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Hi phrontister;

See how long #45 takes you, I considered that a win.

To gain more proficiency I think I need to look into various ways of solving the already-done problems more efficiently

Hey paulb203, sorry for the delay,

Whether anything is physically possible depends on whether the laws of nature can be broken.

We have insufficient information to determine a probability. I speak about probabilities relative to beliefs because there is no proof of which beliefs are true. With roulette mathematics can show what the probabilities are objectively and which beliefs are true or false. There is no mathematics to show whether the Bible is true in its prediction of a second coming, or whether miracles are possible.

Of course it either is or it isn't, they either are or they aren't. But since I cannot demonstrate which it is I talk about both possibilities, conditionally.

Another way to look at it is there is some probability that miracles occur, and some probability they don't - there is some probability a rapture will occur, and some probability it won't - but I do not make claims about what those probabilities are. I just made some general observations about how they may relate to each other or change over time.

**Keep_Relentless**- Replies: 6

I have not had much time recently as I've been moving, but I have scouted some of the easier problems:

Problem 16

2^15 = 32768 and the sum of its digits is 3+2+7+6+8=26. What is the sum of the digits of the number 2^1000?

Problem 20

n! means n * (n-1) * ... * 3 * 2 * 1. For example, 10! = 10*9*...*3*2*1 = 3628800, and the sum of the digits in the number 10! is 3+6+2+8+8+0+0 = 27. Find the sum of the digits in the number 100!.

Problem 25

The Fibonacci sequence is defined by the recurrence relation: Fn = Fn-1 + Fn-2, where F1 = 1 and F2 = 1. Hence the first 12 terms will be: F1=1, F2=1, F3=2, F4=3, F5=5, F6=8, F7=13, F8=21, F9=34, F10=55, F11=89, F12=144. The 12th term, F12, is the first term to contain three digits. What is the index of the first term in the Fibonacci sequence to contain 1000 digits?

Problem 40

An irrational decimal fraction is created by concatenating the positive integers: 0.12345678910**1**112131415161718192021... It can be seen that the 12th digit of the fractional part is 1. If dn represents the nth digit of the fractional part, find the value of the following expression: d1 * d10 * d100 * d1000 * d10000 * d100000 * d1000000

Problem 45

Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

Triangle: Tn = n(n+1)/2 1,3,6,10,15,...

Pentagonal: Pn = n(3n-1)/2 1,5,12,22,35,...

Hexagonal: Hn = n(2n-1) 1,6,15,28,45,...

It can be verified that T285 = P165 = H143 = 40755. Find the next triangle number that is also pentagonal and hexagonal.

Problem 48

The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317. Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.

I notice your questions have gotten trickier.

I'll make a start on the first one.

It's useful to consider the number of different ways you could have three heads.

You could have:

HHHTT

HHTHT

HHTTH

HTHHT

HTHTH

HTTHH

THHHT

THHTH

THTHH

TTHHH

That's 10 possibilities... out of how many in total?

K_R

Problem 10 done on Mathematica.

Problem 11 could be done on Excel but I think it would take annoyingly long if I'm not missing something. I'm going to go ahead and guess that it's the diagonal 89 * 94 * 97 * 87.

Edit: Problem 12 done with Mathematica. Would be very surprised if it could be done with Excel.

Edit 2: Problem 11 done on Excel.

**Keep_Relentless**- Replies: 4

I am now onto Mathematica and not just Excel. The next three:

10. The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.

11. In the 20 x 20 grid below, four numbers along a diagonal line have been marked in red [I have bolded them].

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08

49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00

81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65

52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91

22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80

24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50

32 98 81 28 64 23 67 10 **26** 38 40 67 59 54 70 66 18 38 64 70

67 26 20 68 02 62 12 20 95 **63** 94 39 63 08 40 91 66 49 94 21

24 55 58 05 66 73 99 26 97 17 **78** 78 96 83 14 88 34 89 63 72

21 36 23 09 75 00 76 44 20 45 35 **14** 00 61 33 97 34 31 33 95

78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92

16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57

86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58

19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40

04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66

88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69

04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36

20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16

20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54

01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 * 63 * 78 * 14 = 1788696. What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20 x 20 grid?

12. The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

1: 1

3: 1, 3

6: 1, 2, 3, 6

10: 1, 2, 5, 10

15: 1, 3, 5, 15

21: 1, 3, 7, 21

28: 1, 2, 4, 7, 14, 28

We can see that 28 is the first triangle number to have over five divisors. What is the value of the first triangle number to have over five hundred divisors?

