Russian Math Olympiad Question

[@redfluxbluedawn414]

for those who want to know: 1.005^139 crosses value 2

[@user-nd7sf5ev3i]

1.005^138.9757216107 = 2

[@suneelsingh9580]

What standard is this question. Class 10, 12 or higher level

[@honeyvedant5732]

@@suneelsingh9580 this question is at least class 11 advance level maths...

[@LoisoPondohva]

​@@suneelsingh9580this was featured in 9th and 10th class olympiads.

[@mr.curious1714]

@@suneelsingh9580 this question is very easy. you just need to use log. do this: (1.005)^x = 2. taking log on both sides x = log (2) to the base of 1.005 this can be written as x = (natural log of 2) / (natural log of 1.005). this value comes 138.9

[@DucLe-kg5hx]

I'm not that bad at math but these questions always make me rethink if I'm actually stupid.

[@machtmann2881]

There are math olympiad questions for a reason though! It's not about knowing all the math rules, it's about being really creative with them which is super hard. So rest assured, you are not stupid at math :)

[@freshname]

Don't be hard on yourself. Those type of questions are not part of a school curriculum. They are aimed to find and reward those students who have a certain level of creative thinking and who are much much better then the school curriculum suggests.

[@Asiansxsymbol]

I was on my hs math team that competed with other schools, albeit a long time ago, and this is also a bit confusing to me. 🤣

[@nisbahmumtaz909]

if you are capable of having these introspective thoughts, chances are you've got a decent head on your shoulders

[@ysf-d9i]

I was 1 point away from making the Canadian Math Olympiad and USAMO and had perfect scores on other canada-wide math contests and constantly scored within the top 10-20 in my province for my grade back in my youth and I can't solve any olympiad questions either.

[@Bermatematika]

You can also use binomial theorem: (1+(1/200) )^(200) =1 + 200* (1/200) + something else positive > 2.

[@plasmaastronaut]

yes. this question can be answered by 16-17yo college level maths. Guess this Olympiad is for 12yo or younger.

[@GeekProdigyGuy]

for the engineer-minded, (1+1/200)^200 ≈ e >> 2

[@GeekProdigyGuy]

but seriously, from first principles, it makes more sense to take (1+1/200)x > x+1/200 given x>1 and apply 199 times to get 1.005^200 > 2 directly

[@terryarmbruster9719]

Lol was thinking e or binomial

[@КонстантинСоков-о1ф]

Ряд Тейлора, если не ошибаюсь?

[@HughvanZyl]

A quick, but very rough preliminary test that ended up working in this case, is: we know that the number grows by 0.5% each time, and we know that the smallest amount it grows is 0.5% of 1.005, which is a bit more than 0.005 (I don't think I explained that very well, but it's quite obvious if you think about it). So we know that the number will be more than 1.005 + (0.005 × 200) 1.005 + (0.5 × 2) 1.005 + 1 2.005 So by doing that, we already know that 1.005²⁰⁰ is definitely bigger than 2.

[@monokumannius4522]

Why x200 not x199? You have one for 1.005, the rest should be 199 times 0.005

[@HughvanZyl]

@@monokumannius4522 1.005 is raised to the power of 200, it is multiplied by its self 200 times, so the number grows 200 times by at least 0.005.

[@monokumannius4522]

@@HughvanZyl let's look it at it at simple way. 1.005^2 will be more than 1.005 + (1x0.005) not 1.005 + (2x0.005)

[@HughvanZyl]

@@monokumannius4522 u right, my b. The method still checks out though, we know the answer is greater than 2, instead of 2.005.

[@guerreromanuel]

I liked your form of thinking, very engineer-like. However, this approach does not work if we compare 1.005^150 to 2. Note that 1.005^x starts to be bigger than 2 at x > 139 (aprox)

[@thomasgreene5750]

By inspection, 1.005^200 is approximately 2.71 and therefore greater than 2. You can quickly realize this by noting that the exponential can be written as (1+1/n)^n where n=200. The definition of Euler's number, e, is the limit of this expression as n approaches infinity. Euler's number is 2.718.... By the time n reaches 200, the value of the expression is pretty close to e because it has already reached 2 at n=1 and 2.25 at n=2.

