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Revision as of 18:53, 2 January 2024
Asymptote is fun!
Contents
Problems
See if you can solve these:
1. (Much easier) There is one and only one perfect square in the form
where and
are prime. Find that perfect square.
2. Suppose there is complex values and
that satisfy
Find .
3. Suppose
Find the remainder when is divided by 1000.
4. Suppose is a
-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are
roots, say
. Suppose all integers
ranging from
to
satisfies
. Also, suppose that
for an integer . If
is the minimum possible value of
.
Find the number of factors of the prime in
.
5. (Much harder) is an isosceles triangle where
. Let the circumcircle of
be
. Then, there is a point
and a point
on circle
such that
and
trisects
and
, and point
lies on minor arc
. Point
is chosen on segment
such that
is one of the altitudes of
. Ray
intersects
at point
(not
) and is extended past
to point
, and
. Point
is also on
and
. Let the perpendicular bisector of
and
intersect at
. Let
be a point such that
is both equal to
(in length) and is perpendicular to
and
is on the same side of
as
. Let
be the reflection of point
over line
. There exist a circle
centered at
and tangent to
at point
.
intersect
at
. Now suppose
intersects
at one distinct point, and
, and
are collinear. If
, then
can be expressed in the form
, where
and
are not divisible by the squares of any prime. Find
.
Someone mind making a diagram for this?
User Counts
If this is you first time visiting this page, change the number below by one. (Add 1, do NOT subtract 1)
Doesn't that look like a number on a pyramid?
Answer key & solution to the problems
I will leave a big gap below this sentence so you won't see the answers accidentally.
dsf
fsd
Answer key
Here:
1. 049
2. 170
3. 736
4. 011
5. 054
Solutions
Problem 1
There is one and only one perfect square in the form
where and
is prime. Find that perfect square.
Solution 1
.
Suppose
.
Then,
, so since
,
so
is less than both
and
and thus we have
and
. Adding them gives
in some order. Hence,
.
Problem 2
Suppose there are complex values and
that satisfy
Find .
Solution 1
To make things easier, instead of saying , we say
.
Now, we have
.
Expanding gives
.
To make things even simpler, let
, so that
.
Then, if , Newton's Sums gives
Therefore,
Now, we plug in
.
As we have done many times before, we substitute to get
.
Note: If you don't know Newton's Sums, you can also use Vieta's to bash.
Problem 3
Suppose
Find the remainder when is divided by 1000.
Solution 1 (Euler's Totient Theorem)
We first simplify
so
.
Hence,
Now, you can bash through solving linear congruences, but there is a smarter way. Notice that , and
. Hence,
, so
. With this in mind, we proceed with finding
.
Notice that and that
. Therefore, we obtain the system of congruences :
.
Solving yields , and we're done.