Відео

I got the answer correct.

I didn’t use a calculator.

2^25-2^24-3^2=16777207

4:09 Is there a difference between, x²+3x+1 (x²+3x+1) and (x²+3x+1)(x²+3x+1)? (x²+3x+1)(x²+3x+1)=(x²+3x+1)² x²+3x+1(x²+3x+1)=...?

9⁹⁰⁰(1 - 9) = -8¹9⁹⁰⁰ = -2³9⁹⁰⁰ = -(2¹9³⁰⁰)³

Much easier is to write 148 in binary and read the powers of 2 out. To do this choose the largest power of 2 that is less than or equal to the number you are trying to get to 148 = 128 + 16 + 4 = 2^7 + 2^4 + 2^2 = 2^2 + 2^4 + 2^7 = 2^a + 2^b + 2^c and there you have your answer: (a, b, c) = (2, 4, 7)

you can simplify the calculations by a lot by using (a+b)(a-b) formula, by choosing x the middle of all those numbers, i.e. x = 501.5 answer = a = sqrt(500 * 501 * 502 * 503 + 1) a^2 = (x - 3/2) (x - 1/2) (x + 1/2) (x + 3/2) + 1 = (x^2 - 9/4) (x^2 - 1/4) + 1 use y = x^2 to get a^2 = y^2 - (10/4) y + 9/16 + 1 a^2 = (16 y^2 - 40 y + 25) / 16 a^2 = (4 y - 5)^2 / 4^2 a^2 = ((4 y - 5) / 4)^2 a = (4 y - 5) / 4 a = (4 x^2 - 5) / 4 a = ((2 x)^2 - 5) / 4 a = ((2 * 501.5)^2 - 5) / 4 a = (1003^2 - 5) / 4 a = (1000000 + 6000 + 9 - 5) / 4 a = 250000 + 1500 + 1 a = 251501

Waaaaay tooooo loooooooong You can do it in your mind. Definitely not an Olympiads level question m^2 = 333^2 + 444^2 + 555^2 take the 111^2 common to get m^2 = 111^2 * ((3^2 + 4^2) + 5^2) Now we know the famous Pythagorean triangle 3, 4, 5, so 3^2 + 4^2 = 5^2 and 5^2 + 5^2 = 2 * 5^2 Now take square root of both sides, to get m = 111 * 5 * sqrt(2) = 555 sqrt(2)

I mean sure, it issolved that way, but what's beautiful about it? You are given two big numbers, you substract and get one number, that's still big.

We can assue a<=b<=c (otherwise, we can rename the varabes, so that this is given) Now we can tae out 2^a out of the sum: 2^a*(1+2^(b-a)+2^(c-a))=148=2^2*37 Since a<0b e hae the possibiities a=b and a<b. et us first oo at thhe case a=b: 2^a*(1+2^0+2^(c-a))=2^2*37 2^a*2*(1+2^(c-a-1))=2^1*2*37 Comparison of the exponents of 2 lead us to a=1 and therefor 1+2^(c-a-1)=37 2^(c-a--1)=36 But 36 is not a power of 2, so a=1 is possible (for whhole nummbers a, b and c). So we now a<b<=c and 2^ a*(1+2^(b-a)+2^(c--a)=2^2*37 (1+2^ (b-a)+2^(ca)) is an odd number, so compaisoof exponents of 2 on both sides gives us a=2 and 2^(b-2)+2^(c-2)=37--1=36=2^2*9 Now we can use the same trick:.. 2^(b-c)*(1+2^(c-2-(b-2)))=2^(b-c)*(1+2^(c-b)=2^2*9 Again we oo at thhe 2 cases b=c and b<c. The first gives us no solution, so we get b<c and 2^(b-2)*(1+2^(c-b))=2^2*9 Since 1+2^(c-b)is odd, we get 2^b-2)=2^2 and 1+2^(c-b)=9 b-2=2 and 2^(c-b)=8=2^3 b=4 and c-b=3 b=2 and c=7 So the complete solution is a=2, b=4 and c=7. Chec results in 2^a+2^b+2^ c=2^2+2^4+2^7=4+16+128=148 Why do you rewrite 2^(b-a) to (2^b)/(2^a)? it is unneccesarry. Just tae out 2^(b-a) outt of the su, and ou get 2^(b-a)*(1+2^((c-a)--(b-a)))=36 The first factor is a power of 2, the second factor is odd, so we get 2^(b-a)*(1+2^(c-b)=2^2*9 and than 2^(b-a)=2^2 and 1+2^(c-b)=9 b-a=2 and 2^(c-b)=9-1=8=2^3 b=a+2 and c=b+3 Togetthher withh a=2, we have a=2, b=2+2=4 and c=b+3=4+3=7

I did it in my head.