[@sassora]

Good observation, I did minimal binomial expansion: 1 + 200 * 0.5% + ... . The first two terms added already make 2, the remaining sum will only increase from this

[@thomasgreene5750]

@@sassora Yes. This type of challenge problem is popular on KZbin, and I have noted in the past that the authors usually pick the numbers such that the first two terms of the binomial expansion sum to 2 so that one can use a partial binomial argument. I noticed that when they do that they make the exponential a finite approximation to Euler's number.

[@t_c5266]

That is a good way to do it. But I found the quickest way was to type it into a calculator

[@danielf3623]

Yeah, (1+1/x)^x approaches e=2.71.... , so you just need to see how it approaches. x=1 => 2, x=2 => 2.25, so it's approaching from below, and is always above 2 except for x=1

[@BillRhye]

By inspection💀💀💀

[@HarshaKaujalgi]

This is a very nice and thorough explanation and I appreciate the steps shown here. For this one however, I find it easier to think of it in terms of simple and compound interest. (1 +0.005) ^200 indicates a 0.5% compounded rate 200 times. If this was simple interest, the 0.5 % increase would multiply with 200 to give a net interest of 1 which when added with the principal would give you 2. However since we are compounding, the final value MUST be greater than 2.

[@thomasgreene5750]

Your explanation follows the approach Bernoulli used when he discovered what we now call Euler's number. Euler's number is the amount you get in a year for investing a dollar at an annual interest rate of 100% with continuous compounding.

[@parapunter]

Just consider that "1 + (.005 * 200) = 2" corresponds to simple interest at a rate of 0.5% over 200 periods. "1.005^200" corresponds to compound interest at 0.5% over 200 periods. Compound interest is always greater than simple interest therefore 1.005^200 > 2 .

[@westcoaststacker569]

How I did it in my head also, same thoughts.

[@starpawsy]

Same here. I expanded only the first two terms.

[@leonovgleb8535]

Or, you may know "The second wonderful limit". (1+1/x)^x -> e . Every member of this sequence is greater than 2.

[@pietergeerkens6324]

Yes. A much more interesting question is a comparison to e rather than 2.

[@leonovgleb8535]

@@pietergeerkens6324 It is lower. The sequence mentioned above is strictly increasing.

[@echthroi9]

This is why I love math - explanations like this show how problems can be simple even when they're not easy.

[@timothychinye6008]

This is also why a lot of people hate maths... unnecessarily long answers to seemingly easy questions that no one needed the answers for in the first place.

[@vatsalkhandelwal8046]

​@@timothychinye6008exactly lol..just depends on how we look at it

[@andrewkarsten5268]

There’s a really easy answer using the binomial theorem/Bernoulli’s inequality. Doesn’t need to be this complicated

[@crazychicken8290]

This is so complicated did you see the amount of steps she took

[@iiREYteoii]

Isn't this problem easy by just using a calculator?

[@michiguel]

If we remember that ln(1+x) = x when x approaches zero, then applying ln to both terms we have 200 * ln (1+0.005) > ln 2 200 * 0.005 > ln 2 1 > ln 2

[@tomsxe]

or just use log2

[@legendexists4331]

Me who guessed the correct answer in 0.1 sec based on luck💀

[@DinsAFK]

I mean you can just know that for any number n greater than 1, n^x> n*x

[@Woodside235]

As soon as you wrote 1.005^200 as (1 + 1/200)^200, I figured it would be larger since (1 + 1/x)^x is a form of approximating e -- and that form is always greater than 2 (as long as x is greater than 1).

[@Woodside235]

@@Conserpov You misunderstand. 1.005^139 does not take the form (1 + 1/x)^x. I'm referring to a form where the decimal part of the base is equal to the reciprocal of the exponent.

[@Conserpov]

@@Woodside235 Ah, right. Guess I'm too old for this

[@ИгорьКулик-л2р]

Абсолютно точно. Замечательный предел.

