Last 5 digits of Huge Number

[Guest Mark Orwell]

Find the last 5 digits of a number of the form

 

9^(9^(9^(9^.....9^(9^(9))....)))

 

for 1001 9's.

 

ie.powers on top of each other, if you follow

[Guest tiou-tiou]

Hi,

 

The last 4 digits only (may be generalized with some courage)

 

u(n) = 9^u(n-1) with u(1)=9.

 

We are looking for u(1001) mod 10000.

9^250 = 1 mod 1000

9^50 = 1 mod 250

9^10 = 1 mod 50

9^even_number = 9 mod 10

 

So u(1001) = 9^(u(1000) mod 250) = 9^(9^(u(999) mod 50) = 9^(9^(9)) mod 10000

 

9^9=239 mod 250 so u(3) = 9^11 mod 10000 = 9609 mod 10000

 

So the last 4 digits are 9609.

 

[jordan]

Impressive, but would you be able to explain that to someone like me with only a background of around calc II?

[Alexander]

You know, I'm in Calculus II also. This is the first time I have seen anyone use the concept of modulo. I am a junior in high school now, and I would expect that questions like these would appear on AIME/USAMO tests. Where/when in the sequence of math courses does someone usually encounter the uses of modulo (and these types of problems)? Would this be an analytic number theory course?

[ydoaPs]

how are you taking calc 2 in high school? you lucky *******.

[dryga]

You know, I'm in Calculus II also. This is the first time I have seen anyone use the concept of modulo. I am a junior in high school now, and I would expect that questions like these would appear on AIME/USAMO tests. Where/when in the sequence of math courses does someone usually encounter the uses of modulo (and these types of problems)? Would this be an analytic number theory course?

The concepts of modulo and congruences is textbook discrete mathematics. I got to take a (rather shallow) course on it in high school, and I took it as an elective in my first year of college (I think I took it at the same time as Calc II, actually). It was a very fun course, a lot of first-years took it since it didn't require a lot of knowledge before taking it. And group theory/rings/fields etc is still the most aesthetically pleasing math I've seen

 

I don't know if that answers anything. Anyway, I hadn't heard of the AIME tests or anything before, but fwiw I looked at a few sample questions and it seems to be almost only discrete math stuff. If you have the opportunity I highly recommend taking a course on it

[Alexander]

Dual-Enrollment. My high school only offers through AP Calculus so I have to take my math classes at a local community college. Thanks for your help everyone.

[Guest tiou-tiou]

Impressive, but would you be able to explain that to someone like me with only a background of around calc II?

 

Sorry for very late answer.

 

Finding the last digit of a number N is finding N mod 10.

 

9 = 9 mod 10

9^1 = 81 = 1 mod 10

 

From this we can say that the last digit of 9^(10^(10!) + 1) (just to take a huge power of 9) is 1, without having to do any computations.

 

We can also say that : 9^N mod 10 = 9^(N mod 2 + 1) mod 10

 

I hope it's a bit clearer.

[jordan]

9^1 = 81 = 1 mod 10

I assume you meant 9^2 and not 9^1?

 

From this we can say that the last digit of 9^(10^(10!) + 1) (just to take a huge power of 9) is 1, without having to do any computations.

That's quite a step from the last sentence. But basicly it says that when you're adding up the columns after multiplying, the ones column will always be a one because 9^2=81.

 

We can also say that : 9^N mod 10 = 9^(N mod 2 + 1) mod 10

Now you've lost me. I don't understand how those two are equal. And what does mod 2 mean versus mod 20 or mod 1000?