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**ganesh****Moderator**- Registered: 2005-06-28
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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?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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*Last edited by krassi_holmz (2005-12-26 18:49:07)*

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**ganesh****Moderator**- Registered: 2005-06-28
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krassi_holmz, you get this

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
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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?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Yesterday I solved two of the unsolved problems. When i have time, i'll post the solutions.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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

.

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)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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k+48:

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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here is a picture:

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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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)*

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**ganesh****Moderator**- Registered: 2005-06-28
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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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**krassi_holmz****Real Member**- Registered: 2005-12-02
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**krassi_holmz****Real Member**- Registered: 2005-12-02
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The clock:

*Last edited by krassi_holmz (2005-12-28 21:03:21)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Please post a proof of k+28 Please! It must be more prime than mine!

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**ganesh****Moderator**- Registered: 2005-06-28
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krassi_holmz, you've made some mistake. Check your solution to problem # k + 76 again. The angle is obtuse.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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I'll try different way.

<[BOC]=4*30°+30((60-35)/60)=132.5°

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**ganesh****Moderator**- Registered: 2005-06-28
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You are correct, krassi_holmz! Its much simpler working on degrees than on radians.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Yes, the mistake in pirst proof is from the redians.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Sometimes the degrees are better than the radians...

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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sqr(Phi)=1.272019650...

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
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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?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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K+28-it was very simple!

But my proof is correct, too.

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