### Simple Divisor Problem - Math Puzzle

**Source:**http://www.math.utah.edu/~cherk/puzzles.html

**Problem:**

Prove that for any natural N, 1000^N - 1 cannot be a divisor of 1978^N - 1

Short and Sweet :)

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this article though :P Most of my friends already read an article that I wrote more than an year back - " Speak Up " Here, inspired by the movie, The Beautiful Mind, I give a mathematical analysis of asking a girl out. Nice time it is. Feb 10. No plans for Feb 14 and I am sure this article makes me look even more geekier and all the more reason for me to believe that I will be alone, yet again. But what the hell, lets do it! Note: This is not an independent analysis. There are many "mathematics sites" which does "similar" analysis. @Consultants, correct me if I am wrong in my estimates. :P Why is there a need to be selective? From the age of 15, I guess there are approximately 3,600 girls I have liked (On average days, I don't see new girls. But going outside, I like about 30 girls. Saying that I go out once

Once again.. Another coin puzzle after this and this (I wonder whether puzzles have anything else other than hats, kings and coins :D) I heard this from Rushabh Sheth (Mech Sophie) who got it from Vivek Jha (Elec Thirdie). There are 100 coins on the table out of which 50 are tail-face up and 50 are head face up. You are blind folded and there is no way to determine which side is up by rubbing, etc. You have to divide the 100 coins in two equal halves such that both have equal number of coins with tails face up. (This obviously implies that the two have equal number of coins with heads face up) Second part: There are 100 coins on the table out of which 10 are tail-face up and 90 are head face up. You are blind folded and there is no way to determine which side is up by rubbing, etc. You have to divide the 100 coins in two halves (not necessarily equal) such that both have equal number of coins with tails face up. Update (02/01/10): Sorry for inducing confusion in the system. :

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially)

The short and sweet

ReplyDeleteansweris that 1000ᴺ-1 always has a higher power of three in its prime factorisation than does 1978ᴺ-1, so the former can never be a divisor of the latter.A longer version goes something like this:

1000ᴺ-1 is a 3N-length string of 9s, so always has factors 9 and 111, and is thus always divisible by 27.

1978ᴺ-1 is only divisible by 27 when N = 9n.

However, when N = 9n, 1000ᴺ-1 is divisible by 243.

1978ᴺ-1 is only divisible by 243 when N = 81n.

But when N = 81n, 1000ᴺ-1 is divisible by 2187.

And so on, and in general:

3ʲ | 1978ᴺ-1 ⇒ N = 3ʲ⁻¹n ⇒ 3ʲ⁺² | 1000ᴺ-1

To understand why N = 3ʲn ⇒ 3ʲ⁺³ | 1000ᴺ-1, consider how factors with digit sum 3 are pulled out as N increases:

9 × 111

9 × 111 × 1001001

9 × 111 × 1001001 × 1000000001000000001

...

As for 3ʲ | 1978ᴺ-1 ⇒ N = 3ʲ⁻¹n, I do not actually have an explicit proof but believe it can be proved from elementary results and half a cup of coffee.

The straightforwardness seems plausible from the fact that Wolfram Alpha can calculate it so easily, even for biggish numbers and, at least at time of writing, has a bug for the case N = 1 that it gives no integer solutions to 1978ˣ (mod 3) = 1.

http://tech.groups.yahoo.com/group/mathforfun/message/10836

ReplyDeletecan we prove by using induction?

ReplyDeleteConsider 1978^N-1. For N=4i+1, one's digit is 7. For 4i+2, it is 3. 4i+3, it is 1. 4i, it is 5. And consider 1000^N-1 which has only either 8 or 0 in the one's place. They never match. Hence proved.

ReplyDelete???? 1000^N-1 is of the form 999 or 999999 or 999999999 .... and not as you have described. Actually I also worked on those lines but the solution would not lie there.

DeleteSolution is there in in the link http://tech.groups.yahoo.com/group/mathforfun/message/10836 using group theory .

ReplyDeleteClickable: http://tech.groups.yahoo.com/group/mathforfun/message/10836.

ReplyDeleteVivek: That link is spot on. Thanks. Everything is done "properly". My comment above has the right idea but the details are a bit dodgy, hehe.

Litu: My instinct is that this is one of those problems where the Math you would need for the inductive step is the same that could be used for the general case more easily.

Srikar: Firstly 1000ᴺ-1 always ends in 9, not "either 8 or 0". Secondly, if I understand you correctly then the premise of your proof is mistaken - a number can divide another number even if their final digits do not match.