The one with the `higher power is almost always bigger.

You dindn't had to put +- when cancel √(m² + n²)² because m² + n² is necessarily positive

I pulled out my trusty 50+ years old pocket Dietzgen slide rule, took the logs of 4 & 5 and a couple of flicks to reach the conclusion that 5^(4311) < 4^(5311).

Que zero horrível do caramba, meu irmão. 😄

Nice mental math problem. Answer must be a bit less than 1000 (yielding a hundreds' digit of 9), with a ones' digit of 3 (working mod 10), and (casting out 9's) divisible by 3. That leaves only 993 and 963; but the latter is too small by quite a bit.

16¹⁸

Or just take the log of 20 multiply it by 8 and divide by the log of 2.

Первое число с двадцатью нулями, второе с пятнадцатью нулями. (2*10 и 3*5)

The conclusion x < 625 > 343 -> x > 343 that you are using in the end is wrong. e.g. x=0 : 0<625 (true) -> 0>343 (wrong)

asnwer=5<4 isit

a^b vs b^a If both sides are multiplied by 1/ab, a^(1/a) vs b^(1/b). Differentiating y=x^(1/x) by logarithmic differentiation, we see that it is maximum when x is e(Napier's constant). Therefore, it is larger to the 18th power of 16, which is closer to e.

How do you know the answer will end in a round number before you begin, and that it will end in a decimal?

Alternatively, you can use a complex-geometric approach. Let x=m-1 and y=1. That means, that x and y are 2 complex numbers on the same horizontal line in the complex plane, 1 unit apart, with x on the left and y on the right. Since x^6 = y^6, we know that that they x and y have the same modulus -- 6th root of the modulus of whatever x^6 is. Thus, x and y lie on the same circle centered at the origin. That is, we can think of x as y rotated CCW by some angle t, so that x = y*e^{it}. Since x^6 = y^6, this must mean that 6t is a multiple of 2*pi. There are 5 ways this can happen (technically 6, but the 6th way would imply that x=y), namely t = 2n*pi/6, for n=1,2,3,4,5. These 5 values of t correspond to the 5 answers, which you can solve geometrically using special right triangles.

asnwer=5>7 isit

2024-05-12 tHEOREMA N3 ||||3^3=3*3=3^(1+1)=[3^2]=[2+3]=5 ACADEMIC Marcelius Martirosianas Administrator , ESA ASTRONAUT, MasterASPIRANT EXpert General doctor Nobel, Fermatian Prize Felds Medalist....

ACADEMIC MARCELIUS ADMINISTRATOR ESA ASTRONAUT GENERAL DOCTOR EXPERT k^3=3^3]=[ K=3]=3^3=3^(1+1)=3^2=3+2=5 } Kai k=3 tada reikiniys juda taip ir rezult taip;==[5 ACADEMIC Aspirant Admistrator Marcelius Martirosianas

AcademiC Marcelius Martirosianas Talk edit Filter request protectinon 9 may 2o22 AcademiC HCM Bon 11 auguste ACADEMIC universitadi HUMBOLDT 2o17----2022 17 moksliu atradejas EXPERT General Doctor 2o24 05-23 ADMISTRATOR, et Aspirant, ESA Astronaut |||| 33^44=44^33 =77 ||||

2250.

3 ^ ( m^3 + 1) = 1 ( m + 1) ( m^2 - m + 1) = 0 m = - 1 is only feasible solution

there is a (slight) simplification due to craftfully selected numbers: 3&7, 7&11, 11&15 ( the numbers in pairs differ by four and then 7,11 and 15 also differ by four). Otherwise I could solve it by adding the fractions and essentially waste the same time.

