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

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

**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 10*a*−1, 10*b*−1 and 10*c*−1, is not prime.

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

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**mason777****Member**- Registered: 2007-11-05
- Posts: 44

uuhhh... nice,i guess LOL MAN THATS CONFUSING

Man, what a pain...

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

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

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

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

edit: forgot to hide it

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

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

**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 *x* ∈ *I*, the graphs *y* = f(*x*) and *y* = g(*x*) cannot intersect each other more than twice in the interval *I*.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,552

**igloo** **myrtilles** **fourmis**

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

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

**3.**

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

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**tony123****Member**- Registered: 2007-08-03
- Posts: 189

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

Nice work. (The last part of the proof needs a little bit of touching up, thata 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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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Why did the vector cross the road?

It wanted to be normal.

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

Nice solution, Mathsy.

Actually #3 wasnt exactly mine; it was something I copied from the Web, which I managed to solve myself. Anyway, the following exercise question is mine.

**#5**

A cubic polynomial

has roots . Another cubic polynomial has roots . If the leading term in is , find (in terms ofOffline

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

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

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

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

**#6**

Prove that for all positive integers *n*,

. .

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

**#7**

Solve the following differential equation:

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

**#8**

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

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

#7

integration gives

same as

with solutions:

nice problem btw

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

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

Very nicely done, Kurre!

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

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

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

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

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

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

**#9**

Let *R* and *S* be rings with multiplicative identities 1[sub]*R*[/sub] and 1[sub]*S*[/sub] respectively and suppose f:*R* → *S* is a function which preserves multiplication (i.e. f(*xy*) = f(*x*)f(*y*) for all *x*, *y* ∈ *R*). 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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