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## #326 2005-12-27 14:18:08

ganesh
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### Re: Problems and Solutions

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?

Character is who you are when no one is looking.

## #327 2005-12-27 17:37:47

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### Re: Problems and Solutions

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.

Character is who you are when no one is looking.

## #328 2005-12-27 17:46:54

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### Re: Problems and Solutions

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

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## #329 2005-12-27 18:16:16

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### Re: Problems and Solutions

krassi_holmz, you get this

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## #330 2005-12-28 15:36:38

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### Re: Problems and Solutions

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?

Character is who you are when no one is looking.

## #331 2005-12-28 21:40:59

krassi_holmz
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### Re: Problems and Solutions

IPBLE:  Increasing Performance By Lowering Expectations.

## #332 2005-12-28 21:42:03

krassi_holmz
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### Re: Problems and Solutions

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

IPBLE:  Increasing Performance By Lowering Expectations.

## #333 2005-12-28 21:47:58

krassi_holmz
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### Re: Problems and Solutions

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)

IPBLE:  Increasing Performance By Lowering Expectations.

## #334 2005-12-28 22:35:46

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### Re: Problems and Solutions

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.

## #335 2005-12-28 22:40:17

krassi_holmz
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### Re: Problems and Solutions

k+48:

?

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## #336 2005-12-28 22:48:36

krassi_holmz
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### Re: Problems and Solutions

here is a picture:

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## #337 2005-12-28 23:17:51

krassi_holmz
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### Re: Problems and Solutions

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)

IPBLE:  Increasing Performance By Lowering Expectations.

## #338 2005-12-29 14:53:30

ganesh
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### Re: Problems and Solutions

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 ?

Character is who you are when no one is looking.

## #339 2005-12-29 19:48:44

krassi_holmz
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### Re: Problems and Solutions

IPBLE:  Increasing Performance By Lowering Expectations.

## #340 2005-12-29 19:56:02

krassi_holmz
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### Re: Problems and Solutions

The clock:

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

IPBLE:  Increasing Performance By Lowering Expectations.

## #341 2005-12-29 19:58:08

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### Re: Problems and Solutions

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

IPBLE:  Increasing Performance By Lowering Expectations.

## #342 2005-12-29 20:03:26

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### Re: Problems and Solutions

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

Character is who you are when no one is looking.

## #343 2005-12-29 20:11:01

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### Re: Problems and Solutions

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

IPBLE:  Increasing Performance By Lowering Expectations.

## #344 2005-12-29 20:29:13

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### Re: Problems and Solutions

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

Character is who you are when no one is looking.

## #345 2005-12-29 21:31:30

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### Re: Problems and Solutions

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

IPBLE:  Increasing Performance By Lowering Expectations.

## #346 2005-12-29 21:38:53

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### Re: Problems and Solutions

Sometimes the degrees are better than the radians...

IPBLE:  Increasing Performance By Lowering Expectations.

## #347 2005-12-29 21:39:55

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### Re: Problems and Solutions

sqr(Phi)=1.272019650...

IPBLE:  Increasing Performance By Lowering Expectations.

## #348 2006-01-03 15:27:23

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### Re: Problems and Solutions

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.

Character is who you are when no one is looking.

## #349 2006-01-03 20:34:39

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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?

Character is who you are when no one is looking.

## #350 2006-01-03 20:43:25

krassi_holmz
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### Re: Problems and Solutions

K+28-it was very simple!

But my proof is correct, too.

IPBLE:  Increasing Performance By Lowering Expectations.