Input sqrt(500×501×502×503 + 1) = 251501 Result True 251501

10²⁰-10¹⁵=10¹⁵(10⁵-1)= 99999 * 10¹⁵

another way 2ᵏ = k³² (2⁸)ᵏ= (k³²)⁸ (2⁸)ᵏ= k^2⁸ *k = 2⁸*

2ᵏ = k³² kln2 = 32lnk (1/k)lnk = (1/32)/ln2 (1/k)ln(1/k) = (1/32)/ln(1/2) (1/k)ln(1/k) = (1/2⁵)(1/2³)8ln(1/2) (1/k)ln(1/k) = (1/2⁸)ln(1/2⁸) *k = 2⁸*

9^900-9^901= 9^900- 9^900х9^1= 9^900x( 1-9)= 9^900x(-8)

2:05 - 4:45 99999•10^15= =99,999,000,000,000,000,000

Input 3^15 + 3^12 + 3^9 + 3^6 + 3^3 = 14900787 Result True

Not the only solution k=1.0224

I thought the dot was a multiplication 💀

The answer will be (-9).

m=e^W(ln8) , m=~ 2.38843 , test , m^m=2.38843^2.38843 , 2.38843^ 2.38843=~ 8.00006... , ~same , OK ,

m =-0.5 Now if the exponent had been a ODD number...

Can’t you do 9^11= 9*9^10>>>then take the tenth root of 9^10 & 10^10 9 * 9 > 10 Is this right?

Where are you trying to take us?

if we can use a calculator (like you did) in the end, we can use it at start. Olympiad and maths competitions don't use calculators

n^4=(n-2)^4 n^4-(n-2)^4=0 (n^2)^2-((n-2)^2)^2=0 (n^2-(n-2)^2)*(n^2+(n-2)^2)=0 n^2-(n-2)^2=0 or n^2+(n-2)^2=0 n^2-(n^2-4n+4)=0 or n^2+(n^2+2n+4)=0 4n-4=0 or 2n^2-4n+4=0 4n=4 or n^2-2n+2=0 n=1 or n=1+-sqrt(1-2) n=1 or n=1+-sqrt(-1) n=1 or n=1+-i One real solution (n=1) and 2 complex solutions ((1+i) and (1-i)).

M=16, the other value is not a solution. the only valid value for m is 16.

Whenever e<a<b, then a^b>b^a. It isn't too hard to prove. In particular 333^444>444^333.

333⁴⁴⁴ V 444³³³ |^1/111 333⁴ V 444³ ; (3•111)⁴ V (4•111)³ 3⁴•111⁴ V 4³•111³ |÷111³ 3⁴•111 V 4³ ; 81•111 > 64 => => 333⁴⁴⁴ > 444³³³

A clear explanation. Thanks.

m^3-m=4080 , by faktoring , m^3 |+/-| k*m^2 - m -4080 = 0 , trial k=16 , // 16*255=4080 // , m^3-16m^2 +16m^2-256m +256m-4080 = 0 , -> (m-16)(m^2+16m+255)=0 , m-16=0 , m=16 , for complex , m^2+16m+255=0 , solu. , m= 16 , 8+i*sqrt191 , 8-i*sqrt191 , test with a real root , 16^3-16=4096-16 , 4096-16=4080 , same OK ,

Let manipulate 333/111=3 And 444/111=4 So 333^444 vs 444^333 Is now 3^4. Vs 4^,3 81 vs 64 81>64 So 333^444 > 444^333

Very good

It's easier to do 2x2 22 times than it it is to do 2049x2047

= 2^22

(99^99)/(98^100)=0.02787896816236845386910835026972961216542857538576698302465399525908188870739254622994362028919709741499044295556839530849475965260905202318260110837523472838331266050016934650372565159944633080269772199640391894299030501270893851192314396865628056625944419175702908793065679219903897985499907623116194989413198538748214909240869689028929765964412065178932697076180954230173647161656821970403217934252782986337046316197051013723642025457449726651842045963874618708778041217882100224614685413749387905561425318699340943198829134036134015670833071525115504108660752287735061601523380757199496276369932737804923987757472673261885465673479818183716447159457621284050656031215601316987989966796656108707755829388397823355109894429621609448240456944026388872827258793885908406661270961920349675563158235412944010435174859896648134950745710034926745413126112120828329296283108585385894545889607656179038122058168040715363581517661006524124598991666142023378905012223089945744330063598468282432742386257208007694710400901580486433286536145815196274125351318215442415576376879726621384118745002016403978983910533957185429320066587565521146868677080765725699416211061304396175525640853481404747192710119058612483651480200385194780158634273023886260492644519270122942179034880896952139832507546522690136562900729256895438619064891039735346585321568532531862687121078014989704115494571054427793327284405471886754791418918423163624651839592775081535505351697694312095732053516379212706223055123371780443770658475645523758087525165027216307573845639986851622586298315625…

I did it in my head.

99^99<98^100

98^100=13261955589475318753308980958435182616922905831015351444050494533312509010756375490958520845863725759086387552664739214251402218000635896969959420366494800836445520178642939314417801334276044804325376

99^99=369729637649726772657187905628805440595668764281741102430259972423552570455277523421410650010128232727940978889548326540119429996769494359451621570193644014418071060667659301384999779999159200499899

Good effort, Christian! It is true that x must be a complex number. However, your equation has an infinite number of roots because your -1 can be represented as exp(iπn) where n is any odd integer.~~~~Arthur Ogawa

I did it in my head

Input 2^22 - 1 = 4194303 Result True

you need to learn complex numbers in polar form. then, the solution is 1 line.

Bro, triky and lengthy calculations. ok.good.

10^10/(9^10)=(1+1/9)^10<e<3.< 9^11/(9^10)=9 ie10^10<9^11

interesting indeed!

m^3=3^3 sqrt3(m^3)=sqrt3(3^3) m=3 the easiest asnwer in the world, it's not olympiad equation

Excelente!!!