Russia | Math Olympiad Question | You should know this trick!!

Math Olympiad Problem

[@rchatte100]

Logically, the power is always the most powerful part of a number.

[@OblomSaratov]

It usually is, but it's not always the case. For example, 4^4 > 3^5.

[@GHOST-RIDER-0]

2¹ > 1∞

[@OblomSaratov]

@@GHOST-RIDER-0 that's not true because infinity is not a number and 1∞ is undefined.

[@hafidmostarhfir2245]

Only if u are powering numbers greater than 1 ..I think

@@OblomSaratovthen 1^999999

[@karmakamra]

I am nowhere near fluent enough in math to calculate this problem. I went with 49^51 being larger based on one simple thought.... If you multiply 50 by itself 50 times you get a number that is astronomically huge. If you then multiply 49 with itself 50 times you get a smaller number that is also astronomically huge. But that smaller number will then get multiplied with 49 one more time, which will be a much bigger number. Then, to test my idea I tried smalle numbers. 10^10 is 10 billion. 9^11 is more than 3 times higher.

[@normalone9199]

Same here 🎊

[@alessiosandro123]

The thing is i guess that with higher potencies it gets bigger

[@HashiRa248]

Sometimes the journey is more important than destination.

[@zdenekbina6044]

I thought the opposite. 50 is larger base, so you can calaculte the 1.02 to the power of 50. If its larger than 49, its bigger. If not 49^51 is bigger.

[@zdenekbina6044]

And that is 2.69 so it supports your output. You got it right. My methid confirmed your result.

[@chesfern]

The simplest solution would be to let 50=n, 49=n-1, and 51=n+1. Taking logarithms of both numbers: Log n^n = nlogn and log (n-1)^(n+1)=(n+1)log(n-1). For a large number n, log n is approx. =log (n-1). Therefore, we are left with 2 values n and n+1 and obviously. n+1 is greater than n, which is why 49^51 is bigger than 50^50.

[@jackwilson5542]

The easiest solution is comparing 4^5 vs 5^4. There is no reason why the trend won't continue with higher numbers.

[@wolf5370]

@lson5542 That doesn't match the pattern in the Q. 5^5 vs 4^6 would.

[@konglink3359]

​@@jackwilson5542 but then u have to prove that it's true for all numbers, with concrete evidence

[@tontonbeber4555]

@@konglink3359 only if numbers are > e ... that's the only condition. It doesn't work for 2 and 3 for instance ...

[@grapefruitsyrup8185]

​@@tontonbeber4555 but why e? That's so specific how do you figure the explanation?

[@emreyukselci]

A simpler solution: 49^51 / 50^50 = 49 x (49/50)^50 If take the square root which needs to be over 1 if 49^51 is larger we get: 7 x (49/50)^25 . And (49/50)^25 > 49/50 - (25x1/50) as each power of 49/50 will reduce the number less than 1/50. So (49/50)^25 > 24/50. Then, 7 x (49/50)^25 > 7 x 24/50 > 1. As a result 49^51 > 50^50 .

[@user-qo5jo2qc5q]

Now this is a great solution, the video's solution is less elegant as it depends on students having memorized the definition of e. This solution however only relies in algebraic manipulation. It was a bit hard to follow, I originally thought you were wrong there at the end, but upon further analysis, indeed you have proven it! Well done, thank you for sharing this solution, much better than the video's.

[@young4783]

Good one!

[@dragondompyd7171]

Maybe I don't understand it fully but as you said, we can only use the squareroot while holding inequality if the value (49^51/50^50) is greater than one. And since that is exactly the thing we are trying to prove, we can't assume that it is greater than 1.

[@santoshkumarvlogs3753]

Nice solution

[@alexbayan8302]

This is a very nice "high school" solution. That is to say you will get a high mark if you wrote this solution in a Math Olympiads. But conceptually there are simpler solutions;just binomial expansion.

[@ckshene7212]

No, you still have to prove (1+1/49)^49 < 3. This is because the limit may or may not be monotonically increasing or decreasing. Therefore, you have to prove the limit is monotonically increasing to justify your claim.

[@thanhquenguyen9462]

Agree. Without proving that we can doing the same way as the video to approve 4^4 < 3^5, but it not.

[@teenniche]

kzbin.info/www/bejne/nqK6nIyMacmShrMsi=773t5VKHSSulzbiH

[@vladpetre5674]

You have to analyze the function lnx/(x+1). This has a max value when x is somewhere between 3 and 4 (where its derivative is equal to 0), after that decreasing but staying > 0. So for any x>4 we have lnx/(x+1) > ln(x+1)/(x+2). Making x = 49 we have (ln49)/50) > (ln50)/51, thus 49^51 > 50^50.

[@epevaldon5421]

Oh i got head ache on math. Im so poor on math

[@romain1mp]

Thanks for that! The demonstration on the video is not acceptable (unless if the goal is only to get the most probable answer without caring about how you get there). Using a limit to justify an inequality is not sufficient at all! Your method (even though I didn’t verify it) is more rigourous..

[@theupson]

@@romain1mp the vid is easily rehabilitated; (1+1/49)^49 is immediately less than e because discrete compound < continuous compound (for positive r)

[@lizekamtombe2223]

​@romain1mp That would make the squeeze theorem wrong and also disprove Archimedes, strict inequalities are very useful proofs, but they have to be strict. "Equalities are for children, real men deal with inequalities!"

