Could you make the Singapore math Olympiad team?

[@kingarth0r]

I know someone on the Singapore olympiad team. She had already studied algebraic geometry as a 16 year old. Needless to say IMO kids are very smart.

[@MyOneFiftiethOfADollar]

I know someone on US Olympic team. He studied remedial math long enough to realize he better win a medal so he can monetize his fame.

[@carstenmeyer7786]

7:13: If we let *f(p) := 103 - ⌊103 / p⌋* we can also eliminate most primes by checking intervals such that *f(p)* remains constant: *p ≧ 104: f(p) = 103 ≢ 0 mod p* *103 ≧ p ≧ 52: f(p) = 102 = 2 * 3 * 17 ≢ 0 mod p* *51 ≧ p ≧ 35: f(p) = 101 ≢ 0 mod p* *34 ≧ p ≧ 26: f(p) = 100 = 2^2 * 5^2 ≢ 0 mod p* In each case *f(p)* only has prime factors outside of the interval for *p* , so prime *p* cannot divide *f(p)* . The remaining primes we check directly (intervals are getting small-ish): *p: 2 3 5 7 11 13 17 19 23* *f(p) mod p: 0 0 3 5 6 5 12 3 7* The only two solutions are *p ∈ {2; 3}*

[@JM-us3fr]

Very organized. How did you do your math symbols?

[@carstenmeyer7786]

@@JM-us3fr Search wikipedia for "Mathematical operators and symbols in Unicode" and use copy&paste. I always keep that tabbed^^

[@JM-us3fr]

@@carstenmeyer7786 You can just paste the unicode directly in U+220x symbols?

[@JM-us3fr]

@@carstenmeyer7786 That didn't seem to work

[@carstenmeyer7786]

​@@JM-us3fr Weird, for me it does: U+220x *∀∁ ∂ ∃ ∄ ...* What I do: 1. Mark the symbol 2. Copy to clipboard (CTRL+C) 3. Paste from clipboard (CTRL+V) Do not be tempted to click on the symbol - that does _not_ copy it to clipboard (as one might expect), but rather opens an explanation page.

[@demenion3521]

It's actually pretty easy to spot the solutions p=2 and p=3 just by guessing (obviously that wouldn't prove that these are the only solutions). For p=2 you just sum the first 103 integers and since 103+1 is a multiple of 4, this sum is even. For p=3 you have to sum the first 103 squares, but using the summation formula it's still pretty easy to show that the sum is divisible by 3

[@Dan-cw8xu]

The steps make sense once you see what to do. Utterly amazing!

[@keithmasumoto9698]

98/7=14=0(mod 7) so 7 is also ruled out.

[@ayushbaransen5205]

P=7 does not work as there are 14 terms in expansion which is divisible by 7 and the rest terms are 1mod7 so clearly 7 must divide 103-14=89 which is wrong.

[@bobzarnke1706]

More generally, if the upper limit of summation is N then the above proof can be slightly reformulated to show that prime p is a solution either if p² ≤ N and N ≡ r (mod p) with 0 ≤ r < p and N ≡ r(p+1) (mod p²); or if (p+1) | N and p² > N.

[@goodplacetostop2973]

15:08 Homework 15:17 Good Place To Stop

[@CglravgHRjsksgS]

I really enjoy the video! You had a long time to make such a video...

[@isuckatcodm3638]

It would have been great if u had posted this 2 days later as it will be our national day

[@jamesfortune243]

That is a well posed problem that tests both understanding and problem solving skills.

[@HershO.]

Loving the new animations!!

[@DOROnoDORO]

Very nice problem, loving the visuals!

[@threstytorres4306]

Before answering the Hw in 15:08, i just want to mention that == and ≠≠ notates to congruent and incongruent respectively If p= 7 we have => 103 (==, ≠≠) Flr(103/7) mod 7 => 103 (== ≠≠), 14 mod 7 => 89 (==, ≠≠) 0 mod 7 => 89 ≠≠ 0 mod 7, Because 89 is NOT divisible by 7 EDIT: Thus 7 is not a solution

[@abrahammekonnen]

1:12 Maybe I'm wrong but it seems like a-b is divisible by m? Or is it both a-b and b-a? But that would imply that a and b are multiples of m? What am I missing? Oh I'm actually right. I just need to also consider that a and b can have the same remainders.( I think)

[@abrahammekonnen]

Interesting problem. The p=7 case doesn't work because 103 mod 7 = 5 and the floor of (103/7) = 14 and 14 mod 7 = 0.

[@lucachiesura5191]

very good! i have no other ideas!

[@SuperYoonHo]

This is a Good Place To Start...

[@ibrahimkachal6759]

why is the last term of n that is divisible by p is this floor term

[@General12th]

The overlays are great! My only complaint is I think they're a little hard to follow if what's on screen and what you're saying don't line up. Right now, what you're saying won't be drawn until several seconds later and that throws me off. Maybe other people like it more.

[@sharpnova2]

what are you talking about?

[@General12th]

@@sharpnova2 I call the animation that plays starting at 0:45 an overlay. Do you have a better name for them?

[@caesarinchina]

@@sharpnova2 I think some might not see it depending on how they watch a YT video (if it's in a browser on a phone, i.e. not on an official app, overlays are most probably not shown)...

[@MyOneFiftiethOfADollar]

How do you resolve your complaint on your math channel? One way would be to put all the remainder basics right in the middle of what you are trying to prove. Oh, but that would clutter the screen too much?! I will check your channel where you have obviously solved your concern.

[@playgroundgames3667]

There are 53 prime numbers between 103 & n

[@youtubeuser6581]

Where do you find these questions?

[@haziqthebiohazard3661]

Nice to see my home country being featured ^-^ My answer would be just 2 & 3

[@s4623]

9:44 writing + while saying minus 🤣

[@natepolidoro4565]

I don't think he needs to explain modular arithmetic every video he uses it.

[@leif1075]

Whybwiudl anyone think of dividing up.tje sum.the way he does here..why notnjust divide at the midpoint 103/2 or divide into 1 ton1p2 and then just the last term 103?

[@ojasdeshpande7296]

See what the little theorem says

[@lgooch]

Fermats little theorem

[@leif1075]

@@lgooch still doesn't really explain why he would break it up like that though..if you think about it..

[@lgooch]

@@leif1075 he’s breaking it up like that to separate the different remainders he gets modulus p.

[@isaacringling3823]

he must be smarter than me because i have no idea what he's talking about, im just going to leave and leave a like on my way out because he probably deserves it

[@bradhoward]

no

[@antonryzhov]

p=2 doesn't work, because the sum from 1 to 103 is odd.

[@threstytorres4306]

No, According to Wolfram alpha the sum as n= 1 up to 103 of n^(2-1) is 5,356 which is even

[@redhotdogs3193]

pretty sure that 5356 is even

[@bobh6728]

The sum of 1 to 103 is 5356, so p=2 works. From 1 to 103, you have 52 odd numbers and 51 even numbers. Adding two odds makes an even, so adding 26 pairs of odds gives an even sum.

[@Szynkaa]

dafaq

[@matthewleadbetter5580]

As a guess, you used the Gauss sum formula and didn't realise that 4 divides 104, not just 2.

[@chmjnationalsuperarmygener8564]

？

[@hallohello9885]

Did it in my head in 5 minutes. ezpz

[@aleksandervadla9881]

Ok

[@Tezhut]

Ok

[@ojasdeshpande7296]

Ok

[@yuseifudo6075]

Nobody cares

[@HershO.]

Loving the new animations!!