## Poll

17 votes (50%) | |||

13 votes (38.23%) | |||

5 votes (14.7%) | |||

2 votes (5.88%) | |||

11 votes (32.35%) | |||

3 votes (8.82%) | |||

6 votes (17.64%) | |||

5 votes (14.7%) | |||

10 votes (29.41%) | |||

8 votes (23.52%) |

**34 members have voted**

January 3rd, 2021 at 9:47:17 AM
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Quote:chevy

T = total time = t + 10^100 / (10^10 * 2^(t/10^8)) = t + 10^90 * 2^(-t/10^8)

Set dT/dt=0

1+10^90 * (-1/10^8) ln(2) 2^(-t/10^8) = 0

1=10^82 ln(2) 2^(-t/10^8)

2^(t/10^8) = 10^82 ln(2)

(t/10^8) ln(2) = ln[10^82 ln(2)]

t = 10^8 ln[10^82 ln(2)] / ln(2)

t=271.869 * 10^8 sec.............this looks suspiciously close to e * 10^10........maybe rounding error in my calc?

So total time

T = 271.869 * 10^8 + 1/ln(2) * 10^8

T= 273.312 * 10^8 sec

T = 2.73312*10^10 sec ~ 866.667 years

Comments:

1) it is local minimum since ThatDonGuy said it was....and because this is less than with wait time = 0 and as wait time --> infinity

2) If my algebra is correct and write time reduces to 1/ln(2) * 10^8.....it suggest my algebra was not the most efficient and perhaps an easier simplification in solving for t exists.

Correct.

I get the same value for t: 10^8 (82 ln 10 + ln (ln 2)) / ln 2.

ln (10^82 ln 2) = ln (10^82) + ln (ln 2) = 82 ln 10 + ln (ln 2).

For those of you playing at home, note that if f(x) = a^x for some positive real number a, df/dx = a^x ln a.

Since dT/dt has only one zero point, it is either a maximum, a minimum, or an inflection point.

At t = 0, T = 10^90 > 10^10

At t = 10^11, T > 10^11 > 10^10

Therefore, T is a minumum where dT/dt = 0.

Alternatively, take the second derivative of T:

d^2T/dt^2 = -10^82 ln 2 * (2^(-(10^-8))^t * ln (2^(-(10^-8))

= -10^82 ln 2 * (2^(-(10^-8))^t * (-(10^-8)) ln 2

= 10^74 (ln 2)^2 * (2^(-(10^-8))^t

This > 0 for all t, so dT/dt = 0 is a minimum.

January 4th, 2021 at 2:18:34 PM
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Starting from a come out roll, what is the expected number of rolls for all six points to be won? The points can be won by various shooters

It’s all about making that GTA

January 5th, 2021 at 3:00:12 PM
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Easy math, right?

This year, 2021, is the product of consecutive primes: 2021 = 43*47.

What was the year before this one that was the product of consecutive primes?

What is the next year that will be the product of consecutive primes?

What is the next year that will be the product of twin primes (primes that differ by 2)?

Is every finite group of order 2021 cyclic?

And in the year 1202 (2021 backwards), Fibonacci wrote his famous book Liber Abaci (Book of Abacus or The Book of Calculation), which popularized Hindu–Arabic numerals in Europe and first explained the utility of the Fibonacci sequence for calculating rabbit populations. (thanks Wikipedia!)

In other words, 2021 is very special.

This year, 2021, is the product of consecutive primes: 2021 = 43*47.

What was the year before this one that was the product of consecutive primes?

What is the next year that will be the product of consecutive primes?

What is the next year that will be the product of twin primes (primes that differ by 2)?

Is every finite group of order 2021 cyclic?

And in the year 1202 (2021 backwards), Fibonacci wrote his famous book Liber Abaci (Book of Abacus or The Book of Calculation), which popularized Hindu–Arabic numerals in Europe and first explained the utility of the Fibonacci sequence for calculating rabbit populations. (thanks Wikipedia!)

In other words, 2021 is very special.

Last edited by: teliot on Jan 5, 2021

Poetry website: www.totallydisconnected.com

January 5th, 2021 at 3:24:02 PM
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The last one before 2021 was also the last one (and the first since 899) to be a prime pair:

1763 = 41 x 43

The next one is:

2491 = 47 x 53

The next prime pair is:

3599 = 59 x 61

The only other prime pair before 10,000 is:

5183 = 71 x 73

January 5th, 2021 at 5:18:46 PM
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Yes. And thanks for your 10,000 answer.Quote:ThatDonGuy

The last one before 2021 was also the last one (and the first since 899) to be a prime pair:

1763 = 41 x 43

The next one is:

2491 = 47 x 53

The next prime pair is:

3599 = 59 x 61

The only other prime pair before 10,000 is:

5183 = 71 x 73

What about that cyclic group question? And, if Fibonacci was correct about rabbits and we started with two rabbits in 1202, how many rabbits would we have today?