[@vincenzomassara3840]

THAT’S EXACTLY WHAT I THOUGHT hahahaah solved in 1 second…

[@frentz7]

Excellent point. :)

[@RielMyricyne]

1.005^200 is a fairly good approximation of e. e is defined as the limit of (1+1/n)^n for infinite n.

[@jonsrecordcollection7172]

LHS is (1+1/x)^x evaluated at x=200, while RHS is (1+1/x)^x evaluated at x=1. Just use natural logs to figure out the derivative (1+1/x)^x & show the function is always increasing because the derivative is always positive. Thus, f(200) > f(1), which also implies 1.005^200 is greater than 2.

[@July-gj1st]

beautiful

[@ultronhere4356]

Nice

[@sierpinskibrot]

I like this one

[@XoIoRouge]

Now THIS is a great answer. It'll prove and show knowledge without jumping to default assumptions with "counting the number of fractions."

[@HashanGayasri]

This. And this reaches the limit of "e"

[@manloeste5555]

Very fast solution in 1 second: You can interprete (1 + x) to the power of n (where x < 1) as the rate of return x applied for n years. To calculate this fast, there is a rule called 72-rule. If you divide 72 by the "rate of return" in percent (here 0,5 because x = 0,005 which is the same as 0,5%) you get the approximate number of years until it doubles. 72 / 0,5 = 72 * 2 = 144. Because 144 is much less than 200 it's no problem that it's only an approximation (the exact number of years here is between 138 and 139)

[@manloeste5555]

Sorry, I just saw it's already mentioned in the comments 🙂

[@EvergreenWannabee]

That’s a good one! Not a scientific proof, although the rule of thumb gives the right answer

[@DanBurgaud]

2:00 nice! I like how you created this telescoping fraction! very very nice technique... awesome!

[@davidking4838]

To solve quickly multiple 1.005 * 1.005 one time. You see the result is 1.010 plus a little bit more (a little more than if you simply added .005 to 1.005). This tells you that doing this 200 times will, at a minimum, be more than 2.

[@haltux]

This. It's the first thing I tried and clearly once you see that it's game over. Still have to proprely write the demo but you know the path.

[@KDSBestGameDev]

@@haltux given that the non-compound interest of 1 by 0,5% over 200 periods is 1 + (0.005 * 200) = 2 and 1.005^200 is 0,5% compound interest over 200 periods and compound interest is always bigger than non-compound. 1.005^200 must be bigger than 2. Is a valid q.e.d. proof.

[@thadtheman3751]

Reformulate the question as is (1+1/20)^2^2^2^5^5. (1+20)^2 = 1+1/10+d where d is a small positive number. Rinse three times. Then do a fifth power twice. (1+x)^5=(1+x)*(1+x)*(1+x)*(1+x)(1+x)=(1+2x+d)*(1+2x+d)*(1+x)=(1+4x+d)(1+x)=(1+5x+d). In Each d you throw away is a small positive number so the final number you get (1+200*200) is smaller then the product.. You could even prove this indictively in a treivial way. When I started this problem, I wanted to try a taylor series with epsilon=005 and was fearful that final expresion would be something close to 2. Say just short 1.99998 or just over 2.0001. Iws so dispointed when I started to write out the problem and realized it was not even close. As an olympiad problem this was disappointing.

[@antonzhdanov9653]

Hah. It passes 2 somewhere around ^139. So if I did 2,5 or 3 you would be wrong.

[@davidking4838]

@@antonzhdanov9653 2.5 or 3 would, of course, change my calculations. I was only looking at the problem presented.

[@levistepanian5341]

2 is smaller because 1.005^200 is a representation of lim x->♾ (1+x)^(1/x) which is equal to e or roughly 2.7.

[@Bruh-bk6yo]

Yeah. (1+1/n)^n (or (1+n)^1/n) is an increasing sequence. All the members are in the [2; e] segment. So the 200th number of it is definitely bigger than 2.

[@Cjnw]

​@@Bruh-bk6yoe, the value when approaching infinity.

[@Bruh-bk6yo]

@@Cjnw no shit sherlock.