[@romain1mp]

@@lizekamtombe2223who talked about equalities here? I am just saying that if Lim(f(n)) is smaller than L … when n is close to infinity…. You cannot conclude that f(n=49) is smaller than L without more inputs… For example you need to demonstrate quickly that f is increasing function from a certain level p (i.e. p

[@mingwangzhong117]

We need to compare f = (n+1)^(n+1)/n^(n+2) with 1 for a large n. This expression can be rewritten as f = (1+1/n)^n * (1+1/n)/n. Then using the standard binomial expansion, we have (1+1/n)^n = 1 + {n}*1/n + {n*(n+1)/2}*1/n^2 + {n*(n+1)/3!}*1/n^3 + ... < 1 + 1 + 1/2! + 1/3! + 1/4! + ... < 1 + 1 + 1/2 + 1/4 + 1/8 = 3. Therefore, for a large n such as 49, f should be smaller than 1, i.e., 50^50 < 49^51.

[@nasabdul629]

Take the logarithm of both numbers. For numbers above zero, a > b if log a >log b. Taking log of both sides reduces the problem to 50 * log 50 which is between 84 and 85 51 * log 49 is between 86 and 87

[@michaelhartmann1285]

That was the first approach that crossed my mind, and a great deal simpler than the algebraic thicket the narrator lays out.

[@kanwaljitsingh3248]

Good solution

[@hrvat7770]

But how do you calculate log50 and log49 without a calculator, to come to the conclusion the right side us larger then the left one? I'm sure the point is to solve the problem without a calculator as otherwise you can just calculate both initial terms and see which number is larger 😉

[@justanotherguy469]

I'm not that well versed in mathematics. Is it a proof, though?@@michaelhartmann1285

[@srinathparimi33]

By using properties of log, we can write log50 as log 5 + log 10 and log49 as 2log7. Now, log5 and log7 values can be approximately substituted. log5~0.698 and log7~0.845

[@aakashanantharaman4037]

Thanks for the fun mental challenge, and sharing the lateral thinking and inference thinking that goes with it. I enjoyed watching your video ;)

[@MsLeober]

Нужно показать что 50^50

[@marlanivanovich1828]

Только, пожалуйста, исправьте: квадрат 49 равен 2401.

[@TheSimCaptain]

Its much simpler than the video. You find the 50th root of both numbers. That means you divide each exponent by 50 such that you get 50/50 and 51/50. Then you just calculate 50 to the power of 50/50. And then 49 to the power of 51/50. 50 to the power of50/50 is 50 to the power of 1 or just 50. 49 to the power of 51/50 is 49 to the power of 1.02 which is 52.966. Because the fiftieth root of 49 to the power of 51/50 is bigger, then that number must be bigger.

[@manny2092]

I like this answer already!

[@kaustubhprakash1273]

This is a good answer. However, you would need to show how 49^(1.02) = 52.966. For this you can do it by binomial expansion and it should be easy to do

[@mareshetseleshi2717]

Much appreciated

[@donmoore7785]

How did you calculate 52.966 - a calculator?

[@TheSimCaptain]

@@donmoore7785 Yep.

[@user-hz5ne2rl5e]

You need to set up an axillary function to analyse the rate of the fuction at a point. 50^50 ? 49^51 ln(49)/(49+1) ? ln(50)/(50+1). Then axillary f(x)=ln(x)/(x+1). Find the derivative and approximate extremum point. Maximum of f(x) is at x on (3, 4). f(x) at x=49 is monotonic and decreasing. Hence, ln(49)/50 > ln(50)/51 49^51 > 50^50 .

[@TheSoteriologist]

Finding an unnecessarily complicated, inelegant and difficult solution is not a sign that one should be a mathematician.

[@tassiedevil2200]

Since you appeal to recognising Euler's number as the limit as n-> infinity, you need to show that it approaches that limit from below to use it as a bound for finite n=49 case.

[@phajgo2]

which is actually quite obvious..

[@user-gk3on7xp7e]

@@phajgo2 No, it isn't obvious! Unless you have bounds of the accuracy. On the other hand, the result is correct because it is well known that the sequence (1+1/n)^n is strictly increasing.

[@phajgo2]

@@user-gk3on7xp7e exactly this is why I think it is obvious. For n=1, (1+1/n)^n = 2, for n=2, (1+1/n)^n=(3/2)^2=9/4=2,25. and at infinity we know it is e so it's approaching from the left. Isn't that enough?

[@tassiedevil2200]

@@phajgo2 - I guess that provided one adds @user-gk3on7xp7e insight and says "the limit as n-> Infinity is Euler's number and it is well known that the sequence is monotonically increasing" you'd deserve the marks? This rather makes it a general knowledge test in my opinion. One can use calculus to prove that (1+1/x)^x is monotonically increasing for x>0. I'm not convinced that listing the first few terms of a sequence is a proof, although you seem actually in furious agreement with @user-gk3on7xp7e and the "it's well-known" proof by assertion. I looked at raios of successive terms and didn't see a quick proof of monotonicity. Just my 2c.

