

\[9^{9^9}\equiv ? \ \text{mod} \ 100\]so after this we should remember that \[9^{10}\equiv 1 \ \text{mod} \ 100\]so\[9^{10}\equiv 9\times 9^9 \equiv -99 \ \ \text{mod} \ 100\]\(9^{9}\equiv -11\equiv 89 \ \text{mod} \ 100\) since \(\gcd(9,100)=1\)

on the other hand \(9^9=10k+9\) since \(9^{9}\equiv-1 \ \text{mod} \ 10\)\[9^{9^9}\equiv 9^{10k+9}\equiv 9^{10k}\times9^9\equiv(9^{10})^k \times 9^9 \equiv1\times89 \equiv89 \ \ \text{mod} \ 100\]

Answer 31

as i am new to solving these types of problems, i am having a hard time to clearly understand each step. But i really appreciate the time and explanation u gave for this, @SqueeSpleen and @mukushla
where did u guys learn this ?
any material where i can get this in more lucid and detailed manner starting with simple problems ?

Answer 32

I learned how to work with modular algebra in Algebra II course, to bad I have horrible hand writting so I don't keep my notes after finishing.

Here's a couple of links about modular algebra, the external links (sources) may have some adittional information.
http://en.wikipedia.org/wiki/Modular_arithmetic
http://en.wikipedia.org/wiki/Fermat%27s_little_theorem
http://en.wikipedia.org/wiki/Chinese_remainder_theorem