[@diegosanchez894]

Alternative approach: the derivative of x²⁰⁰ around 1 is 200. Using linear approximation, 1.005 should be equal to 2. However, since the function is convex, this is an underestimation. Therefore, 1.005²⁰⁰ >2

[@notsoancientpelican]

Just use the Old Rule of 72. That is: divide the interest rate or rate of growth into 72 and that is the number of years or periods or iterations which it will take to double the starting amount (approximately, of course…but close enough…). In this case the rate of growth is 1/2 of a percent, so dividing 72 by 1/2 means that the original amount of 1.005 will double after about 144 times-well short of 200. So, since the only question is which is larger, clearly 1.005pwr200 is larger. than the integer 2.

[@Abitibidoug]

That's a smart way to solve this problem. Why didn't I think of that?

[@TNMJAD]

This was my solution as well! Thanks high school Econ teacher.

[@Funtyao]

Вообще то через 139.

[@ILoveMaths07]

What? What's with the 72? What's going on?

[@ivanvanechok1679]

The Rule of 72 is a simplified formula that calculates how long it'll take for an investment to DOUBLE in value, based on its rate of return. (we are comparing 1 and 2 so its perfect for this case) The Rule of 72 could apply to anything that grows at a compounded rate. Ex. If the gross domestic product (GDP) grows at 4% annually, the economy will be expected to double in 72 / 4% = 18 years. In this case the growth is 0.5%. 72/0.5 = 144. Its way under the 200. Since 1.005^144 = 2 that means 1.005^200 > 2

[@DerangedAussieMan]

Always start off with a simpler version of the problem, and work your way towards the original problem. Which is larger: 1.5^2 or 2? Answer: 1.5^2 = 2.25 which is larger. How about 1.25^4 or 2? Answer: 1.25^4 = 625/256 which is like 2.4ish which is larger than 2. How about 1.1^10 vs 2? Well 1.1 multiplied by itself is already 1.21, and since it's "jumping" more than 0.1 each time, with 10 "jumps", it must cross over 2. So following the pattern, 1.005^200 must be more than 2. In fact, following the pattern even more, we see that (1+1/n)^n only gets larger as n gets smaller. And what happens as n approaches infinity? (1+1/n)^n approaches e, which is like 2.7ish.

[@DerangedAussieMan]

Mistyped: meant to say (1+1/n}^n only gets larger as n gets LARGER***

[@biscuit_6081]

This equation computes e for large numbers of n in (1+1/n)^n, precisely when n approaches infinity. Granted 200 is not a "large number" it should still be more than enough to be bigger than 2 since e≈2.71

[@sauzerfenicedinanto]

In reality it can be demonstrated that the sequence that defines the Euler number is always increasing, therefore the value increases as n grows and the minimum value is 2 when n is equal to 1; therefore given that the term n=200 of the sequence, appears in the problem it can be said that it will certainly be greater than 2.

[@Nikioko]

You don't need large numbers. Any n > 1 (and also non-natural numbers) leads to results > 2 for (n + 1/n)^n.

[@GeekProdigyGuy]

​@@sauzerfenicedinantoproving that sequence increasing would be the meat of the proof, so there's not much point other than having memorized it already

[@sebmata135]

You can use the taylor series expansion of x^200 centered at 1 to approximate 1.005^200. The first two terms are 1+1, and the remaining terms are positive and make the sum more accurate therefore it converges to a number greater than 2

[@spiderjerusalem4009]

Or just an application of bernoulli's inequality (which can easily prove whether the sequences (1+¹/ₙ)ⁿ and (1+¹/ₙ)ⁿ⁺¹ increase or decrease monotonically) (1+x)ⁿ⩾1+nx for all nonnegative integers n for all reals 1+x⩾0 Pf: It is obvious for x⩾0 by binomial expansion. Assume -1⩽x

[@Tryha4d]

True we were told this in highschool physics class as a way to approximate the higher power and small bases and I just used it lol

[@timerertim]

Now I am not bad at maths and achieved quite high rankings in my country when participating in competitions, but I would have probably never thought about this that way.

