Math Is Fun Forum / Problems and Solutions

ganesh · 2005-12-27 14:18:08

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?

ganesh · 2005-12-27 17:37:47

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 · 2005-12-27 17:46:54

Last edited by krassi_holmz (2005-12-27 17:49:07)

ganesh · 2005-12-27 18:16:16

krassi_holmz, you get this

ganesh · 2005-12-28 15:36:38

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 · 2005-12-28 21:40:59

krassi_holmz · 2005-12-28 21:42:03

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

krassi_holmz · 2005-12-28 21:47:58

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 23:12:15)

krassi_holmz · 2005-12-28 22:35:46

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 · 2005-12-28 22:40:17

k+48:

?

krassi_holmz · 2005-12-28 22:48:36

here is a picture:

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krassi_holmz · 2005-12-28 23:17:51

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 23:19:16)

ganesh · 2005-12-29 14:53:30

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 ?

krassi_holmz · 2005-12-29 19:48:44

krassi_holmz · 2005-12-29 19:56:02

The clock:

Uploaded Images

Last edited by krassi_holmz (2005-12-29 20:03:21)

krassi_holmz · 2005-12-29 19:58:08

Please post a proof of k+28 Please! It must be more prime than mine!

ganesh · 2005-12-29 20:03:26

krassi_holmz, you've made some mistake. Check your solution to problem # k + 76 again. The angle is obtuse.

krassi_holmz · 2005-12-29 20:11:01

I'll try different way.
<[BOC]=4*30°+30((60-35)/60)=132.5°

ganesh · 2005-12-29 20:29:13

You are correct, krassi_holmz! Its much simpler working on degrees than on radians.

krassi_holmz · 2005-12-29 21:31:30

Yes, the mistake in pirst proof is from the redians.

krassi_holmz · 2005-12-29 21:38:53

Sometimes the degrees are better than the radians...

krassi_holmz · 2005-12-29 21:39:55

sqr(Phi)=1.272019650...

ganesh · 2006-01-03 15:27:23

Proof : Problem # k + 28

n²! can be expressed as n!(n+1)(n+2)...(n+n)(2n+1)(2n+2)....(3n)(3n+1).....(kn)(kn+1)........(n²).
In the denominator, we have (n!)^n.
It can be seen that (n+1)(n+2)......(2n) is divisble by n!, (2n+1)(2n+2)...(3n) is divisible by n! and so on.
This is because the product of any n consecutive natural numbers is always divisible by n!.(Remember, for n>r, nCr is always a natural number).
Thus for every n! in the denominator, there is a term in the numerator which is divisible by it.
Therefore, n²! is divisible by (n!)^n.

ganesh · 2006-01-03 20:34:39

Problem # k + 77

A piece of equipment cost a certain factory $ 600, 000. If it depreciates in value, 15% the first year, 13.5 % the next year, 12% the third year, and so on, what will be its value at the end of 10 years, all percentages applying to the original cost?

krassi_holmz · 2006-01-03 20:43:25

K+28-it was very simple!

But my proof is correct, too.

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