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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?

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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.

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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

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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?

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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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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.

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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
- Posts: 13,258

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.

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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.

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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?

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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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