[@Triggered_Carefree]

What would you have thought

[@endhalf]

"fuck"

[@filippocernuschi6715]

probably because there is an easier and faster way, that is to write it in the form of (1+1/x)^x equal (in this case) to (1+1/200)^200... since lim(x->inf) (1+1/x)^x = e>2 and since 200 is a good enough approximation of infinity (cause its a big number) you can guess that 1.005^200 > 2...

[@ptewans]

I don’t believe that you achieve high rank in your country without ability to see that LHS = e

[@mr.p1759]

@@ptewans they said they would have never thought of solving this problem with this method, doesn't mean they don't know other methods to solve this

[@TheFinalRevelation2]

simply exand it as 1^200 + 200 (1/200)+ other terms (using binomial expansion) Or since each 1/200 is multiplied by a 1 and addeded 200 times .

[@robertobokarev439]

Chinese and Russian walk into a bar and start discussing the math problem... **video begins**

[@peterdecupis8296]

A basic topic in any Calculus Course is the study of the succession converging to the Neperus constant e, i. e. S(n) =(1+1/n)^n, with n Natural. The first part of the convergence proof is relevant to its strictly crescent behavior; since for n=1 we obtain S(1)=2, for any n>1 we have S(n)>2: in our case we have to evaluate S(200), i. e. n=200>1, then S(200)>2.

[@georgelang5113]

You can log both sides and log(1+x)~= x for x in a small neighborhood of zero. So the log of LHS is around 200*0.005=1, meaning that LHS is around e>2.

[@nihalbinu]

I was thinking the same. Taking log would make the solving time lesser.

[@Aleblanco1987]

its even easier to use the property log (x^n) = n log (x)

[@rasmusturkka480]

It's even easier to use a calculator

[@OlegLecinsky]

As a physicist I didn't bother myself even with that - my hunch was that the first number was larger, so that was my working hypothesis, and it was proven later to be the correct one!

[@Cline3911]

Came for the equation, stayed for the beautiful handwriting.

[@dicerson9976]

The way I thought about it is that each multiplication is a .5% increase of whatever the value is at that point. In video games, it is often called a multiplicative or exponential increase- with the contrasting form of increase being an additive increase (where all the "multipliers" are added together as a single flat multiplier. If you have two 2x multiplicative multipliers, that ends up being an overall 4x increase since each multiplier is applied separately one after the other; but if they are additive multipliers than you would add the increased together before multiplying anything at all- since a 2x multiplier is an *increase* of 100%, two additive 2x multipliers would be an increase of 200% or a 3x multiplier). Point is to say that for any given set of multipliers, if they are applied multiplicatively (/exponentially) the end result is higher than if they are applied additively. In this particular case, .005 happens to be precisely .5%, or 1/200. So if you were to apply 1.005^200 as an additive multiplier you'd end up with a +100% increase (or a 2x multiplier). Thus, since it is trivially true that any given set of multipliers applied multiplicatively is higher than the same set applied additively, it is also trivially true in this case that 1.005^200 is higher than 2- since ^200 is an exponential operation and an additive form of that operation would be precisely 2. If the question were something like 1.004^200, or 1.005^150 or something like that, then the above logic would not work and it'd be a much more difficult question for me since i'd have to dig through my brain for more "proper" mathematical principles rather than relying on common game knowledge xd

[@nayer2060]

Sounds like something a PoE player would say

[@checkybox]

whole ass essay

[@imlimecake]

LHS is the element of the sequence (1+1/x)^x, which is increasing and converges to e. It is enough to see that at x=2 we get (1+1/2)^2=9/4>2, and at larger x the sequence will only increase, so that 1.005 is greater than 2

[@July-gj1st]

Using something someone used earlier, consider the function f(x) = (1 + 1/x)^x and notice that this function is e as x tends to infinity. Therefore, for all x > 0 as x tends to infinity it will yield a better approximation of e which is greater than 2. Thus, since 2 = (1 + 1/1)^1 = f(1) and 200>1, f(200) should be closer to e than f(1) is. Therefore, f(200)>f(1).

[@gregoryk_lite]

It's hard to make this a proof though, because the definition of limit doesn't guarantee that the convergence will be fast.

[@andrewkarsten5268]

@@gregoryk_litecorrect. To make it better, you should show the series (1+1/n)^n is monotonically increasing, and is already greater than 2 at n=2, and hence the 200th term must be greater than 2.

