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#1 2007-10-31 07:23:33

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Jane’s exercises II

I thought I’d start a new exercise thread since nobody seems to be reading the first thread any more. neutral

1.
Let a[sub]1[/sub], a[sub]2[/sub], … a[sub]k[/sub] be k consecutive terms of an arithmetic progression with common difference d (where the a[sub]i[/sub] are integers). Show that if gcd(k,d) = 1, then k divides a[sub]r[/sub] for some r = 1, …, k.

2.
Let a, b, c be three consecutive integers. Prove that at least one of 10a−1, 10b−1 and 10c−1, is not prime.

Last edited by JaneFairfax (2007-12-30 06:53:00)


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#2 2007-11-05 14:37:52

mason777
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Registered: 2007-11-05
Posts: 44

Re: Jane’s exercises II

uuhhh... nice,i guess what lol LOL MAN THATS CONFUSING faint


Man, what a pain...

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#3 2007-11-05 23:59:27

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

Yeah, I made both problems up myself. On reflection, I now realize that #1 is a bit on the tough side. Sorry. dizzy

Last edited by JaneFairfax (2007-11-06 00:25:32)


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#4 2007-11-19 09:57:40

Kurre
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Registered: 2006-07-18
Posts: 280

Re: Jane’s exercises II


edit: forgot to hide it hmm

Last edited by Kurre (2007-11-20 04:13:50)

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#5 2007-12-07 04:06:43

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

3.

Let f and g be real-valued functions which are twice differentiable on an interval I. Show that if f″(x) ≠ g″(x) for all xI, the graphs y = f(x) and y = g(x) cannot intersect each other more than twice in the interval I.


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#6 2007-12-08 05:50:49

John E. Franklin
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Registered: 2005-08-29
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Re: Jane’s exercises II


igloo myrtilles fourmis

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#7 2007-12-08 06:11:33

JaneFairfax
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Re: Jane’s exercises II


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#8 2007-12-23 15:04:13

JaneFairfax
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Registered: 2007-02-23
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Re: Jane’s exercises II

3.

a, b, c, d are real numbers such that 0 ≤ abcd and a + db + c. Show that adbc.


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#9 2007-12-24 08:09:17

tony123
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Registered: 2007-08-03
Posts: 189

Re: Jane’s exercises II

a, b, c, d are real numbers such that 0 ≤ a ≤ b ≤ c ≤ d and a + d ≤ b + c. Show that ad ≤ bc.



let


 

 

a+d≤b+c

a+a+z≤a+x+a+y

z≤x+y



=a(z-x-y)-xy≤0

ad ≤ bc

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#10 2007-12-24 21:57:39

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

Nice work. smile (The last part of the proof needs a little bit of touching up, that’a all.)

4.

Let a, b, x, y be rational numbers. Consider the following statement.

(i) Give a counterexample to show that the above statement, as it stands, is not true.
(ii) What condition must be imposed on the above statement to make it true?


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#11 2007-12-25 04:05:07

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Jane’s exercises II

I had another method for 3 half-formed in my mind yesterday.
After a good night's sleep I've fortified it, so I thought I'd post even though I'm late.

Let p and q be the mean and half-range of a and d, and define r and s similarly for b and c. That is:

p = (a+d)/2
q = (d-a)/2
r = (b+c)/2
s = (c-b)/2

Then it is a simple matter to show that:

p-q = a
p+q = d
r-s = b
r+s = c

Therefore ad = (p-q)(p+q) = p²-q² and bc = (r-s)(r+s) = r²-s².

Further, using the fact that 0≤a≤b≤c≤d shows that c-b≤d-b≤d-a and so s≤q.
Also, since a+d≤b+c, then p≤r.

Hence p²-q² ≤ r²-q² ≤ r²-s² and thusly ad ≤ bc, WWWWW.


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It wanted to be normal.

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#12 2008-01-07 01:44:27

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

Nice solution, Mathsy. smile

Actually #3 wasn’t exactly mine; it was something I copied from the Web, which I managed to solve myself. Anyway, the following exercise question is mine. big_smile

#5

A cubic polynomial

has roots
. Another cubic polynomial
has roots
. If the leading term in
is
, find (in terms of p, q and r) the coefficient of the term in
in
.


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#13 2008-02-22 03:08:17

Kurre
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Registered: 2006-07-18
Posts: 280

Re: Jane’s exercises II

5#
There is probably noone looking here, but anyway:P:
lets call the requested coefficient t.
according to vietes identity:


we have that p=a+b+c, q=ab+bc+ac, r=abc, so that gives:


if we take (a+b+c)^3 we get:

therefor

thus

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#14 2008-02-22 13:01:31

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

You could also make use of the following very useful identity: smile


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#15 2008-02-28 13:20:12

JaneFairfax
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Registered: 2007-02-23
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Re: Jane’s exercises II

#6

Prove that for all positive integers n,

.               .


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#16 2008-04-13 22:24:35

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

#7

Solve the following differential equation:

                 


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#17 2008-05-12 13:58:36

JaneFairfax
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Registered: 2007-02-23
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Re: Jane’s exercises II

#8

     


     


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#18 2008-05-16 01:20:35

Kurre
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Registered: 2006-07-18
Posts: 280

Re: Jane’s exercises II

Using the inequality between arithmetic and geometric metdium yields:



equality holds if and only if ab=2bc=4ca

letting ab=1/c, bc=1/a, ca=1/b, yields
1/c=2/a=4/b -> b=2a and c=a/2
abc=1 ->a*2a*a/2=1 -> a=1, which gives that b=2, c=1/2

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#19 2008-05-16 01:42:30

Kurre
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Registered: 2006-07-18
Posts: 280

Re: Jane’s exercises II

#7


integration gives

same as

with solutions:

nice problem btw smile

Last edited by Kurre (2008-05-16 01:59:06)

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#20 2008-05-16 06:52:57

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

Very nicely done, Kurre! up


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#21 2008-05-24 12:34:18

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

#8

Suppose the roots of the cubic equation

are all real and non-negative. Prove that

                         

What condition is necessary and sufficient for equality to occur?


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#22 2008-05-25 04:18:59

Kurre
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Registered: 2006-07-18
Posts: 280

Re: Jane’s exercises II

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#23 2008-05-25 04:41:47

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

welldone2.gif

Last edited by JaneFairfax (2008-05-25 04:52:21)


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#24 2008-05-25 05:25:29

Kurre
Member
Registered: 2006-07-18
Posts: 280

Re: Jane’s exercises II

smile
Its also possible to use the rearrangement inequality directly on the first inequality. However, I spotted an error on my solution to the second one:

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#25 2008-06-30 14:40:21

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Jane’s exercises II

#9

Let R and S be rings with multiplicative identities 1[sub]R[/sub] and 1[sub]S[/sub] respectively and suppose f:RS is a function which preserves multiplication (i.e. f(xy) = f(x)f(y) for all x, yR). Show that if f does not map all the elements of R into the zero element of S and S is an integral domain, then f(1[sub]R[/sub]) = 1[sub]S[/sub].
­


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