Problems and Solutions

Ricky · 2005-12-23 18:15:05

Last edited by Ricky (2005-12-23 18:15:23)

Jai Ganesh · 2005-12-23 20:44:00

Correct, Ricky!

krassi_holmz, the solution to problem # k + 47 iis:-

1 - (8/9)^n - (5/9)^n + (4/9)^n.

A number is divisible by 10 if it has an even number and a 5 amongst its digits. The probability of no 5 is (8/9)^n. The probability of no even number is (5/9)^n. The probability of no 5 and no even number is (4/9)^n. Hence the result is as given above.

mathsyperson · 2005-12-23 22:34:54

With nothing but intuition, I guess

.

Jai Ganesh · 2005-12-23 23:59:15

Guessed it right, mathsyperson
Either A throws more heads than B, or A throws more tails than B, but (since A has only one extra coin) not both.
By symmetry, these two mutually exclusive possibilities occur with equal probability.

Therefore the probability that A obtains more heads than B is ½. It is perhaps surprising that this probability is independent of the number of coins held by the players.

mathsyperson · 2005-12-24 00:43:28

Yes, I was thinking along the same lines, I just hadn't formed it into an expanation.

Jai Ganesh · 2005-12-25 16:18:33

Right, Mathsy!

Problem # k + 73

Is it possible that the three sides of a right-angled triangle are in Geomtric Progression? If so, what would the lenght of the hypotenuse be?

krassi_holmz · 2005-12-26 05:24:44

Let the sides be

, and . Then

Let . Then

, p>0 so

Then the hypotenuse is

Last edited by krassi_holmz (2005-12-26 05:38:21)

mathsyperson · 2005-12-26 05:57:25

The other case is when

, which happens when b<1.

In this case,

, but the hypotenuse is just .

Last edited by mathsyperson (2005-12-26 07:08:21)

krassi_holmz · 2005-12-26 06:46:56

krassi_holmz · 2005-12-26 06:53:54

In mathsy's case b = sqrt((sqrt(5)-1)/2)q=0.7861513778...q

mathsyperson · 2005-12-26 07:09:10

Whoops, you're right. I was so busy trying to get the LaTeX right that I got the maths wrong.

krassi_holmz · 2005-12-26 07:43:56

#k+38-bertrand's postulate. I think there isn't a simple proof. I wrote something with the chebishev's theta function.

Jai Ganesh · 2005-12-26 15:18:08

Solution to problem # k + 73 is correct! Well done, krassi_holmz!
I'll check the solution to Problem # k + 28 and Problem # k + 38 later and post.

Problem # k + 74

The Greatest Common Divisor (or Highest Common Factor) and Least Common Multiple of two numbers are 3 and 84 respectively. If one of these numbers is 9 greater than the other, what's the smaller number?

Jai Ganesh · 2005-12-26 18:37:47

krassi_holmz, I am not fully convinced by your proof (solution to Problem # k + 28), although you seem to be on the right track. The reason, I think, is I am unable to follow what you intend conveying. Maybe, you have got the full proof in your mind but had not posted it. I shall wait (for a week) for amendment from you, or proof by someone else before I post the solution.

krassi_holmz · 2005-12-26 18:46:54

Last edited by krassi_holmz (2005-12-26 18:49:07)

Jai Ganesh · 2005-12-26 19:16:16

krassi_holmz, you get this

Jai Ganesh · 2005-12-27 16:36:38

Problem # k + 75

X,Y,and Z take 20, 30 and 60 days respectively to complete a job independently.
They set out to complete a job together . However Y leaves after 4 days and Z leaves after another 6 days .
How many more days will it take for X alone to complete the job now?

krassi_holmz · 2005-12-27 22:40:59

krassi_holmz · 2005-12-27 22:42:03

Yesterday I solved two of the unsolved problems. When i have time, i'll post the solutions.

krassi_holmz · 2005-12-27 22:47:58

k+28
My last proof wasn't correct at all.
In it i proved only that
LCM(1^n,2^n, ... ,n^n)/n^2!
But LCM(1^n,2^n, ...n^n) {!=} n!^n

Then I proved that
(n-k)^(n+k)/n^2!
but this wasn't enough.

I found a beautiful number theory proof with prime numbers, floors and an numberic theory theorem:
Let write n! as

, where p_1 , p_2 ... ,p_k ∈ P(prime numbers) and p_k = max(p ∈ P : p <= n). Then
.
Let write n^2! ia the same way:
.
To proof that n!^n/n^2! it is enough to proof that

...

Now we will use a theorem from the numeric theory:
The biggest power of the prime number p that divides n! is exactly

,
where [x] means Floor(x).
Now we are using it:

Now, at last, we will prove that

We use the following:
If n ∈ N and x ∈ R then
[nx]=[n([x]+{x})]=[n[x]+n{x}]=n[x]+[n{x}] >= n[x];

so =>
<=>
=> .

Last edited by krassi_holmz (2005-12-28 00:12:15)

krassi_holmz · 2005-12-27 23:35:46

And for k+47
Not I understood that the product of the numbers must be divisible by 10. When i wrote the my post i though that the SUM must be divisible by 10. My mistake.

krassi_holmz · 2005-12-27 23:40:17

k+48:

?

krassi_holmz · 2005-12-27 23:48:36

here is a picture:

krassi_holmz · 2005-12-28 00:17:51

Triangle MHO is simular to triangle OHB.
Let yr=h. then hr=yr²=x. But
x²=y²+h²
(yr²)²=y²+y²r²
Let f = r². Then
f²=1+f
f²-f-1=0
D=1+4=5
f= (1+sqrt(5))/2
r=x/h=2x/2h=AB/BC=sqrt((1+sqrt(5))/2)

Last edited by krassi_holmz (2005-12-28 00:19:16)

Jai Ganesh · 2005-12-28 15:53:30

krassi_holmz, your solution to problem # k + 75 is correct. Well done.
Regarding the solution to problem # k + 48 and the proof, I shall tell you after putting the answers under a microscope. I had a different proof in my mind, similar to yours though.

Problem # k + 76

What is the angle between the two hands of a clock when the time is 02.35 ?

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