[@July-gj1st]

@@andrewkarsten5268 This, and yes, my bad.

[@TheSecurityAgency]

The methodology chosen in this video recalls me an event, back in 80th some dude tried to replicate in Basic my code written in Assembly.

[@hippophile]

(1+a)(1+a)(1+a)... (1+a) 200 times is going to expand to 1 + 200a + some smaller terms. If a = 0.005 that is 1+(200*.005)+ smaller terms = 2+ smaller terms. Or you could simply use the binomial expansion if you can remember it!

[@pradpradprad1]

What was the expansion called man... I remember reading it somewhere years years ago

[@sdaiwepm]

Lovely clear handwriting. Incredibly ingenious solution!

[@itz_killer_queen]

As someone who was taught that . means thousand and , means decimal, I see this as an absolute win.

[@adriani9432]

Indonesia moment

[@NazarikShev]

Russia moment

[@pekkapaco5333]

"I don't care about the math" answer: (0.005 x 200) + 1 = 2. With compounding, it's greater.

[@govindgupta3805]

For those who want to know 1.005^200 = 2.7115

[@scramptha5949]

I used the investing rule of 72. 72 / 0.5 is roughly 140, meaning that 0.5% growth rate will double in roughly 140 years, so 1.005^140 ~ 2. Therefore 1.005^200 is definitely larger than 2.

[@evenh4210]

When expanding to the 200 factors of (1+1/200), you could realize that you could write it out by first multiplying all the 1s, (obtaining 1), then strategally using all except a single 1, and obtain 1/200. You have 200 combinations to acquire 1/200, which yields 1. The rest of the terms are smaller, but positive, so you end up with 1 + 200* 1/200 + (some positive number), which gives more than 2. (This would be easier to illustrate as a video, bit the point is that you have a shortcut.)

[@blymark83]

Well said!

[@geraldvalverde]

The real question here is: if I don't like math, suck at it, and has nothing to do with my job, why am I watching this video on a Thursday at 1am?

[@Abhismash]

You explained it beautifully. I wish I had a math teacher like you in my school

[@SillyPerrin]

I agree, but I don’t really like the accent tho

[@scottpeters1847]

You're joking right?

[@olivergroning6421]

A simpler way in my opinion is to use the binomial theorem according to which (1+1/n)^n=1+n*1^(n-1)*(1/n)+additional_terms=1+1+additional_terms>2 because all the additional_terms are positive and non zero (for n>1). For n=1 there are no additional_terms and accordingly (1+1/1)^1=2

[@johnterry4595]

A simpler answer is to write 0.005 as delta then perform a Taylor's series in delta to second order. The first term in the expansion is 1 and the second term in the expansion is 200!/1!/199!*delta = 1. The third term gives 200!/2!/198!delta^2. Just by tracking the sign you can see that it will give a positive contribution

[@orangenostril]

Simpler as in fewer steps I guess, but this is just canceling fractions which I think most people find easier than Calculus lol

[@kreslokkd]

Bro, just for your info, this question is meant for 7 graders, maybe 8 graders... I studied at math school in Saint Petersburg, we prepared a lot for different olympiad problems and we had to participate in every possible math olympiad and I personally recognized this question and instantly got Vietnam flashbacks from like 7-8 grade. So I guess Teylor's series wasn't the solution designed for this problem by the authors. I guess they would still count it, but they certainly didn't rely on the fact that 7-8 graders would know that.

[@thedarksideofnature1221]

I solved the problem in 10s using my calculator. I'm an engineer.

[@ЧувакИзКосмоса]

The 1.005^200 is ≈2.7 by the way

[@hobocraft0]

Oh damn, that's actually one of the formulas for e. It should be equal to approximately 2.7 .

[@Avighna]

Wow, that's honestly a really cool explanation. I thought of two methods: using logarithms (assuming I'd be allowed to use a log table of sorts) or noticing that the left hand side was an approximation of Euler's constant.

[@dsspiegel]

Also, the first expression is pretty close to e ~ 2.718, which is > 2. In general, (1 + 1/n)^n is always > 2 for n>1.

