Re: Maths problems (interesting ones)
 

(#58 in this thread)

I wrote a program (see attachement) for a brute force algorithm and I only got one solution: 381654729. With a very high likelihood ;-) there is no second solution.

Carsten

Re: Maths problems (interesting ones)
 

Hi

Thats what i got, i narrowed it down to about 12 possible results, and then tried all of these and that was the only one that i got too work.

oljomo

Re: Maths problems (interesting ones)
 

Ok well I know that there has been debate on the Door issue.

What you are looking at are too completly different events. There are two children ( any thing can be substituted for children typically coins as they are a favorite in most statistics books. Ok now forget about the door you have to people each has a 50% chance of being male and a 50% chance of being female. Thus there is a 25 % chance of both being male 50% chance of one being male and ne being female and a 25% chance of both being female ( the favorable outcome) Thus the answer would be 25 % if the question asked what are the odds of both the door opener and their sibling being female. How ever because gender is completely independent you know that their is a 50% chance that the first is feamle and a 50% chance that the second is female thus .50*.50=.25 (see earlier explination) once you know the gender of the first child the problem becomes 1.00*.5=.5 or 50%.

To disprove the answer being 33% you have to view it as the two children being the door opener and the non opener (as said earlier) rather than being the elder and the younger with the door opener being an additional variable

Example:
opener/non opener
Girl/Girl
Girl/boy
Boy/Girl
Boy/Boy
The last two are invalid because the Opener must be female.

Example2:
There are three combinations of children
2 Girls
1 of Each
2 Boys
Since one must be female the final solution is invalid thus it must be one of two solutions or 50%

What it appears is happening is that those who are arguing 33% dont truly understand probability and are instead repeting the answer to a simmilar problem that is part of Algebra 2 as well as Bio 1 and Genetics
There are 2 children what are the odds that they are both female if both are not male thus there are 3 possible results elder/younger
male/female
female/male
female/female
only one is favorable thus there is a 1/3 or 33% chance

This is the law of independent statistics or something to that effect (I dont remember the name)

This is also what says If you play powerball (Lottery game in which 6 numbers out of 60 are pulled, jackpot is won if all 6 match) the odds are 60*59*58*57*56*55 or 1 in 632,385,600 how ever if you lose the first 632,385,599 times your odds of winning the last time are still 632,385,600

Re: Maths problems (interesting ones)
 

Hi

There is an easier way, tha finds all the answers for certain

take the equations from the question
P+3=n(j-3)
j+n=3(P-n)
rearrange the first, substitute it into the second and you end up with
j+n=3nj-9n-9-3n
j-3nj=-13n-9
j(3n-1)=13n+9
3j(3n-1)=39n+27
3j(3n-1)=13(3n-1)+40
(3j-13)(3n-1)=40
as j and n have to be integers this allows you to work out all solutions which are 1,2,3,7

oljomo

Re: Maths problems (interesting ones)
 

The first and last ones need to be counted twice, because there are two ways to assign the opener and non opener. That is why the answer comes out in thirds.
But only half of the families with the second combination meet the requirements to be included in the count. Again, the answer comes out in thirds.
I believe you're thinking about conditional vs. unconditional probabilities. However, the problem as stated is definitely about conditional probabilities, asking "Given that x is true, what is the probability that y is true?"

Re: Maths problems (interesting ones)
 

Oops, of course you are right there is only one solution! I did this one a long, long time ago, so please understand me ;-) (I am getting old...)

The solution is 381654729 as stated before!

Re: Maths problems (interesting ones)
 

No one has gotton 21 (1000 factorial) correct, but lets get some new questions going. These are not that complex, but we need some new material. This girl/boy debate has gotten a bit out of hand (The answer is 1/2, but I do not wish to debate it).

23) What number comes next: 77, 49, 36, 18...

24) What is the maximum number of bishops you can put on a chess board so than none of them can take another in one move.

Re: Maths problems (interesting ones)
 

24) 14. Line up each of the end rows with bishops on all but one corner of each row. This is assuming all bishops can take all other bishops. If we go white can't take white, then correct answer is 64: put bishops on all squares of the board.

Re: Maths problems (interesting ones)
 

Ok let me try again; 2 and 5 are the prims of 10. The factorial of 1000 can be divided by 2 (1000/2)=500 times. It can also be divided by 5 200 times. therefore 1000! can be by 10 divided 200 times.
Knowing this 1000! has 200 zeros on the end.
Tell me if I am wrong.

Re: Maths problems (interesting ones)
 

I think you're wrong. :)

Every 2nd number is divisible by 2, every 2nd one of those has a second factor of two, every 2nd one of those a third, and so forth. The total number of factors of two, then, is 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 994.

Similarly, the number of factors of 5 is 200 + 40 + 8 + 1 = 249.

If you had as many factors of 5 as factors of 2, you would have 994 zeroes. But there are only 249, so the answer (unless my analysis is way off) is that 1000! has 249 trailing zeroes.

Re: Maths problems (interesting ones)
 

I believe that one's 8.

Re: Maths problems (interesting ones)
 

You made your point. At least the answer is based on my idea... :)

Re: Maths problems (interesting ones)
 

Yes. Thank you for the hint. :)

Re: Maths problems (interesting ones)
 

Yup, all three are right:

21) 249
23) 8
24) 14

I am starting to get low on problems. But, here is one more:

25)
'SEND
+MORE
MONEY

What numbers do each of the letters stand for? (' is a place holder)

Re: Maths problems (interesting ones)
 

why 8?