Відео

The more difficult version is the set of such sequence of 1's is never a perfect prime power

If n-m must be multiple of 100 (i.e. n-m≥100) and also m≥1, doesn't that mean that min(n+m) is for n=101 and m=1, thus min(n+m)=102? And obviously 101=n≥3, so that also checks out in this case, so isn't the answer 102 instead of 106 or am I missing something?

Лемма Титу.

Keep in mind "an Asian" Is solving this!

Given the condition that y≥4 at 2:43 then the corresponding solution at 3:30 should indeed be (y,z)=(2t,3t), where for t∈ℕ, but only for t>1 (for t=1 => y=2<4). But one might then think "but (x,y,z)=(1,2,3) is obviously a solution". Yeah, but it's a result from the solution found at 6:46, not the one at 3:30

I think was a very easy imo problem.

The expansion of (a(x+1)+b)(cx+d) at 5:30 should have a bcx term instead of the bdx term shown in the video

do i have any problem for this solution? since proven f is 1-1 and f(x+1987)=f(x)+1987, then f(x+1987)-f(x)=1987, hence f is a linear function (since the difference between every set of (x,x+1987) must be a constant), which x>0. Then, since f(f(x))=x+1987, m(mx+b)+b=x+1987, which we can get m=1,-1(rejected) and b=1987/2. considering that previously we know x>0 and f defined in Z>0, then looking back to the f(x)=x+1987/2, which firstly f is not integer for integer inputs, and it is also not defined in [0,1987/2). Therefore, we can claim that no such function.

I've created 2 years ago a shortcut for such math problems: √[n (n+1) (n+2) (n+3) + 1] ------> for 4 consecutive numbers + 1 5n + (n -1 )^2 --------------------- 5 * 1000 + 1000000 - 2000 + 1 = 1000000 + 3001 = 1003001

LHS must be even assuming it's not an even prime (but p=2 works too) so then RHS is odd, meaning it's congruent to 0 mod 4. Therefore p^3 + 1 = 0 mod 4 so p must be 3 mod 4, and primes must be 1 or 5 mod 6, so then p must be 11 or 7 mod 12. The only primes generated by this are 7,11,19,23,31, and 43 and checking yields p = 7,11 (it's annoying if you don't have a calculator but very doable with some mod checks + divisibility checks, eg if it's congruent to 0 modulo by n and not n^2 where n is not a perfect square then it cannot be a perfect square)

I've the easiest and shortest method. It can be solved in a minute or two.

How about using weirstrass inquealitiy considering a2a3......an=1 but its not given that a2a3........an<1

bro is insane

ua-cam.com/video/7-GkLcIZqiQ/v-deo.html outstanding!!!

Where is 1/659?

I didn't grab a finer point: how you went from 2^(x-1) congruent to 0 mod 8 to y+1+2^(x-1)=7*2^(2x-3). Can someone explain why k must be 2x-3, not 2x-1 or 2x+7 and so on?

Another approach without using modulo and log (this is "no-pen-no-paper" method): 1. the last digit of x must be 9, because only 9 goes in cycles (9,1) with the 13th power being 9. 2. 100^13 is 27-digit number, so x must be 2-digit number ----> 26(digits) : 13 = 2. 3. x can't be 99, because the given 26-digit number is too small (it's obvious). 4. 8 = (2^3)^13 = 2^39; let's calculate 2^40 5. 2^10 = 1024, let's use "1000" 6. 2^40 = approx. 8^13 = approx 1000 ^4 7. 80^13 = approx 1000 ^4 * 10^13 = 10^12 * 10^13 = 10^25 8. It is less than the given 26-digit number 9. therefore x = 89

How would you choose the squares if you got something like m^4 + 3m^3 + ... with the cubic term

I got lost on the sumation part (note im finishing calc 1 so xd)

Two cool facts about 13th powers is that a) the answer is approximately 13 times the length of the number; and b) that the last digit is always the last digit of the number we’re after… Given this, we can confidently say this is a two digit number ending with 9 (26 digits long, very convenient!), given 100^13 has 27 digits! The digit sum is not divisible by 3 or 9, so that leaves 39, 69 and 99 out… Another trick to know is that for 13th powers the second to last digit (which we can call x) can be used to determine the second to last digit of the root… in the case of a digit ending with 9, the value of the root’s second to last (technically first) digit would be 7(x-2) mod 10… equal to 7 x (6-2) mod 10 = 8… that would mean in our case that the 13th root of our number is 89…

Can someone help me with where 2p<=q+6 comes from? Thank you!