Poetry website: www.totallydisconnected.com

January 5th, 2021 at 5:49:51 PM
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Quote:teliotWhat about that cyclic group question? And, if Fibonacci was correct about rabbits and we started with two rabbits in 1202, how many rabbits would we have today?

Group theory is a little above my math pay grade.

As for Fibonacci, given that his problem specified that rabbits breed every month (except for their first month, I get, at the end of the 9828th month:

61,663,903,718,616,908,525,832,683,988

,551,651,393,900,081,202,751,092,380,950,215,566,662,244,207

,284,750,440,519,578,393,588,447,614,735,878,375,794,896,673

,274,514,678,510,451,799,646,640,898,584,844,359,660,921,952

,068,794,519,569,421,864,294,836,756,628,011,880,354,423,856

,116,428,594,717,113,792,571,646,410,720,910,229,221,446,096

,672,177,466,398,165,302,139,176,956,673,875,869,873,856,661

,936,473,679,366,046,684,037,573,614,161,599,244,469,508,447

,929,172,932,399,012,382,057,061,668,036,598,817,830,201,665

,224,884,876,468,036,003,712,273,382,911,548,687,051,790,739

,514,476,747,916,403,971,270,195,378,786,500,815,293,965,865

,801,209,875,431,666,331,747,346,115,806,258,701,975,985,829

,260,476,870,971,415,698,776,733,543,961,383,983,620,459,615

,669,012,191,317,339,037,600,121,121,255,637,261,858,612,792

,926,221,936,783,685,806,225,716,254,054,667,093,530,428,914

,267,379,047,100,176,392,176,251,489,991,745,056,866,698,332

,425,126,200,016,294,309,522,479,701,327,300,042,515,792,950

,302,083,721,160,766,866,074,277,218,879,169,563,234,968,285

,429,439,015,220,721,475,010,428,876,777,937,770,681,685,843

,188,428,926,710,450,827,786,132,382,379,071,494,507,578,169

,572,746,899,190,405,997,109,942,054,345,947,212,298,054,489

,717,275,675,760,015,895,418,376,483,190,974,576,419,527,316

,764,651,703,858,243,325,729,124,337,830,723,048,331,973,802

,417,852,046,917,567,028,503,180,837,023,707,802,339,694,972

,150,401,284,793,060,942,917,981,803,968,272,470,750,650,466

,544,449,421,119,090,825,825,272,008,137,267,940,410,700,812

,956,708,987,685,722,577,025,503,176,516,424,364,051,583,655

,446,346,356,838,042,979,799,030,620,396,270,123,765,302,092

,051,949,907,917,257,557,788,999,094,170,729,236,153,624,878

,257,045,670,580,700,482,856,717,944,986,952,247,405,838,218

,761,175,992,545,972,752,326,472,603,300,539,634,321,155,949

,430,567,074,223,211,355,252,938,944,678,182,463,179,356,178

,824,187,261,456,390,158,424,491,991,202,585,308,836,850,984

,649,147,783,401,368,105,827,180,488,876,804,419,833,439,459

,883,001,408,424,459,974,600,815,056,554,525,031,418,869,901

,435,095,440,868,684,957,323,255,719,478,934,220,478,001,045

,522,785,742,630,470,022,349,192,569,155,075,955,919,898,677

,808,665,046,634,221,443,717,722,483,277,957,817,586,380,981

,534,033,239,133,697,074,464,904,345,390,789,063,092,335,323

,886,031,743,098,106,868,678,548,564,110,177,926,345,923,719

,592,298,939,624,002,204,984,926,913,610,626,006,668,864,521

,808,553,050,825,192,928,223,905,603,275,231,246,492,130,162

,498,040,633,542,103,528,194,102,611,732,895,205,837,649,890

,738,420,684,184,123,291,262,112,187,863,160,333,730,632,712

,748,899,685,743,507,353,078,611,474,006,633,102,733,937,136

,354,476,686,142,902,541,336,014,399,319,029,954,775,873,529

That's 2054 digits

January 5th, 2021 at 5:52:33 PM
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Love that you did this to the exact rabbit.Quote:ThatDonGuyGroup theory is a little above my math pay grade.

