Easy math puzzles
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43 members have voted
February 1st, 2022 at 4:38:23 PM
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Quote: Ace2
Now that we know the mean value of x (on one side) is (2/π)^(.5), we can calculate the variance by taking the integral from zero to infinity of:
(2π)^(-.5) * e^(-.5x^2) * (x - (2/π)^(.5))^2 dx
I don't believe there's a way to get an exact antiderivative for this function, so I used an integral calculator to get the answer of: (π - 2) / (2π)
We can expand (x - (2/π)^(.5))^2, (x - (2/π)^(.5))^2=x^2-2*(2/π)^(.5)*x+2/π and we can calculate the integral of each term multiplied by (2π)^(-.5) * e^(-.5x^2) separately.
The integral of (2π)^(-.5)*e^(-.5x^2)*x^2 dx from x=0 to x=∞ is 1/2 of the integral of the same function from x=-∞ to x=∞, which is the integral expressing the variance of the normal distribution, so the integral from x=-∞ to x=∞ is 1 and the integral from x=0 to x=∞ is 1/2.
We have already calculated that the integral of (2π)^(-.5)*e^(-.5x^2)*x dx from x=0 to x=∞ is 1/(2π)^(.5), so the integral of (2π)^(-.5)*e^(-.5x^2)*2*(2/π)^(.5)*x dx from x=0 to x=∞ is 1/(2π)^(.5)*2*(2/π)^(.5)=2/π.
The integral of (2π)^(-.5)*e^(-.5x^2) dx from x=0 to x=∞ is 1/2, it is just the probability that a standard normal random variable is positive.
Putting everything together, the value of the integral is 1/2-2/π+1/2=1-2/π=(π-2)/π.
February 1st, 2022 at 6:24:59 PM
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Thanks again, GM. It took me a minute to grasp that step, but once I did, getting the antiderivative of the function was easy. I suspected there was a simple way to get there since (pi - 2) / pi is such a “clean” answer…using only 2 constants from the original functionQuote: GM
The integral of (2π)^(-.5)*e^(-.5x^2)*x^2 dx from x=0 to x=∞ is 1/2 of the integral of the same function from x=-∞ to x=∞, which is the integral expressing the variance of the normal distribution, so the integral from x=-∞ to x=∞ is 1 and the integral from x=0 to x=∞ is 1/2.
My calculus skills are quite basic, especially when it comes to rearranging/simplifying functions. I rarely use calculus so I don’t get much practice
Last edited by: Ace2 on Feb 1, 2022
It’s all about making that GTA
February 2nd, 2022 at 8:17:15 AM
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This puzzle is rather wordy so read carefully...
The number N represents the first 6 digits of a special number.
N consists of three prime numbers put side by side. These three prime numbers come xth, yth and zth on the list of primes, where x, y and z are themselves three consecutive primes (for example, x, y and z could be 3, 5 and 7, in which case we'd be looking at the third, fifth and seventh prime numbers).
In addition, if N is split in the middle into two separate numbers, the prime factors of the left part of N add up to its right part.
What is N? And what is the special number whose first 6 digits it forms?
The number N represents the first 6 digits of a special number.
N consists of three prime numbers put side by side. These three prime numbers come xth, yth and zth on the list of primes, where x, y and z are themselves three consecutive primes (for example, x, y and z could be 3, 5 and 7, in which case we'd be looking at the third, fifth and seventh prime numbers).
In addition, if N is split in the middle into two separate numbers, the prime factors of the left part of N add up to its right part.
What is N? And what is the special number whose first 6 digits it forms?
Have you tried 22 tonight? I said 22.
February 2nd, 2022 at 8:54:15 AM
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Quote: GialmereThis puzzle...What is N?...link to original post
First one lists the first few primes being 2,3,5,7,11,13,17,19,23 and then list the 2nd, 3rd, 5th etc primes. The latter gives 3, 5, 11, 17, 31, 41, 59, 67, 83, 109. Then the six-digits numbers can be 11-17-31 thru 59-67-83. What is interesting is 314159 has a lower second half (159) than first half (314) and 314 = 157*2. Thus N is nearly Pi*100000! (I'm guessing this is what the last part refers to.)
February 2nd, 2022 at 8:10:51 PM
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Quote: charliepatrickFirst one lists the first few primes being 2,3,5,7,11,13,17,19,23 and then list the 2nd, 3rd, 5th etc primes. The latter gives 3, 5, 11, 17, 31, 41, 59, 67, 83, 109. Then the six-digits numbers can be 11-17-31 thru 59-67-83. What is interesting is 314159 has a lower second half (159) than first half (314) and 314 = 157*2. Thus N is nearly Pi*100000! (I'm guessing this is what the last part refers to.)
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Correct!!
The numbers 31, 41, and 59 are the 11th, 13th, and 17th primes where 11, 13 and 17 form three consecutive primes. So the number we get by putting them together, 314159, fulfills the first part of our requirement for the number N.
Now split 314159 into the two numbers 314 and 159. The prime factors of 314 are 2 and 157, which add up to 159. So 314159 also fulfills the second requirement — hence N=314159 is the number we are looking for. It also represents the first six digits of π.
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This puzzle is by Aziz Inan, Professor of Electrical Engineering at the University of Portland. His own name, when written in capital letters, is a geometric word puzzle.
Now split 314159 into the two numbers 314 and 159. The prime factors of 314 are 2 and 157, which add up to 159. So 314159 also fulfills the second requirement — hence N=314159 is the number we are looking for. It also represents the first six digits of π.
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This puzzle is by Aziz Inan, Professor of Electrical Engineering at the University of Portland. His own name, when written in capital letters, is a geometric word puzzle.
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Thank you for your undivided attention.
Have you tried 22 tonight? I said 22.
February 2nd, 2022 at 9:45:13 PM
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Your jokes are the bestQuote: Gialmere
Thank you for your undivided attention.
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It’s all about making that GTA
February 3rd, 2022 at 10:49:38 AM
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Two dice are rolled until either event happens:
A) A total of 2 is rolled.
B) A total of 7 is rolled twice consecutively.
What is the probability the total of 2 occurs first?
A) A total of 2 is rolled.
B) A total of 7 is rolled twice consecutively.
What is the probability the total of 2 occurs first?
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
February 3rd, 2022 at 12:13:17 PM
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From the starting state, one of four things will happen:
Only three things need to be considered:
(a) A 2;
(b) A 7 followed by a 2;
(c) A 7 followed by a 7;
(d) anything else, which returns to the starting state, so only (a), (b), and (c) need to be considered.
(a) has probability 1/36
(b) has probability 1/216
(c) has probability 1/36
P(a + b) / P(a + b + c) = 7/13
February 3rd, 2022 at 2:34:15 PM
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Quote: ThatDonGuy
From the starting state, one of four things will happen:
Only three things need to be considered:
(a) A 2;
(b) A 7 followed by a 2;
(c) A 7 followed by a 7;
(d) anything else, which returns to the starting state, so only (a), (b), and (c) need to be considered.
(a) has probability 1/36
(b) has probability 1/216
(c) has probability 1/36
P(a + b) / P(a + b + c) = 7/13
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I agree. I solved it with a simple Markov chain, but your way looks easier to explain.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
