Art of Random Solving
by SomeonecoolLovesMaths, Jan 27, 2025, 10:29 PM
Here are a few of my random solves:
Quote:
In a grid of unit squares, a 
is defined as a pair of unit squares which share a vertex but do not share a side. For which pairs of integers 
can an 
rectangular grid of unit squares be tiled with cucumbers?
is defined as a pair of unit squares which share a vertex but do not share a side. For which pairs of integers 
can an 
rectangular grid of unit squares be tiled with cucumbers?
.
rows and 
.
rectangle and 
rectangle. By induction both can be filled using cucumbers and so can the whole rectangle. For the base case of 
many 
squares, which is possible since 
are odd. So there are 
squares. Each cucumber will fill 
squares and hence it cannot fill odd number of squares and hence we are done! 
.
.
number of blue squares. Each cucumber must occupy atleast one red and one blue square. Thus if the board is tiled using cucumbers then we get that number of red tiles 
number of blue tiles. A contradiction! 
Quote:
Determine all finite nonempty sets 
of positive integers satisfying 
is an element of for all 
and 
(not necesarily distinct in S)
of positive integers satisfying 
is an element of for all 
and 
(not necesarily distinct in S)
.
which gives 
is in the set 
be the largest odd number in the set. Then 
is in 
and is the least even element in the set after 
.
which is our original claim. 
is even 
(or else 
)
.Quote:
The numbers 
and 
are prime and satisfy
![\[\frac{p}{{p + 1}} + \frac{{q + 1}}{q} = \frac{{2n}}{{n + 2}}\]](//latex.artofproblemsolving.com/7/3/5/735ec9416a654f40121021bdd1233ffa86e42495.png)
for some positive integer 
. Find all possible values of 
.
and 
are prime and satisfy![\[\frac{p}{{p + 1}} + \frac{{q + 1}}{q} = \frac{{2n}}{{n + 2}}\]](http://latex.artofproblemsolving.com/7/3/5/735ec9416a654f40121021bdd1233ffa86e42495.png)
for some positive integer 
. Find all possible values of 
.
Ofcourse 
And expanding gives 
Ofcourse 
Since 
and 
.
can be 
.
.
work.Quote:
Let 
be a sequence of primes such that 
and for 
is the largest prime factor of 
. Prove that 
for any .
be a sequence of primes such that 
and for 
is the largest prime factor of 
. Prove that 
for any .
,
,
.
such that 
has its largest prime divisor as 
.
.
,
.
.
thus 
.
for some 
,
.
But 
,
denotes the largest expononent of ISL 1977 wrote:
Let 
be two natural numbers. When we divide 
by 
, we the the remainder 
and the quotient 
Determine all pairs 
for which 
be two natural numbers. When we divide 
by 
, we the the remainder 
and the quotient 
Determine all pairs 
for which 
.
then the maximum value of 
will be 
.
.
.
. By Jacobi's two square theorem,
.Quote:
Find the greatest possible value of 
.
.
Thus, 
Quote:
For how many value of the 
is a perfect square number.
the 
is a perfect square number.
gives us, 
.
.
is a perfect square too!
.
.
.
and 
and 
.
and 
.
.Quote:
Find all natural numbers such that for each of them there exist 
primes such that these terms satisfy.
and 
are primes.
natural numbers such that for each of them there exist 
primes such that these terms satisfy.
and 
are primes.
.
.
.
.
and 
is a prime hence 
.
.
.
and 
.
and 
are both primes.
which means that 
.
.
or 
,
.
.
i.e 
,
Thus, 
or 
This post has been edited 8 times. Last edited by SomeonecoolLovesMaths, Mar 6, 2025, 2:58 PM
March 2025
February 2025