[@goaway4991]

For a bit more of a simple approach, when you write 1.005^200 as (1+1/200)...(1+1/200), it's just a sum of every binary choice you can make-- so you suggestively choose 1 for every number to get 1+trash, then you can select 1 for all choice aswell as a single 1/200 term to get 1+1/200+trash. There are 200 other ways you can do this so you can be sure that the number is at least 1+200*1/200.

[@kenji2787]

Why 1 multiplies itself never increases, 0.5 multplies itself always decreases, but 1.5 multiplied itself always increases?

[@mab9316]

A beautiful solution. Thanks.

[@LuisRodriguez-vh6fg]

Really interesting! This was my solution: e = (1 + 1/n)^n when n tends to infinity. Then, for any n > 0, e > (1 + 1/n)^n. For n = 1, (1 + 1/1)^1 = 2. For n = 200, 2 < (1 + 1/200)^200 < e. Finally 2 < 1.005^200.

[@matteoFalsitta]

As a Math Olympiad Question, solving time matters. I think that the intended solition is to recognize that (1+ 1/x)^x is e as x tends to infinity, recognize that 200 is big enough for the result be bigger than two. If you have any concept of e, you should know that 200 is enough because 2 is enough (1+1/2)^2 is already 2.25). This is no brainer and a problem that could be answered straight away, with no calculation at all.

[@spiderjerusalem4009]

yes, just an application of bernoulli's inequality (which can easily prove whether the sequences (1+¹/ₙ)ⁿ and (1+¹/ₙ)ⁿ⁺¹ increase or decrease monotonically) (1+x)ⁿ⩾1+nx for all nonnegative integers n for all reals 1+x⩾0 Pf: It is obvious for x>0 by binomial expansion. Assume -1⩽x⩽0 (1+x)ⁿ-1 = x 𝚺ⁿ⁻¹ₖ₌₀ (1+x)ᵏ ⩾ x 𝚺ⁿ⁻¹ₖ₌₀ 1 = nx (since 0⩽(1+x)ᵏ⩽(1+x)⩽1, for any k>0, and x

[@mkostya]

1 is enough, it’s (1+ 1/1)^1=2

[@jamesLucas221]

for whoever didn't notice 1.005^200 = (1 + 1/200)^200 is euler's number with fixed lengths after the dot e = lim (n->inf) (1 + 1/n)^n just fun little thing i noticed they probably done on purpose

[@matejuhlir680]

You can also just rewrite the left side as (201/200)^200 = 201^200/200^200 and the right side as (2*200^200)/200^200. When you then compare the numerators, 201^200 is definitely greater than 2*200^200, making the 1,005^200 greater as well.

[@OptimusPrime-vg2ti]

201^200 is definitely greater than 2*200^200 -- *Why?* Why is the left hand side definitely greater? The entire question is about proving this statement, which you have assumed inside your answer.

[@dermeister1744]

KZbin recomendations overvalues my preferences

[@cicik57]

the number squared is 1.005 + 0.005... > 1.01 and the next step is 1.015 + something , what grows larger. So making it power 200 would be 1 + 0.005*200 + = 2 + > 2

[@Dekross]

Clever

[@cicik57]

@@Dekross thanks, basic power trick

[@Nicolas-gw4lw]

Short solution: the sequence x_n = (1+1/n)^n is monotonically increasing and 2 = x_1, 1.005^200 = x_200. An analytical solution: it's easy to show that (1+x)^n > 1+xn for x > 0 using derivatives. Now, x = 0.005 and n = 200 yields the solution.

[@VoidloniXaarii]

Wooow, this was mind blowing! Thanks a lot! Reminded me of being a kid, Olympic first in county in ro at physics and yet being humiliated into humblety by not even being able to complete page 2-3 of Russian math competition book Smirnov .. Thank you for this epic vid ❤

[@borismarinkovicgutierrez2349]

I solved the problem this way (with arithmetics and a logarithm table) : 1.005^200 2 log 1.005^200 log 2 200 log 1.005 log 2 200 x .002166 0.30103 0.43321233 > 0.30103 1.005^200 is larger

[@kylejohnston2271]

I just knew that if it were A x 1.01, A would increase by 1%, and even adding 1% of the original value 100 times would be twice the original value. 1.005 is adding 0.5% so it would need to be 200 times. And given the fact that you start with a value greater than 1, you will most definitely cross 2.