[@mikaelhakobyan9363]

@@phajgo2 No, it isn't. A function can increase at first, then start to decrease, and after that move to it's limit.

[@Arunmsharma]

Use compound interest logic, you’ll do this much earlier. 50/49 is 1.02 approx raised to 50, will be much below 49, think of it as getting a 2% interest on your bank deposit. It will take maybe 25-30 years to double, and would max be 4 times in 50 years - hence 50/49 raised to 50 is def below 4, and when you divide by 1/49, it’s clearly below 1.

[@user-qo5jo2qc5q]

That's not a mathematical proof, you can have a gut feel (if you have worked a lot with interest) that your answer is right, but that's not what this question is about.

[@iviewthetube]

Thank you for explaining such a terrifying problem so calmly.

[@user-hb1vf6lo7p]

Spasibo. If you can't wait, quickly in python: print(len(str(50**50)), len(str(49**51)))

[@user-ec8ru7je7b]

why len? int же

[@user-hb1vf6lo7p]

@@user-ec8ru7je7b привет. len посчитает количество символов в каждом из представленных(чтобы не выводить большие числа). Какой из вариантов больше, будет более очевидно.

[@madankundu6035]

that's cheating... ha ha

[@gibbogle]

You didn't prove that (1 + 1/m)^m < lim (1 + 1/n)^n as n -> infinity. It's true, but you assumed it without proof. You need to show that (1 + 1/n)^n < (1 + 1/(n+1))^(n+1) for all integers n > 0.

[@tharock220]

That's a good point. Maybe take the derivative and show it's always positive? Either way it's a good explanation and doesn't require a calculator.

[@thomasdalton1508]

​@@tharock220The derivative is rather messy, so I don't think it will be easy to prove it is greater than zero, but I expect it is possible.

[@bumbarabun]

@@thomasdalton1508 I believe that derivative is n * ( 1 + 1/n ) ^ (n-1) and it cannot be negative for positive n

[@thomasdalton1508]

@@bumbarabun Why do you believe that? You are differentiating with respect to n, so you can't use the rule for differentiating x^n with respect to x. The variable we are differentiating with respect to appears in both the base and the exponent, so it is a complicated differentiation. It's like differentiating x^x, but worse.

[@bumbarabun]

@@thomasdalton1508 you are right, my mistake

[@karthik999x-narrowone8]

I don't care about what your teaching. I just came here cuz your voice is soothing and your handwriting is crystal clear.

[@zeroun92]

I took the natural log of both sides and saw that it was 50 ln 50 vs 2450 ln 49. This showed a clear difference. The other method suggested is a more general approach that I didn't think of. Works better in the long term.

[@texasaggiegigsem]

I did the same thing, but I like her approach without having to find the natural log, which I reached for my calculator for.

[@AbhishekChoudharyB]

Shouldn't it be 51 ln 49 How did u get 2450?

[@mda7763]

And if there’s anyone who knows a harder way to do this, the ball is in your court now

[@RikMaxSpeed]

Spot on, this was way over-complicated.

[@88kgs]

😂😂

[@Ben-pw3qe]

By using Log you come to the answer in a few Seconds; Log 50^50 = 50 Log 50 = almost 85 Log 49^51 = 2 x 51 x Log 7 = almost 86 So 49^51 is almost 10 times bigger than 50^50 😊

[@neiljohnson7914]

But you need to use a calculator. This problem asks that you come to a solution according to math principles, not raw calculations

[@user-bo5fi7os2p]

I knew this without even calculate anything lmao

[@chaplainmattsanders4884]

i don’t understand that, but I believe you!

[@vandemaataram2600]

Yes. Your solution is similar to mine.

[@teenniche]

kzbin.info/www/bejne/nqK6nIyMacmShrMsi=773t5VKHSSulzbiH

[@supergaga]

My intuitive solution is: Let x = 50, and apply natural logarithm on both sides, we have x*ln(x), and (x+1)*ln(x-1). Now look at two terms: (x+1) / x and lnx / ln(x-1), they are just gradient of functions y=x, and y=ln(x) respectively, with dx = 1. Obviously y = x has constant gradient of 1, while y=ln(x) has decreasing gradient (always < 1) approaching to zero. Thus we have (x+1) / x > ln(x) / ln(x-1), therefore, x*ln(x) < (x+1) / ln(x-1). so we have proved 50^50 < 49^51 in a very simple way.

[@stem2-orgayafelizardoiiiy.9]

By Modular Arithmetic for divisibility of 50 50^50 ___ 49^51 0 ____ (-1)^51 0 ___ (-1) 0 ___ 49 (since the remainder must be positive) 0 < 49 therefore 50^50 < 49^51

[@jarl3434]

I ended up here on an insomniac night (very counterintuitive) and this is just the perfect voice I needed to calm down and have a good night. Thx!

[@tonybantu1681]

Should have applied natural logarithm rule. Very easy... ln( 50^50 ) ...?... ln( 49^51 ). Take any log of both sides. 50x (ln 50 ) ...?... 51x ( ln 49 ) ---> 50/51 ...?... ln 49 / ln 50 ---> 0.98... < 0.99... (Reduced to numeric 2-decimal places on both sides). Or 50/51 < ln 49 / ln 51. Therefore: 50^50 < 49^51 No need for advanced formulas or calculus. Just logarithm rule and basic arithmetic. Someone was close to this but complicated the simplicity of basic power rule of logarithms with functions, extrema, etc. All exponential power comparisons are pre-calculus algebra or arithmetic. Variations, infinitesimals or their limits are unnecessary.