Vieta jumping and we done

And why did you add u? When you can easly 1²-2×1×√2+1+(√2+1)²

ทำไปแล้ว

Hk?

(3,4,5) and (3,5,6) and (4,3,5) and ( (4,5,3) and (5,3,4) and (5,4,3) is Solution. => n! Is Solution for n= 3, 3!=6 => For one Solution => 6 add. Total 30 solutions combined

Couldn't A1 = B ? Because if it could, then we don't have a contradiction.

My solution : If you subtract 1 from a multiple of seven , that number will either be odd or factor into an odd * even . As 2^n is always even * even we can never have this relation .

My solution : If you subtract 1 from a multiple of seven , that number will either be odd or factor into an odd * even . As 2^n is always even * even we can never have this relation .

This solution is not good.

I just learned that we can think "critically" in 6 minutes. In school, critical thinking was often associated with "long thinks".

It is not correct and your english is very bad. You did not solve this problem.

An ancillary result you "proved" here was that the order of an integer mod n divides. Phi(n). Nice job. People will consider subscribing to your channel if they like the content. No need to beg a the beginning and end for subscriptions.

Discriminant is so overpowered on math

B cannot be equal to a due to them being integral Hence root b less than root a 2 root b less than root 2009 B less than 2009 / 4 Now u do the thing done in the video and u get a = 2009 + b - 2 root 2009×b Hence for (a) to be integer b = 41 k sq 41k sq less than 2009/4 Hence k = 0 1 2 3 So here we get all solutions in which a greater than b Same we do for b greater than a and get all the 8 solutions

pourquoi 1979 ne divise pas b ?

Here's a homework: a^b + b^c + c^a = (abc)^abc

wow! this is rizztastic!

Does my solution work? (a+2b+3c+4d) is less than or equal to (a+b+c+d+a+b+c+d+2d)=(2+2d). Since 2d is at most 2*0.25, (a+2b+3c+4d) is less than or equal to 2+0.5=2.5. Now (a^a * b^b * c^c * d^d) is less than or equal to (abcd)^d since a smaller exponent would increase a number between 0 and 1. Also note that, (a^a * b^b * c^c * d^d) is less than or equal to (a)^(a+b+c+d)=a. Now, this is exactly equal when a=d or d=0.25. So now we can consider (abcd)^d with d=0.25 because that is when it is largest as per the previous inequality. By the AM-GM inequality, (abcd)^0.25 is less than or equal to (a+b+c+d)/4=0.25. Finally 2.5*0.25=0.625 so it's less than one.

wow pairing the terms up is genius

4:23 ish I think you have to consider the fact that the lim as t approaches 0 u approaches + OR MINUS infinity. Don't think this matters because that limit would still approach 0 after evaluating the other case so I think the result is still 0 but I think you need this to be rigorous.

were you an IMO candidate?

Since it's only integers, calculate f(f(n+1)) and get f(n+1)-f(n)... Who knows what you could possibly get... something recognisable, by any chance ?

14 mins just to study that number in 3 different mods, kind of a waste of time

Wow thats so cool.

I dont get this isnt there a function as in f(x)= x+993.5 ? f(f(x)) would be x+993.5+993.5=x+1987 ?? Please explain this to me and thanks

Isnt it possible to first ignore the irrationality condition, then find all functions for all reals such that f(ab) = f(a+b). The only such function satisfying this is f(x) = c, a constant. So it also satisfies for any irrational a and b since the irrationals are a subset of the reals. ?

Very good

"so yay we are done" - every math olympic when they finish a problem