As for Fibonacci, given that his problem specified that rabbits breed every month (except for their first month, I get, at the end of the 9828th month:

61,663,903,718,616,908,525,832,683,988

,551,651,393,900,081,202,751,092,380,950,215,566,662,244,207

,284,750,440,519,578,393,588,447,614,735,878,375,794,896,673

,274,514,678,510,451,799,646,640,898,584,844,359,660,921,952

,068,794,519,569,421,864,294,836,756,628,011,880,354,423,856

,116,428,594,717,113,792,571,646,410,720,910,229,221,446,096

,672,177,466,398,165,302,139,176,956,673,875,869,873,856,661

,936,473,679,366,046,684,037,573,614,161,599,244,469,508,447

,929,172,932,399,012,382,057,061,668,036,598,817,830,201,665

,224,884,876,468,036,003,712,273,382,911,548,687,051,790,739

,514,476,747,916,403,971,270,195,378,786,500,815,293,965,865

,801,209,875,431,666,331,747,346,115,806,258,701,975,985,829

,260,476,870,971,415,698,776,733,543,961,383,983,620,459,615

,669,012,191,317,339,037,600,121,121,255,637,261,858,612,792

,926,221,936,783,685,806,225,716,254,054,667,093,530,428,914

,267,379,047,100,176,392,176,251,489,991,745,056,866,698,332

,425,126,200,016,294,309,522,479,701,327,300,042,515,792,950

,302,083,721,160,766,866,074,277,218,879,169,563,234,968,285

,429,439,015,220,721,475,010,428,876,777,937,770,681,685,843

,188,428,926,710,450,827,786,132,382,379,071,494,507,578,169

,572,746,899,190,405,997,109,942,054,345,947,212,298,054,489

,717,275,675,760,015,895,418,376,483,190,974,576,419,527,316

,764,651,703,858,243,325,729,124,337,830,723,048,331,973,802

,417,852,046,917,567,028,503,180,837,023,707,802,339,694,972

,150,401,284,793,060,942,917,981,803,968,272,470,750,650,466

,544,449,421,119,090,825,825,272,008,137,267,940,410,700,812

,956,708,987,685,722,577,025,503,176,516,424,364,051,583,655

,446,346,356,838,042,979,799,030,620,396,270,123,765,302,092

,051,949,907,917,257,557,788,999,094,170,729,236,153,624,878

,257,045,670,580,700,482,856,717,944,986,952,247,405,838,218

,761,175,992,545,972,752,326,472,603,300,539,634,321,155,949

,430,567,074,223,211,355,252,938,944,678,182,463,179,356,178

,824,187,261,456,390,158,424,491,991,202,585,308,836,850,984

,649,147,783,401,368,105,827,180,488,876,804,419,833,439,459

,883,001,408,424,459,974,600,815,056,554,525,031,418,869,901

,435,095,440,868,684,957,323,255,719,478,934,220,478,001,045

,522,785,742,630,470,022,349,192,569,155,075,955,919,898,677

,808,665,046,634,221,443,717,722,483,277,957,817,586,380,981

,534,033,239,133,697,074,464,904,345,390,789,063,092,335,323

,886,031,743,098,106,868,678,548,564,110,177,926,345,923,719

,592,298,939,624,002,204,984,926,913,610,626,006,668,864,521

,808,553,050,825,192,928,223,905,603,275,231,246,492,130,162

,498,040,633,542,103,528,194,102,611,732,895,205,837,649,890

,738,420,684,184,123,291,262,112,187,863,160,333,730,632,712

,748,899,685,743,507,353,078,611,474,006,633,102,733,937,136

,354,476,686,142,902,541,336,014,399,319,029,954,775,873,529

That's 2054 digits

And a little bit of Fibonacci trivia, the largest Fibonacci number that is a perfect square is 144.

Poetry website: www.totallydisconnected.com

January 5th, 2021 at 6:15:46 PM
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Quote:teliotAnd a little bit of Fibonacci trivia, the largest Fibonacci number that is a perfect square is 144.

Also, the largest one that is a triangular/triangle (I always called them triangle numbers, but I am under the impression that triangular is the proper term) is 55. This was actually a question in this year's Old Farmer's Almanac, but it has the wrong answer (it says 21).

January 8th, 2021 at 8:31:01 PM
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Here is the Mensa logic/math puzzle for January 2021.

Amy and Bob play 20 rounds of rock paper scissors. There were no ties.

Amy played, in no particular order, rock 3 times, paper 6 times, and scissors 11 times.

Bob player, in no particular order, rock 6 times, paper 5 times, and scissors 9 times.

How many games did each player win?

Amy and Bob play 20 rounds of rock paper scissors. There were no ties.

Amy played, in no particular order, rock 3 times, paper 6 times, and scissors 11 times.

Bob player, in no particular order, rock 6 times, paper 5 times, and scissors 9 times.

How many games did each player win?

It's not whether you win or lose; it's whether or not you had a good bet.

January 8th, 2021 at 8:52:30 PM
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Amy wins 3 rock against scissors

Amy loses 6 paper against scissors

Amy wins 5 scissors against paper

Amy loses 6 scissors against rock

Total

Amy wins 8, Bob wins 12.

(Because no ties, Bob's 9 scissors must be against Amy's rock and paper (3+6=9). Leaving Amy's 11 scissors against Bob's rock and paper (6+5=11))