[@Aeyo]

Your voice was soothing and gave me peace while my mind was screaming inside

[@Tomaslyftning]

Divide both sides with 49^50. 50/49 is about 1.02 therefore we can compare 1.02 ^50 with 49. Rule of thumb: to double 1.02 we have to multiply it 35 times so 1.02^50 is less than 4. Since 4 is less than 49 we get the same answer as the video.

[@herotb221091]

Wow. I love it

[@Aut0KAD]

nice, using the rule of 72?

[@mujtaba21]

That's how I thought of it. You wrote it concisely 👏🏽

[@mujtaba21]

​@@Aut0KADyes

[@Btitude]

I found my think-a-like buddy. I thought exactly the same way applying the rule of 72. I thought no one else would be smart enough. You are really smart, mate.

[@Skaahn]

My simple approach to guessing, just simply the problem as (50)^1 = 50 and (49)^2 = 2401, so RHS will be bigger

[@crannogman6289]

Write 49^51 as (50-1)^51 and write out the first few (5 should do it) terms of the expansion. Use combinatorics to find the coefficients. Rewriting it in terms of 50^50 gets you to see its about 18*50^50.

[@larswilms8275]

I rewrote 50 as 49+1, since this gives only positive terms. But same reasoning. end up with it being less than 2*49^50 so the fraction would be less than 2/49 which is less than 1 so 49^51 is greater

[@chris8535]

Yea this is super obvious with just logic. 50^50 is less than (50-1)^51. Like 2 seconds

[@MauuuAlpha]

I did something similar

[@GarryBoyer]

The binomial expansion would be a good solution if it were more of a close call. But if you have done enough problems like this you can quickly recognize the limit to e and come up with an answer more quickly without doing much math.

[@EdwardCurrent]

I just started with small numbers and gradually incremented them. By the time you get to 5^5 vs. 4^6, the 2nd evaluation is outpacing the first, so you can safely assume the trend will continue for numbers up to 50.

[@JH-pe3ro]

Since I've lost access to my logarithm skills I took the approach of graphing the first terms in the series to observe convergence. At first it's unclear with 1^3=1 vs 2^2=2, but 2^4=16 narrows with 3^3=27, and again 3^5=81*3=243 vs 4^4=16*16=256. Intuitively, as n grows, the relative effect of the larger exponent will overtake the base value - 2^64 is much more than 64^2. Since the problem doesn't call for a more specific answer, it can end after plotting an estimate of the crossover. I plugged it in a graphing caculator to check and it's very obvious that before n=5 you've already crossed over.

[@billj5645]

maybe simplify this greatly- which is smaller- 3 to the power of 3, or 2 to the power of 4? (or 4 to the power of 2). 3 to the power of 3 is the smallest so must be a minimum point if you graphed these

[@catalinx7301]

It's simple. The exponent has a bigger influence than the base, so 49^51 is bigger 🤭

[@UltraStarWarsFanatic]

Well not necessarily, since 3^2 > 2^3... but in this case yeah, the answer is obvious.

[@landpro28]

Exactly! Took me 10 seconds to conclude that

[@catalinx7301]

@@UltraStarWarsFanatic 3 and 2 are small numbers. Exponential starts to grow after a while, so my logic is for numbers a little bigger than 1.

[@vandemaataram2600]

Take 'log', then the problem wil become too easy. 👍👍👍 But I think, the problem is of arithmatic. That's why we are having these complicated solutions.

[@RikMaxSpeed]

That looked like a very long complicated approach: I took a log on both sides and approximated ln(50) = ln(49)+1/49 (ie: first order derivative and taylor series). Way simpler!

[@thomasdalton1508]

If you are being rigorous, you would need to put bounds on the error in the Taylor approximation and show that they can't change the conclusion. You can certainly do that, but it gets a little messy. (Doing it the way in the video, you need to prove that (1+1/n)^n is monotonically increasing, which also isn't straightforward.)

[@atulyaroy8962]

You can see by binomial approximation that 49^51 is greater. Though I believe the gap is big enough that the error wouldn't matter. Since LHS will have factor of around 2 and RHS will have factor of 49 which is quite large.

[@sorinturle4599]

When the numbers are close, the exponent (power) beats the base (the number). In fact, the bigger these numbers are, the smaller number with the bigger exponent (only by one unit) can go lower and lower from 50% of the higher number and the exponentiasion will be higher. Of course, this is only the answer, not the demonstration. More rigurous, but still simple is using the logarithms.

[@atulyaroy8962]

@@sorinturle4599 well even at near x=0 there is crazy separation for exponential graph with higher bases so it makes sense differing by 1 in base doesn't matter as much differing by power by 1

[@MonsterERB]

Move everything to one side and you're asking "Which is bigger: (50^50)/(49^51), or 1?" Then that simplifies to [(50/49)^50] versus 1... which is "number very close to e, divided by 49" versus 1. Pretty obvious then than 49^51 is larger.

