Art of Problem Solving: 2020 AMC 10 A #22

[@zhushiroll]

can you give me your eyebrows please :(

[@sipdip238]

"i like throwing away fractions" - Richard Rusczyk(while wiggling eyebrows)

[@RedOctober_]

poor fractions what did they ever do

[@IJustAteAChineseThrowingStar]

@@RedOctober_ They’re foul creatures that are annoying..

[@aldricbenalan4755]

Nice Work Mr. Rusczyk!!

[@yong-hwanpark749]

I am watching this in South Korea. Thanks.

[@sallyxu4668]

One day this going to be a chapter mathcounts CDR problem...

[@NobleNeon111]

How many t-shirts does the grand mathematician have now?

[@gobbledegook2327]

Really enjoy your work. Rusczyk.

[@danielwang8833]

How am I suppose to get all of this during the test?

[@yashuppot3214]

Ideally u shouldnt have to test to see such an obvious pattern

[@RedOctober_]

1. get a time machine 2. kidnap mr ruszcyk from the future 3. tell him he has to tell you all the answers or else you will make him do 100! by hand 4. get 150/150 flawless

[@firstlast7994]

@@yashuppot3214 well no, you have ample time to test a little then find the answer. That's what I did

[@vinh8022]

@@yashuppot3214 What's the ideal method in your opinion?

[@RaZorasiangamer]

@@vinh8022 a bit late here but with enough experience, you can quickly see that if the expression is in between two consecutive multiples of n, then they will all round down, thus creating a multiple of 3.

[@某中二且沙雕的闪电君]

Bro Rusczyk pls come back...

[@derekmiller823]

Cool Solution, Rusczyk

[@rp1968]

Nice example of using inductive reasoning

[@patrick-cc4yu]

Anybody know what the cutoff will be?

[@alexwang982]

google it.

[@newkid9807]

@@alexwang982 pokyu

[@spacehead1au936]

Love the videos!

[@ericj4094]

So it seems that illegal contraband AMC 10As are out already. (Hasn't been 24 hours yet in my time)

[@ariwang8613]

It's AoPS. They're not going to post an AMC 10 related video until its discussion is allowed.

[@williamhu9567]

i wish he were my math teacher (the eyebrows are fire)

[@AlexandrBorschchev]

I didnt understand the end.

[@johnfabrycky4332]

For a number with prime factorization x^a*y^b, to choose a factor, we can choose x^c where c=0, 1, 2,..., a and y^d where d=0, 1, 2, or..., b. There are a+1 choices for exponents for the X factor, and b+1 for the y factor. So, in total there are (a+1)*(b+1) factors. Hope that helps.

[@eric4334]

HI RICHARD U ARE THE BESTTTT

[@wallyx4541]

geeky captain america