huh
by batteredbutnotdefeated, Apr 10, 2010, 11:51 PM
9.
Prove that there does not exist an integer which is doubled when its initial digit is transferred to the end.
Prove that there does not exist an integer which is doubled when its initial digit is transferred to the end.
= 
,Teh Double Facepalm
by batteredbutnotdefeated, Apr 8, 2010, 10:55 PM
I double face-palmed at the trivialness of this problem:
8.
Given distinct integers
such that 
has an integral root 
, show that 
DOUBLE FACEPALM http://www.offresonance.com/wp-content/uploads/2009/04/double-facepalm.jpg
8.
Given distinct integers
such that 
has an integral root 
, show that 
are all distinct factors of 
.
,DOUBLE FACEPALM http://www.offresonance.com/wp-content/uploads/2009/04/double-facepalm.jpg
A helpful theorem
by batteredbutnotdefeated, Apr 7, 2010, 11:25 PM
7.
Let
be positive integers which are pairwise relatively prime. Show that there exist 
consecutive integers 
such that 
divides 
for 
.
Let
be positive integers which are pairwise relatively prime. Show that there exist 
consecutive integers 
such that 
divides 
for 
.
.
such that:
are all consecutive, the 
are a set of r+1 consecutive integers each of which that satisfy the necessary conditions.Lesson: Problem Solving Principle
by batteredbutnotdefeated, Apr 6, 2010, 11:03 PM
6.
Show that the algebraic sum of any number of irreducible fractions whose denominators are all relatively prime to each other cannot be an integer.
In this proof, I will demonstrate the use of the problem solving principle "solve a smaller version of the problem":
Show that the algebraic sum of any number of irreducible fractions whose denominators are all relatively prime to each other cannot be an integer.
In this proof, I will demonstrate the use of the problem solving principle "solve a smaller version of the problem":
:
such that:
must equal some multiple of 
and 
must equal some multiple of 
,recursions
by batteredbutnotdefeated, Apr 5, 2010, 3:44 PM
5.
Let
denote the number of regions formed when 
lines are drawn in the Euclidean plane in such a way that no three are concurrent and no two are parallel. Show that 
.
too many induction problems.
Let
denote the number of regions formed when 
lines are drawn in the Euclidean plane in such a way that no three are concurrent and no two are parallel. Show that 
.
the recursion is fulfilled, so this will be our base case. Now for our inductive hypothesis, we assume that the recursion is satisfied by 
.
is true. To do this, we will split the recursion:![$ P_{n+2} = [\frac{(n+2)^2+(n+2)}{2}+1]$](http://latex.artofproblemsolving.com/5/1/3/5137397092d2f7e1bb96d1ce01a546296a8c4b99.png)
(the bracketed part is from repeatedly using the recursive formula)
- 
,
.too many induction problems.
Strong Induction
by batteredbutnotdefeated, Apr 4, 2010, 3:45 PM
4.
Prove that every positive integer greater than 1 can be expressed as the product of primes and 1.
Prove that every positive integer greater than 1 can be expressed as the product of primes and 1.
.
.
and some positive integer 
which is less than 
.
and 
.A lesson on induction [problem 3 in disguise]
by batteredbutnotdefeated, Apr 3, 2010, 10:11 PM
Anyways, although I fail at math I'm going to give a brief description of induction and its application in mathematical proofs. Hopefully you'll find this useful.
Basically, the principle of induction is simple:
Let
be an integer and 
be a statement about for each greater than or equal to .
If:
(1) P(a) is true
(2) And for each integer
[greater than or equal to ], 
implies 
true, then is true for all integers [which are greater than or equal to ].
This is somewhat of a difficult way to put it, so here's induction in its bare bones:
(1) Say some proposition P(a) is true for some arbitrary integer a [for example, let a be 2 and let P be
(2) Then assume P(k) is true, and attempt to manipulate P(k) to show that P(k+1) is true
here's an example:
Prove that the equation
has a solution in positive integers (x, y, z) for natural number values of .
Basically, the principle of induction is simple:
Let
be an integer and 
be a statement about for each greater than or equal to .If:
(1) P(a) is true
(2) And for each integer
[greater than or equal to ], 
implies 
true, then is true for all integers [which are greater than or equal to ].This is somewhat of a difficult way to put it, so here's induction in its bare bones:
(1) Say some proposition P(a) is true for some arbitrary integer a [for example, let a be 2 and let P be
(2) Then assume P(k) is true, and attempt to manipulate P(k) to show that P(k+1) is true
here's an example:
Prove that the equation
has a solution in positive integers (x, y, z) for natural number values of .
has a solution in positive integers. Now multiply both sides of the equation by 
to get: 
.
has a solution in positive integers. Again, multiplying by 
.
January 2012
January 2011