[@robertoguerra5375]

The ratio of the 2 numbers would be approaching n/2.72 Less than 1 for n>2

[@GoodChemistry]

I love these problems, great mental exercise! Thanks.

[@hectormata449]

I’m glad she’s not my introductory algebra teacher or I’d go insane. I may be ignorant on this convoluted mathematical solution but I just assumed the following which gave me the correct “guess” to the problem given. I worked out a simpler similar equation in my addled mind: of 4 squared vs 3 cubed, answer: 16 vs 27, therefore 49 to the 51st power is larger. 😱 👀 ⁉️ 🤔

[@wolf5370]

But its an Olympiad Q - you need to show it, not just be content you know the answer. Indeed, 4^4 > 3^5 which does not follow the seeming general rule n^n < (n-1)^(n+1) - and in you sample 3^3 vs 4^2 does not match the pattern in the Q: you have n^n (n+1)^(n-1).

[@browntigerus]

I had this question 35 years ago while living in Soviet Union. Did not use Euler, just natural logarithm. The same exact result.

[@thatonenoname]

Натуральный логарифм это логарифм по основанию е, так-что почти одно и тоже

[@oo_rahbel_oo]

Well done. It's pretty clever how you've managed to prove it

[@rahuldwivedi4758]

If you needed to apply Euler’s, why did you need to extract (1+ 1/49)^49* (1+1/49) Wouldn’t that hold true directly for (1+1/49)^50?

[@curaticac5391]

This is flawed like many such math "solutions" on KZbin. The fact that the limit of a sequence is < 3 does not guarantee that the terms of the sequence are equally so. Like other viewers pointed out.

[@mrcleanatemywife7045]

I guessed it was the 49^51 because powers mean more than single digits, and found both the answers on a calculator and subtracted them to prove it in about 10 seconds.

[@Chawlas57]

Your voice is very sweet to listen... Loving and enjoying your voice

[@TheThrakatuluk]

6:53 How did one < times one > times one < ended up as one < symbol? Does it only bother me?

[@varunkotwal2427]

One more trick is you can see what 5^5 is less than 4^6 by a good margin. With that logic a bigger number such as 50^50 would be less than 49^51

[@michaelguth4007]

Just don't make the mistake and use even lower numbers to make your point: 2^4 and 3^3

[@RAINBOWbelongsTOtheKIDS]

Math is exact science(c) 😎

[@user-qo5jo2qc5q]

Yup...if it was just so easy...look at this: 3+2 = 5, 7+2 = 9, 11+2 = 13...see? It looks to me that every number I choose when I add 2 I get an odd number...so n+2 is always odd!! I just proved it by this guy's logic :) @@RAINBOWbelongsTOtheKIDS

[@wavrekordz]

Ok this could be an intuitive answer, but it is so obvius to me that 49^51 is higher simply because there is one more multiplication in it. I’m pretty sure that 49^50*10 would be also higher.

[@thomasdalton1508]

You are correct, but that's just a lucky guess. 49^50*2 also has one more multiplication, but isn't higher.

[@wavrekordz]

@@thomasdalton1508 Intuition is not a lucky guess. I didn’t say that any multiplication is enough here. That could be *0.1 too, which is also a multiplication and obviously wrong. You can say that this is not an acceptable proof, but the luck has nothing to do with.

[@ashton.m]

It does seem intuitive cuz of that one additional multiplication. But it's not true for 3^3, and 4^4. From 5^5, the latter expression is larger So Is 50^50 > 49^51. Answer: No. Cuz m^m < m-1^m+1, where m=/>5 👽

[@thomasdalton1508]

@@Selendeki But, as Ashton pointed out, increasing the exponent by one doesn't always do the job - it only works for 5 and greater. I don't think it is at all intuitive that the threshold is 5. That requires doing the calculations. Just because your intuition gives the right answer doesn't mean it is good intuition. The exact same intuition would have led you astray for different numbers.

[@thomasdalton1508]

@@Selendeki Intuition certainly has its uses and developing your intuition is a very important part of learning mathematics. There are two problems with the OP's comment. First, it is bad intuition - there is no intuitive reason that multiplying more numbers should get a larger result than multiplying larger numbers. It depends on how many more and how much larger. And second, it is completely wrong to say something is "obvious" based on intuition. That just isn't how you do maths. You don't guess and then say your guess is obviously correct.

[@Gbhmagic]

It sucks how quickly you forget math that you don't use all the time 😢

[@SkydivingSquid]

I simplified this. Take the 50th root of both numbers.. 49^(51/50) vs 50... Simplified: 52.966 vs 50. 49^51 is bigger.

[@thelearningmachine_]

In school I was too lazy to do math, but I had a good sense of logic So I usually cheated the calculations with small numbers to have a guess what was going on with the math Something like this: 2^4 < 3^3 (16 < 27) 3^5 < 4^4 (243 < 256) 4^6 > 5^5 (4096 > 3125) inverted gap 5^7 > 6^6 (you could stop here because you already have a proof what is going on) 49^51 > 50^50 "teacher, I don't know how to do the math, but 49^51 > 50^50 for the logic reason listed above, the gap inverted and keeps increasing". "A" 😂. I did so much of this when I was a kid. I still remember there was a 5 question exam once and I did all 5 questions without a single math, only writing sentences explaining why the answer would be X or Y / True False/ how many how much. Teacher told me I "cheated" but still gave me an "A" because he had never seen a math test done correctly without any calculations, only with pure logic. Good memories

[@DesertObserver491]

Nice going. I wouldn't call it cheating. Rather it's unconventional solving, like MacGyver. Did you follow the syllabus? No. Did you nail the concept and solve the problem? Yes. I'd love to know what you ended up doing as work or hobby using these skills.