In the end he used a calculator

Maybe 15s to realise the solution is 3^8, then I lost interest

Without so much ado: 5 E (m + 6) = 6 E (m + 5) (m + 6) lg 5 = (m + 5) lg 6 | lg = log to the base of 10 (m + 6 ) / (m + 5) = lg 6 / lg5 | lg 6 / lg 5 = a ~ 1,113 (m + 6) / (m + 5) = a m + 6 = a * (m + 5) m + 6 = am + 5a m - am = 5a - 6 m (1 - a) = (5a - 6) m = (5a - 6) / (1 - a) | ~ 3,827 5 E (m + 6) ~ 7.397.860,423 6 E (m + 5) ~ 7.397.860,423

I think he’s paid by the minute! Or by the anti calculator league. It’s all well and good for these contrived examples, but it would be better to show a technique that works for any numbers. So here you end up with 17787/266805k=150. Thus k=2250.

what's funny about the trick to this solution is that you can arrive at k/15 just as quickly if you had simply combined the 3 fractions under a common denominator

x={0,sqrt(2)}

And what is the "nice olympiad trick"? Programmer know the powers of 2. So 2^16=65536. Subtract 1 to get the result.

Very nice. You are a transformational master! 😅

If a calculator is allowed this one-liner would work as well: log(log(log(512,2),3),4)

88^89 will always be bigger. I'm guessing any pair of numbers above "e" will qualify.

Excelente, pela lógica, eu acertei, mas o seu método foi mais elegante.

Basically every developer knows how much is 2^16

It really is not that difficult to multiply 257 x 255. No need to break up like you did. 257 x 255 _________ 1285 1285 514 ------------- 65575

Simpler method will be Express 33^44 ,= 3^4=81 And 44^33 as 4^3= 64 So 81 > 640 So 33^44> 44^33 >

What a waste of time. All you did was make it more complicated than it started.

More elegant method If we can get both quantities with the same exponent, we can just judge which has the larger base. Therefore- Expand: 18^16 vs 16^2*16^16 Expand: 9^16 * 2^ 16 vs 16^2 * 8^16 * 2^16 Cancel Out 2^16: 9^16 * vs 16^2 * 8^16 Restate: 9^16 * vs sqrt(2)^16 * 8^16 = (8*sqrt(2))^16 Root 16 both sides: 9 < 8*sqrt(2)

easy 111110 in base 9

for the first time i was able to solve a problem lol :)

Clever.

asnwer=16>18 isit

That's a lot of calculation and difficult to plan. I replaced every 9^n with (10-1)^n And then opened everything with newton binomial (I don't know if that's the right name, English is not my first language) and I just had to sum all factors. And then we have just factors of power of 10 which is almost the answer

Sum all the pascal triangle coefficients for (10-1)^n for n=1 to n=5 (sign alternates) then do the sum of powers of ten. Way faster

9 - 7

We are stupid. We don't know the rules of exponents

this is your best question yet up until now! truly awesome

Nice trick, but the faster method is by analyzing qube root of 991026973. We know that the searched number must be 3-digit number, because 1000^3 (4-digit #) has 10 digits. Make 3 segments of 991026973 : 991 | 026 | 973, and let the searched number be in the form ABC . (ABC) ^3 -----> the last digit (C) must be 7, because the last digit of 7^3 is 3. C = 7 The first segment is 991, we are looking the perfect cube less than 991 -----> it must be 9, because 9^3 = 729 A = 9 so our whole number has the form like this : 9B7. What is the middle digit B? Ok, let subtract from the last segment of given number (973) the cube of 7 (343). 973 - 343 = 630, than consider only the middle digit 3. Now let's triple our C^2 (square of C): 3 * 7^2 = 3 * 49 = 147 than we are looking some factor x, which gave us the number with the last digit 3 (our considered earlier the middle digit 3). 147 * x = _ 3 7 * 9 = 63 So the middle digit B = 9 ABC = 997 * BTW, we should immediately know after analyzing A and C that B of our serached number must be 9, because 991026973 is very close to 10^9 and 1000^3 = 10^9.

3^8 = 3^4 x 3^4 3^4 =81 It is not that difficult to multiply 81 x 81.

997^3=991026973

9^5= 9*9*9*9*9= 55719 9^4= 9*9*9*9= 6451 9^3= 9*9*9= 729 9^2= 9*9= 81 9^1= 9= 9

Bullshit to the power of 9! The solution is 0.7833… But neither you nor I can calculate that without a calculator. You simply transformed the question into another question. And is it more simple? I don't think so. If I define simplicity by the number of keystrokes I need on my calculator, your 'solution' is more than twice as complicated as the original question. (5 vs 11) 9 1/x ENTER Shift y^x btw to understand the first sentence: 9!=362880