| Author | Maths and Physics (Problems) |
|---|
Just for fun - Tough problems of Maths and Physics! (Not necessarily high level concept-wise, but tough) :)
If a person wants then he can put a prize money on his problem hehe ^_^ |
| Whats 9+10? |
| I am an odd number. Take away one letter and I become even. What number am I? |
| Seven. |
| seven |
| Or better, Make 7 even by just 1 step. And please, don't mess with the "s" :P |
Ok here's one from me.
What is the probability that the product of 5 natural numbers chosen randomly is divisible by 5? |
| Is the answer 61/63? |
100%
all natural number is divisible by 5 as long as it allow decimal :P |
Here is one
There are three tombs and all tombs must be paid leaving fee
However there is no limit on how much you have to paid however all tombs will only accept same payment.
To help you, at the enterence of all tombs is helper who would double your coins *no matter what*.
How much coins you should start and how much should you paid for leaving fees?
PS. Don't say that "Oh! I won't enter any tombs" or something like that :P and also I carry nothing and paid nothing is not allow. |
What is the probability that the product of 5 natural numbers chosen randomly is divisible by 5?
67.232%
Thanks for an interesting problem. |
I think you answer based on
4/5 chance not to land 5 (or 10)
so total
4/5^5
so to land at least one with 5 will be
1-(4/5)^5= 1- 0.8^5=1-0.32769= 0.67232 = 67.232% |
my exams jsut got over and u guys created a quiz forum?
wew |
| ROFL, Answer all questions, No choice :P |
Post 10: How much coins you should start and how much should you paid for leaving fees?
I would only need one coin to start with because I would double it at the entrance to get two coins. Charon usually takes a coin for the trip leaving a tomb. It never hurts to have an extra coin in the unlikely event that I might be able to take the trip back. |
Yea, My bad.
You cann't also leave last tomb with any coin in hand. |
| can't I mean, but who care . |
Yea answer to my problem is 1 - (4/5)^5, as answered by Superkrypton and Lord STB. You're welcome mate, I have a nice stock of even more interesting ones :)
#10 - I assumed that the question meant that we need to start with a certain sum such that in the end we are left with nothing..please correct me if wrong.
Using Reverse engineering,
In the last tomb if you paid x coins, so the last tomb helper doubled from x/2. That means before exiting 2nd tomb, you had 3x/2. Before that, 3x/4. Before exiting first tomb, 7x/4. Before entering it, 7x/8. Thus you should start with multiples of 7/8. Or to make it an integer, atleast 7 coins. The cost on each tomb will be 8/7 times the starting money.
Another question.
Find the coefficient of x^70 in (x-1)(x^2 -2)(x^3 -3)(x^4 - 4)...(x^12 - 12). Challenge from my side is to do this within 2 minutes, that is the approx time it took me.
Cheers mates! :) |
| Oh! my, coefficient, it was left 9 or 10 years ago my study. I won't study it back. XD |
Haha well it's easier than what it looks like ;)
Firstly I would like to say thank you everyone for the awesome response, this thread got so many posts already :)
Btw, my unanswered question - Make 7 even in one step (apart from removing the "s") ^_^ |