[@GolldLining]

You proved nothing in the rambling

[@yasserahmed-bg7qj]

​@@GolldLiningyes but if he continued with what he was doing and learnt mathematical induction he would've proved it

[@DesertObserver491]

Exactly. He was on the right track and in a multiple choice test, he would have got the correct answer.

[@lakshay3745]

Bro can you give an example of questions you did without solving which shocked your teacher

[@keanming99]

That’s weird. I had a hunch 49^51 is bigger than 50^50 because exponent is always much bigger than a base, without doing all the arithmetic 😂

[@Stepan_H]

But without proof, it's crystal ball gazing. Try 3^3 vs 2^4 ... 🤓 By your logic, the higher number should be the one with the higher exponent, but it's not... 😱

[@pc6985]

​​​​@@Stepan_HIt's better to have proof instead of assumptions, but it's also true that generally higher exponents mean high numbers overall. Yes, 3^3 > 2^4, but even just raising each base by 1 will already make the no. w/ higher exponent larger with 4^3 < 3^4. Keep increasing the bases by 1 after that and the number raised to 4 will always be larger. Even if you had a difference of 2 for the bases, with 4^3 > 2^4, increase the bases by 1 you'll still get 5^3 > 3^4, yes, but increase it one more time and everything starting with and after it'll always be 6^3 < 4^4 or n^3 < (n-2)^4 as long as n > 5. Proof is good, but we're in the comments, not a math comp. Most people here just want the answer, so might as well let them know a "trick" even if it isn't always true. These kinda problems aren't really encountered by the general public often anyways, and anytime they do it'll involve huge numbers with little differences in bases and exponents, at which point the number with the higher exponent is larger 99.9% of the time. By the very small chance it isn't, well it's just youtube so does it really matter? Haha

[@carolharris2401]

That's how I looked at it too. I didn't know how to prove it. But using common sense I figured if decrease the base by 1 and raise the exponent by 1. Then the 1 with the raised exponent is larger

[@jackmclane1826]

Higher power wins for all numbers > 5^5 49^51 is actually almost 20 times larger than 50^50. (bruteforced it by excel)

[@OblomSaratov]

Higher power doesn't always win. 4^4 > 3^5.

[@Delhi_Guy]

It is simple if you start comparing the series and see the difference growing.

[@banjo4us1]

Your solution is not suitable for Olympiad. You need to attempt it without using Limits.

[@PARPROX777]

It depends of year of olympiad. In USSR limits was part of school program.

[@banjo4us1]

@@PARPROX777 Nice ... Which year calculus is introduced. Russians make some great mathematicians

[@IGAgames]

we studied limits in 9th grade, it was 2015, but it is mathphysics school

[@user-zp3fm9ot8v]

​@@IGAgamesIn Russia every non math school teach limits in 10 grade

[@frakekera415]

i just compared 10^3 and 9^4. since 9^4 is bigger i thought 49^51 would be bigger

[@FranCarreira]

Exactly… I did something similar. 50^50 is n^n and 49^51 is (n-1)^(n+1) and then, as you did, changed the n for a much smaller number, so I could easily do the calculations, I chose n=2 so I had 2^2 in one side and 1^3 on the other and it appears that the second part is not bigger, but as n grows over 2, you go getting bigger numbers each time on the (n-1)^(n+1) side

[@arthurhairumian7179]

I solved the problem in 2 seconds with my intuition ...and the answer was correct, so it's true that imagination is more important than the knowledge - Einstein

[@skhadka2466]

Its so simple left side become (49+1)^50 then on simplification it becomes (49^50)+1^50 direct compare to righ side which is greater than left side.

[@Alhamdulillah_muslim313]

Aa..nice way👍🏻 I can solve it in 2 steps 🙂

[@antoniojunior936]

Eu assisti em outro idioma e entendi, por isso eu amo a matemática ❤

[@mikojan85channel]

So basically to know to solve this you should already at the first moment suspect, that the numbers chosen (49 and 50 and 51) in the question were similar to each other with a reason. The one who created this question had to chose similar numbers to make the question solvable.

[@TravelingMooseMedia]

Pretty easy. It’s easy to calculate in your head that 4^6 > 5^5. 4096 > 3125. So for any number greater or equal to n = 5, n-1^ n+1 > n^n.

[@septone]

I don’t know the proofing method I’d use but.. if I was asked what the comparison between the two number is, I would’ve gotten the same answer by a simpler means. The base number and exponent is off by one but one is exponentially more valuable. Therefore it will weigh more heavily. 2^3 < 3^2

[@EricPerreault]

Disclaimer: not a math person. Issue I see is that the left hand side is 50^50 (same same), so 2^3 vs 3^2 isn't a relevant pattern. Should be 3^3 vs 2^4, or 4^4 vs 3^5, which would give the wrong answer here. I think only starting from 5^5 vs 4^6 is left hand side lesser.