You have been busy!

I got problem 9 with a mixture of Mathematica and Excel. Is it considered cheating if you use Mathematica? You don't have to understand how the maths is done to use it. But you've got to start somewhere, it's useful incentive to learn to use Mathematica better. I will consider it as partial credit, haha, because I got formulas for a and c in terms of b and I have no idea how they were derived, but I was then able to hunt for answers systematically.

I will consider it full credit if you understand how the computer did it (or at least some theory that would let you arrive at the same conclusion), and partial credit otherwise. So I have labeled my excel for problem 9 "partial credit".

But by this criterion, we would be allowed to get the answer directly often, because there's often a simple iterative process. By this criterion, if I understand the sieve of eratosthenes, to cross out the multiples of every prime, I have a (laborious) method I could potentially carry out on pen and paper. So it is not really a mystery how the computer arrives at the answer.

I'm trying to justify myself in using M so we can go on to problems 10-12! I'll just go ahead and post them.

In case there's a way of explaining what M spit out regarding problem 9, I will share its result:

Mathisfun has a page on tree diagrams: https://www.mathsisfun.com/data/probability-tree-diagrams.html

I would say flipping a coin three times yields a simple tree, of independent events.

I don't know a specific book to recommend, sorry. Maybe someone else does. Your guess would be as good as mine; but there must be many suitable books.

Good post Bob.

But I have some doubts that we can't say anything at all.

Can we generalise this to say there is no way of determining the probability of an event that violates the laws of nature?

It seems to me like the probability should be much higher if you believe the sources that claim miracles have happened (in connection with Jesus).

And much lower given that many experiments have been done in which the laws were not violated.

If you believe the Bible 100%, then the probability of a rapture occurring is 100%. The only question is when.

If you don't believe the Bible 100%, then I would expect the probability of a rapture to be getting lower with every day that passes without a rapture, and every day that passes with no independent confirmation of miracles.

Am I over-surmising here?

K_R

These ones are trickier. It might be time to actually start coding.

They say anything is possible because there is no logical contradiction if it were to happen. There is no logical contradiction in a miracle occurring. No scientific theory can prove a miracle could never happen. It can only show that it is very improbable - since it has not occurred in many repetitions/trials. But that does not get the probability to zero, it just gets it very small.

Of course if a naturalistic view is correct where the laws of nature are absolutely inviolable then the probability is zero. But I don't know how you could establish with literally 100% confidence that this is the case.

So, basically, anything that is conceivable is possible. That's how it looks to me at least.

Suppose we guess 0.001 (with three zeros) is the smallest number. We can find a smaller one: 0.0001 (with four zeros). And yet a smaller one: 0.00001 (with five zeros). This process can continue forever; there is no limit to the number of zeros we can add. So there is no smallest positive number.

My explanation is not as fancy as the others, but it works for me.

Yep looks right

I think the probability is small but positive, but the probability drops the later in the day it is.

It is 101/125, Europe just accidentally wrote 101/25.

K_R

Interesting, any chance you can explain how you worked it out? We all had a go and didn't get there.

K_R

Yes looks right.

Since there are a number of ways any of them could take the same train, I would instead do the probability that none of them chose the same train.

Amy picks a train: 5/5

Bobby picks a different train: 4/5

Chris picks yet another train: 3/5

Dusty picks another different train: 2/5

Multiply those together, and you get the probability they all chose a different train.

Take this probability away from 1, and you get the probability any of them chose the same train.

I think that's right?

K_R

Not sure whether problems 7 or 9 can be done on Excel in a reasonable amount of time.

A way to do problem 8 occurred to me. Let's give it a go.

Edit: Problem 8 done. Still no coding used.

Wow, you just used a calculator for #5? I made columns checking divisibility by every number up to 20 and searched for a good while. Found 46512 is divisible by 16-19, checked its multiples. Found a multiple divisible by every number 1-20 except for 11. Found the smallest one that fits 11 too.

But it was not as quick and easy as using a calculator!

I don't really know what I'm doing, but I made it this far.

Edit: If you want, you can make an account on the website Project Euler which has hundreds of these problems, and a forum for each problem you can view after you complete it where people share their code etc. These are the easy problems copied directly from that site.