# The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.

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MTGAP
Daedalian Member

 Posted: Sun Oct 12, 2008 11:19 pm    Post subject: 1 I was thinking the other day, "What is the last digit of [url="http://everything2.com/e2node/Graham%2527s%2520Number"]Graham's number?[/url]" In base three, it would obviously be zero. But it's trickier in base ten. It turns out that 3^3^...^3^3 any number of times always ends in a seven. I also noticed that each digit has a cycle. 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6 = 729. The ones digit repeats. It goes 1, 3, 9, 7. You can see why. 1x3 is 3, 3x3 is 9, 9x3 ends in 7, and 7x3 ends in 1. This type of pattern works for every place value. The tens place has a 40-digit cycle, the hundreds place has a 100-digit cycle, and the thousands place has a 500-digit cycle. I was unable to calculate any cycle above that. So what I'm wondering is, is there a predictable pattern to the increasing values of these cycles? Here's a number that's simpler than three: 5. The ones digit is always 5, the tens is always 2. The hundreds cycles through 1 and 6. The thousands cycles through 3, 5, 8, and 0. The cycles get much longer after that. So what's the pattern?_________________This statement is false.
ralphmerridew
Daedalian Member

 Posted: Mon Oct 13, 2008 12:10 am    Post subject: 2 First, the tens place has a 20-digit cycle. So for the special case of powers of 3, modulo 10^i, the cycle length is a simple GP. More generally, define phi(n) as the number of numbers 1 <= i <= n such that gcf(i,n) = 1. If p_1, p_2, ... p_j are distinct primes, and n = p_1^e_1 * p_2^e_2 * ... * p_j^e_j, then phi(n) can be calculated as phi(n) = (p_1-1)*p_1^(e_1-1) * (p_2-1)*p_2^(e_2-1) * ... * (p_j-1)*p_j^(e_j-1) = n * (1-1/p_1) * (1-1/p_2) * ... * (1-1/p_j). IIRC, the period of any base mod n will be a factor of phi(n)
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The text contains a mathematical question about the last digit of Graham's number in base ten and a discussion about patterns in the cycles of digits in powers of numbers, specifically powers of 3 and 5.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.

Author Message
MTGAP
Daedalian Member

 Posted: Sun Oct 12, 2008 11:19 pm    Post subject: 1 I was thinking the other day, "What is the last digit of [url="http://everything2.com/e2node/Graham%2527s%2520Number"]Graham's number?[/url]" In base three, it would obviously be zero. But it's trickier in base ten. It turns out that 3^3^...^3^3 any number of times always ends in a seven. I also noticed that each digit has a cycle. 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6 = 729. The ones digit repeats. It goes 1, 3, 9, 7. You can see why. 1x3 is 3, 3x3 is 9, 9x3 ends in 7, and 7x3 ends in 1. This type of pattern works for every place value. The tens place has a 40-digit cycle, the hundreds place has a 100-digit cycle, and the thousands place has a 500-digit cycle. I was unable to calculate any cycle above that. So what I'm wondering is, is there a predictable pattern to the increasing values of these cycles? Here's a number that's simpler than three: 5. The ones digit is always 5, the tens is always 2. The hundreds cycles through 1 and 6. The thousands cycles through 3, 5, 8, and 0. The cycles get much longer after that. So what's the pattern?_________________This statement is false.
ralphmerridew
Daedalian Member

 Posted: Mon Oct 13, 2008 12:10 am    Post subject: 2 First, the tens place has a 20-digit cycle. So for the special case of powers of 3, modulo 10^i, the cycle length is a simple GP. More generally, define phi(n) as the number of numbers 1 <= i <= n such that gcf(i,n) = 1. If p_1, p_2, ... p_j are distinct primes, and n = p_1^e_1 * p_2^e_2 * ... * p_j^e_j, then phi(n) can be calculated as phi(n) = (p_1-1)*p_1^(e_1-1) * (p_2-1)*p_2^(e_2-1) * ... * (p_j-1)*p_j^(e_j-1) = n * (1-1/p_1) * (1-1/p_2) * ... * (1-1/p_j). IIRC, the period of any base mod n will be a factor of phi(n)
 Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year by All usersMTGAPralphmerridew Oldest FirstNewest First
 All times are GMT Page 1 of 1

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The text contains a mathematical question about the last digit of Graham's number in base ten and a discussion about patterns in the cycles of digits in powers of numbers, specifically powers of 3 and 5.

TYPE: SINGLE_QUESTION_WITH_ANSWER