29,879 reputation
32794
bio website brilliant.org
location San Francisco, CA
age 29
visits member for 1 year, 7 months
seen 21 hours ago

Calvin Lin is the Math Challenge Master at Brilliant. He was born in Singapore, represented his home country at the International Mathematical Olympiad in 2001 and 2002, and trained the Singapore IMO team in 2005. Calvin studied economics and mathematics at the University of Chicago and graduated with a joint BA-MA in Mathematics in four years. While he was a student at the U of C, he continued training bright young mathematicians as an instructor for the Young Scholars Program for four years.


Jul
25
comment Last 7 digits of 7th powers
@res Thanks for filling in the details, I have removed my down vote. The original solution wasn't clearly expressed and I didn't understand the meaning of "But being $ϕ(10^7)$ coprime with 7-th every invertible element mod $10^7$ is a perfect 7-th power".
Jul
24
comment Mediteranean Mathematics Olympiad 2014 number theory problem.
FYI, I think that is a good start. I was continuing your chain of thought.
Jul
24
comment Mediteranean Mathematics Olympiad 2014 number theory problem.
@AlexanderVigodner There is a unique (up to permutation) solution to your system of equations. A necessary and sufficient condition for the solutions to be integers is that the discriminant is a perfect square.
Jul
24
comment Last 7 digits of 7th powers
-1 This explanation doesn't make any sense at all.
Jul
21
comment I am searching for an unusual real-valued function.
-1 I don't think you can "assume $f(xy) = f(x) f(y)$", but otherwise this is nice.
Jul
21
comment Websites for math tests/quizzes
You can check out Brilliant.org's practice section. We are building out the Geometry and Calculus sections, which should be out by the end of the month.
Jul
21
comment Find odd numbers $(o_1,o_2,o_3,o_4)$ such that $o_1^2-o_2^2=2(o_3^2-o_4^2)$ such that $o_1>o_2$ and $o_3>o_4$
There are solutions, like with $ p+q = r+s = 10$ and $ p-q = 4, r-2 = 2$, which gives us $p = 7, q = 3, r = 6, s = 4$.
Jul
20
revised N points and N perpendiculars
added 48 characters in body
Jul
20
answered N points and N perpendiculars
Jul
20
comment N points and N perpendiculars
@Semiclassical Potential gap is that "parallel" doesn't invert to "concentric". Also, you only go from a line to a circle if the center of inversion is on the line.
Jul
20
comment N points and N perpendiculars
How is it possible that "no two such perpendiculars intersect". Are these line segments, or lines? Otherwise, the perpendicular lines are not parallel, so much intersect.
Jul
20
revised Help with complicated functional equation
added 870 characters in body
Jul
20
answered Help with complicated functional equation
Jul
20
answered Finding the value of this product
Jul
20
comment Is this graph connected
Are you allowed to use negative integers, or is this graph only on positive integers?
Jul
20
answered Jordan chain when matrix has only one eigenvalue.
Jul
20
comment Nonlinear system of equations / factoring two-variable cubic over $\mathbb{R}$
@dmk Edited.......
Jul
20
revised Nonlinear system of equations / factoring two-variable cubic over $\mathbb{R}$
added 148 characters in body
Jul
20
answered Nonlinear system of equations / factoring two-variable cubic over $\mathbb{R}$
Jul
20
comment Nonlinear system of equations / factoring two-variable cubic over $\mathbb{R}$
Why can't we have $ u=v=w$ equal to some other constant? Since you said you used barycentric coordinates, do you want those to sum up to 1?