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http://online.wsj.com/article/SB123119236117055127.html
Hi Mathisfun;
From Real Dummy to Real Member in only nine days, woof! Love being here. Thanks.
Hi MathisFun;
I didn't think of that. Chats can get obnoxious. If you want to upgrade me to Real Member that would be fine. I appreciate that.
bobbym
Hi;
Could we get a chat box for this forum?
Hi smiyc86;
solution to #3
solution
Hi smiyc86;
Found this page:
http://nz.answers.yahoo.com/question/index?qid=20080620225029AAeW6Uh
with this proof:
We know that Iterative Sum of Digits = n mod 9( also called the digital root)
D = (1999^1999) mod 9 = (1999 mod 9)^1999
We know:
1) 1999 mod 9 = 1
2) 1^f = 1, for any value of f.
D = 1^1999 = 1
This is what I have been saying all along.
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I was afraid that you would ask that. I did the calculation using a computer and added every digit up. This is as far as I know, and that wikipedia site agrees, the only way to get the digital sum. The digital root can be done by congruences.
Hi smiyc86;
This process:
A=1999^1999
B=29656
C=28
D=10
E=?
is the definition of the digital root. That is computed by 1999 is congruent to 1 mod 9 so 1^1999 =1. This saves all the computing. E=1. How was that?
Your welcome, glad it helped.
There are ways of getting that minimum without calculus but if I mention any of them I would be leaving myself wide open to much deserved criticism. There is a book by Ivan Niven that covers this particular subject.
Thanks for looking at the post.
See nurshodiqs good idea below!
Thanks for providing me with another way to do it.
smiyc86;
You are absolutely right. I was confusing the digital root with the digital sum.
I got the following from:
http://en.wikipedia.org/wiki/Digit_sum
Digital root of 199 =1 (199 is congruent to 1 mod 9)
Digital sum of 199 =19 but still not 10 (No easy way to compute. Just sum the digits)
Digital root of 84001 = 4 (84001 is congruent to 4 mod 9)
Digital sum of 84001 = 13 (No easy way to compute. Just sum the digits)
Based on what I think you require, I am stuck on following the rules for digital sums so:
A=1999^1999
B=29656
C=28
D=10
Thanks a lot for listening, please respond.
Hi ganesh;
Answer
Answer
Hi ganesh
for (2)
Hi mickeen;
Also the inequality is not true for negative numbers.
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Hi Mickeen;
If p,q, and r are greater than 0 than there is a longer proof that only uses algebra and a little calculus.
Expand the left hand side:
(A)
subtract
from both sidesDivide both sides by
group the left hand side like this
Now look at the first bracket. It is the sum of a fraction and its reciprocal. We can think of it as
Try to prove that this function is always >= to 2. Take the derivative and set it to 0
solve the equation:
disregard the negative root and plug the 1 into
You get 2. Determine that x=1 is a minimum by the second derivative or by graphing the function. So now we know that the first bracketed term is >= 2. We can do the same thing for the remaining 2 bracketed terms. We have 3 terms all >= 2: The sum of them is >= 6. We have proven this:
This implies (A) for p,q and r >0
Hi bitus,
For #4 I think you mean
Hi smiyc86;
Thanks for responding.
The sum of the digits of 199 = 19 = 1 because 199 is congruent to 1 mod 9, The sum of the digits of B = 1999^1999 = 29656 (by expanding this number, I assumed you didn't want to sum it again else B=1 C=1 D=1) and C = Sum of the digits of B = 28. D = The sum of the digits of C =10. I assumed you would reduce this 10 down to 1. This is as far as I follow the problem please provide more guidance.
Hi ganesh;
Answer for (1)
Hi ganesh;
A smothered mate is delivered by a knight to a king that is totally surrounded by pieces or pawns. This is a short example:
1.e4 c6 2.d4 d5 3.Nc3 dxe4 4.Nxe4 Nd7 5.Qe2!? Ngf6?? 6.Nd6 mate.
I looked at this post and liked the explanation so much I didn't even see the arithmetic error.
Hi quittyqat;
Yep, I have it highlighted.
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