[@gatedscs]

I didn't understood step at 6:28 where you take 1/6 instead of 1/49

[@air9music]

I think it was totally unnecessary though it was just to make the inequality simpler by canceling using a multiple of 3. She should've just calculated (3 x 50)/(49 x 49) which isn't difficult and very evidently much smaller than 1.

[@Kiran2101]

Because (1 + 1/49)^50.... Is smaller than 3x 50/49x1/49. So,if that thing Is smaller,It has also to be smaller than 3x50/49x1/6.

[@Legendaryking1234]

The guy who solved it through hard vs the guy who guessed out of two options.

[@steveftoth]

I love how the proof is based on transcendental numbers.

[@jamesdoughty5530]

You are good, but it becomes so convoluted I give up before the answer. Is there an short cut to do the problem?

[@MathsMadeSimple101]

That’s math for ya

[@leonadordavinci2703]

There's no route only for nobles in math

[@ScorpioHR]

Why simply not divide on with the other and see what happens? (50/49)^50 * 1/49 and see if it's greater or less than 1 ? If it's less than one, then 49^51 is greater than 50^50. Now, eyeballing this, 49 is by 1 lesseer than 50, which is 2%, so 50/49 is probably around 1,002 -1,003. Do I believe 1.003^50 is greater than 49? I don't believe it's even greater than 2, so if I was a betting man, I'd say 49^51 is definitely greater than 50^50. (Since 1,003^3 / 49 is certainly lesser than 1) Now I can go and see what you did here in 8 minutes...

[@ScorpioHR]

@@English_shahriar1 OMG! I think you've just helped me realize I'm in an early stage of Alzheimer's :( But thanks, I guess....

[@TFKiller1]

Even if your answer ls right, You are wrong in some some aspects, so be careful in the future. 50/49 it is a difference of 2% SO it would be around 1,02-1,03. Then 1,03^50 is not greater than 49 but it is greater than 2

[@joshuavasquez9019]

freaking mic drop at the end there. sheesh. Super cool video, thanks for making it!

[@ducngoctd]

Chứng minh được mệnh đề tổng quát, (bằng phương pháp quy nạp toán học): n^n > (n+1)^(n-1). Với n = 50 là bài toán mà bạn nêu ở trên..

[@lagautmd]

This is much simpler to analyze. Make it easily envisioned by reducing the bewilderingly large powers. Get them down to human scale. 50^1 compared to 49^2. 50^1=50. 49^2 is bigger than 50 by inspection. Therefore, by induction, n^n < (n-1)^(n+1).

[@aniruddhshandilyak3289]

A good analysis But this works only when n>=5

[@user-zo1bv7hk1l]

right hand side is increasing at a slower rate than the left hand side, so at some point when n increases, left hand side should overtake the right hand side

[@OblomSaratov]

As a calculus fan, to solve this I analyzed the function y=(50-x)^(50+x), where y(0)=50^50, y(1)=49^51. Its derivative is y'=(50-x)^(50+x) (ln(50-x)-(50+x)/(50-x)). We only need to consider x on the interval [0; 1]. Now, to find the sign of y', let's do following estimations: 1) 0≤x≤1 => 50≤50+x≤51 and 49≤50-x≤50; 2) consequently, ln49≤ln(50-x)≤ln50; 3) e log(3; 49) ln50

[@user-pk3oq7ow4k]

I looked at both numbers and could immediately see that that 49^51st is larger. Visualize 50^50th as 50x50... all the way to the 50th one. Visualize 49^51th as 49x50th ... x the 51st 49. Clearly that will result in 49^51st being larger. You can even use a simpler model to prove it. 50 ^ 3rd vs 49 ^ 4th; 50x50x50 = 125k; 49x49x49x49 = 5,764,801. Notice that the difference between 50 and 49 is only 1. That seems to be true for any two consecutive numbers starting with the number 3.

[@k_research605]

I think you mean 5 and higher. (n^2 > (n-1)^(n+1))

[@GetMeThere1]

Thanks for doing this for "n." So (for future reference) we know that n^n < (n-1)^(n+1) for relatively large n. What is the lower cutoff (using integers), where the inequality sign switches? 4^4>3^5 but 5^5 < 4^6

[@zahariastoianovici8590]

This is an excellent problem and great way to resolve , did learn a lot

[@muntahajamil]

Excellent!! Your writing system of "9" is totally exceptional 😮

[@user-bh4rt9ng1q]

How did you reduced 49 as to 6?

[@louisgrateau]

She chose 6 to make it simple, as long as it was smaller than 49 it would have worked. In a sense, she gave it a try, to see if the whole equation would be smaller than 1 this way, if not, she could have tried bigger denominator, if still smaller than 49, because anyway, the whole reasoning with e is based on not having to be precise if the equation is smaller than 1. If it was bigger, equal to 1, or really close to 1, thing that we are not supposed to know at that moment, then the reasoning would be probably no conclusive, but it was worth giving it an easy try.

[@benji1775]

glad to hear someone pronounce Euler as I did in the past lol

[@bumpypants3241]

That's the correct way to pronounce it 😅

[@rasikparray5575]

I think there's no need for such a laborious approach... We can get this done in two steps Step one divide both terms and multiply and divide by 50 Step two (50÷49)^51 /50 is final expression which is clearly< 1

[@jwac3io]

Time management is also part of the test. The answer is C.

