With the widespread use of calculators and computers doing the actual calculations for us so often, it's hard to come up with a mathematical puzzle that actually is difficult to do. It strikes me, that to do so requires that we find something that cannot be simply done with those tools. Here's a try.

What is the last digit of this number?

2,452,081,165,297 ^ 284,518,006,713

That's one big number "A" to the power of another big number "B", for those of you unfamiliar with the conventional use of the ^ symbol - we're multiplying A times itself B times here.

The result is no doubt a huge number, but we just want to know what is its last digit.

zero?

I figured out a way to solve the puzzle I think, but im not sure how to do it on paper.


nevermind :D


I did find a page that shows you how to solve this problem on paper, but Im not good enough with math to do it and it wouldnt solve this problem.

Nevermind. My lack of sleep made me think it was scientific notation even though you clearly stated otherwise.

Sleep is good...

I will be an odd number. I am guessing 1

Todd

Not too hard...

It's 4 more than the first digit

:D

What is the last digit of this number?

2,452,081,165,297 ^ 284,518,006,713

on the first number, only the last digit is interesting (it could have been 7^284,518,006,713 , the result would be the same)

7^0 = 1
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
...

so we can see 4 series :

a) 0, 4, 8, 12... -> last digit = 1
b) 1, 5, 9, 13... -> last digit = 7
c) 2, 6, 10, 14... -> last digit = 9
d) 3, 7, 11, 15... -> last digit = 3

a) & c) only have even numbers, 284,518,006,713 is an odd one

so the result can only be either 7 or 3

Un = U0 + n*r

we're looking for n : n = (Un - U0)/r

for d) Un = 284,518,006,713 U0 = 3 r = 4

-> n = 71129501677.5 (the right answer can't be this one)

for b) Un = 284,518,006,713 U0 = 1 r = 4

-> n = 71129501678 (that's the good answer)

so the last digit of 2,452,081,165,297 ^ 284,518,006,713 is 7


Sorry if there aren't many explanations, but my english isn't good enough to explaint that :)

Fine job, BreizhManu/ I don't quite follow all your work, but you are correct to pick up in the sequences...

Actually, any number when continuously multiplied by itself will yield a last digit the same as its own last digit multipllied by itself the same number of times. So you're right - this puzzle works out just like taking 7 to the same power.

It's also true that taking 7 to increasing powers ends up in a four-digit repeating cycle: 1,7,9,3,1,7,9,3,.. So all you need to do with this puzzle is determine which stage of the cycle you're on with the power shown - in this case, it's back to the original 7.

Turns out that every digit has a comparable 4-digit cycle- and so when you think it through, any number taken to that power (or to any multiple of four, plus one) will have as its final digit the last digit of the base number.

Feel free to experiemnt for yourself...

Originally posted by QuikSand
Feel free to experiemnt for yourself...

Ok.... ;)

I had a math geek who was over here watching the KU-Az St game and he took one look at this and solved it in about 30 seconds and muttered something about modulus math or something like that.

SI

Yup, it's simple modulus math, actually.

Sort of like the way that anyone with a reasonably facile mind can do the old parlor trick of telling the day of the week for any given date. People who don't comprehend modulus math see this as staggeringly impressive... those who do see it as banal.

Wow, I was actually on to something I Just didnt know what to do.

I'm reading a book on computer cryptography and that involves a lot of math like this. One method of encryption involves taking a 150-digit number to a 150-digit power.... using the straightfoward methods (ie, x^3 = x*x*x) to compute that would require all the world's computing power and more time than exists in the history of the universe. Using modular math they can do it pretty quickly and get the info that counts. Pretty interesting stuff with a lot of practical value..