[@egogh6055]

What age group is targeted?

[@banjo4us1]

12 years

[@phajgo2]

@@banjo4us1 no way! at least 14-15. There is no way you can have that limes part at 12

[@banjo4us1]

@@phajgo2 Maybe you are right. However most Olympiad competitors start early. In our school, we started early coaching by 12. Most competitive exams prep starts by 10 to 12. I am talking about India and how most successful candidate crack exams.

[@phajgo2]

@@banjo4us1 I'm from Poland so there may be difference in programs :) we also start olympiads early (around 10-12 as you say) but they usually do not go beyond the program of math classes for the given age as the intention for children is not to incentivize learning extensive material before you're supposed to but it is rather to find smart solutions within the knowledge you have. Still I'm curious to know what age was that question intended for because I found it quite difficult :)

[@banjo4us1]

@@phajgo2 I agree with you. Most problem in these Olympiad could be solved either using higher theory or rudiment maths. I remember once there was a problem of a bird catching a fish and then perching on a tree. The problem could have been solved using Snell's law, but it could also be solved using Similar Triangles, albeit it is a longer solution. I always studied upto 3 years ahead so that I could solve these kind of exams. Even in IIT papers here in India.... most questions has higher maths solutions.

[@MathsMadeSimple101]

Great explanation

[@d8ngdeld8ng]

Same answer result. But it glaringly shows how Mathematician and an Economics and Finance pips answered this numerical logic query step by step relative to their learned principles.

[@GunjanSharma-nf2ce]

I went with binomial expansion by splitting 49^51 into (50-1)^51. Then in its expansion we have 50^51 with the additions of many more digits.

[@thenamedoesnotmatter]

I used simple intuition. It makes more sense to me that the (49x49 ... x49) falls behind (50x50... x50), however the extra instance of multiplying x49 accounts for all of the previous distance between those equations. If you are repeating something 50x, and 1x49 is just 1 less than 50. We haven't gotten far enough exponentially to create more than 1 digit of a gap. We know from multiplication rules it will go around 2x10^20 for either equation, but it just intuitively makes sense that 49^51 > 50^50.

[@bo-dg3bh]

when n goes to infinity, that thing becomes e, so when n is equal to 49, it’s smaller than 3. real good logic

[@Palisade5810]

Since 1/49 will diminish rapidly in the binomial expansion of (1+1/49)^50 you can approximate it to 1+50/49= 2+1/49 to the first order so (50^50)/(49^51) will be (2*49+1)/(49^2)=99/2399

[@sudeeptobaidya6558]

I am an indian and i did binomial expansion and did it in like 10 seconds.Just equate (50/49)^50 to 49 first and 50/49 is 1.02 ,which can be written as (1+0.02).After that just multiply 50 to 0.02 so tha answer will come 1+(50×0.02)=2 which is less than 49. So 50^50 is way way smaller than 49^51.

[@kgfes]

To compare which number is larger between 50^50 and 49^51, we can calculate their values. 50^50 ≈ 8.881784 × 10^69 49^51 ≈ 1.319507 × 10^77 Comparing the two values, 49^51 is larger than 50^50.

[@neiljohnson7914]

Her voice is so soothing i couldnt stay awake until the end of this video

[@maksimivanov5417]

The limit being smaller than 3 doesn't imply the specific term for n=49 is smaller than 3. The key part of the solution seems missing.

[@javanautski]

Interesting. If you use 4^4, that's > 3^5, but starting with n=5, we have n^n < (n-1)^(n+1). I think the left hand is Lambert's w-function?

[@notray2445]

It’s easy, man. For all x,y,a,b > 1: x^a > y^b if a>b AT ALMOST CASE

[@GMBeaulac]

I’m not sure why you back away from the numbers at the end. Sure it’s less than 1, intuitively one might have guessed that a small difference in powers will have more of an impact than a small difference in bases. But without that simplification you’d have an idea of how large the difference actually is. 3x50 (150) over 49 squared isn’t that bad; square 50 (2500) then subtract 50 to get 50x49 then subtract 49 to get 49 squared, or 2500-99 or 2401. It’s still not exact by any means, since the 3 is an approximation, but you can easily tell at a glance then that 49^51 is between 15 and 20x larger than 50^50

[@TheThaLime]

You can just see it by looking at the numbers, that's how exponentials work Just like you can intuitively see that 1045 is larger than 983.

[@JagatK_]

I knew the answer but i couldn’t stop the video in between because of that voice. That damn voice ❤

[@pengzhang-sy6zw]

wa oh! your voice is so good, It feels like I am hearing asmr. And it helps me to sleep.

[@somnath3986]

Just take log directly and then take the difference, (50log50-51log49) = (51log50 - log50) -(51log49) It'll give , 51log(50/49) - log50 The first part is so small as 50/49 is nearly equal to 1 thus ignore it. Then ans is negative of log50 which is negetive no. Thus 50log50 was smaller and 50^50 was smaller

[@user-vo2zo9jh3o]

I wrote the equation as: 50^50 / ( 50-1)^50 = 50-1, then, replace 50 by X. But still cant solve, it any advice?