[@marcushendriksen8415]

This also follows immediately from Bernoulli's inequality: if n is an integer greater than or equal to 1 and x is a real number greater than or equal to -1, then (1+x)^n is greater than or equal to 1+nx. Substitute 200 for n and 1/200 for x and you get (1+1/200)^200 >= 2. Otherwise you can simply look at the first two terms of the binomial expansion, which are 200C0×1/(200^0) and 200C1×1/(200^1). 200C0=1 and 1/(200^0)=1, so the first term is 1; likewise, 200C1=200 and 1/(200^1)=1/200, so the second term is also 1. All the other terms are strictly positive, so the result must be greater than 2

[@mktsp2]

1.005^200 has to end in the digit 5. So they are not equal. So > holds

[@thomasdalton1508]

The question is basically asking you which gives you more money, compound interest or simple interest. I think most people know that compound interest gives you more money, so 1.005^200 must be greater than 1+0.005*200=2

[@t_c5266]

This video took over 3 minutes to answer the question. My calculator took 5 seconds

[@ashutoshpanda1031]

Brilliantly explained.thank you ma'am.

[@MadaraUchihaSecondRikudo]

I went (1 + 1/200)^200 roughly equals e, which is greater than 2 :D Also, first two terms of binomial expansion are 1+1*1/200*1C200 = 2, and there are more terms after that

[@yosedu8061]

what about such solution: lim (n---->infin ) (1+1/n)^n=e >2 just when n=1 it is already 2 . when n=2 it is 2.25 so far as lim is existing it must be increasing

[@karolkurek9201]

The same. f(x) = (1+1/x)^x is increasing, but probably you can't use that at early stage of education (no derivatives).

[@emmk4992]

For anyone interested. It comes out 2.7... . So first one is greater

[@jebbush3130]

before watching the video, I would do it like this, with the binomial expansion (1 + 0.005)^200 = 1 + 200 * 0.005 + ... + 0.005^200 = 2 + ... + 0.005^200 > 2 seems pretty easy

[@wavingbuddy3535]

way i did is to consider the function 1.005^x. By setting it equal to 2 we get 1.005^x = 2 (take logs on both sides) ln(1.005^x) = ln2 (simplify using log rules) x ln(1.005) = ln2 => x = ln2/ln(1.005) ~ 138.98 => 1.005^200 > 2

[@kaybei9771]

IF you don't know what to do or where to start, you can also try looking for pattern: write a few (1+5/1000)*(1+5/1000)*(1+5/1000). Evaluate and you will begin to notice that there will be 200 terms of 5/1000 and some extra stuff which is very small. So we get 1+200*5/1000=2. And then there is some extra small stuff to be added. So it must be bigger. Pattern searching is great in any situation when you dont know the tricks or shortcuts that many olympians know, you can discover one for yourself.

[@thedempsus9234]

then there's me who didn't forget their calculator

[@jeromegarcia5396]

I use this method to adjust my cereal bowl slightly to the left when I eat so I can say I've actually used this in life 🤭

[@bluematter435]

this is a very neat trick, but also man i wanted it to be smaller than 2 so badly, because that would have been cool

[@some_personn]

As a Russian, this is one of the easiest problems in our Olympiads

[@RandomPerson-hj8fq]

It is very easy for any Olympiad level indeed

[@some_personn]

@@RandomPerson-hj8fq I'm agree, but other problems in hardest Russian olympiads are more difficult, if you are interested in it you can check out final stage of "All-Russian Olympiad in Mathematics" or something like this, idk how to write the name of this olympiad in English

[@Rahul-pf5kw]

This is a very basic question that we science students deal within the first lecture of binomial theorem. Very easy ti solve ...

[@Sathishkumar-pm8tu]

As a physicist, my first thoughts on this problem that I came across accidentally is to apply "Error Analysis" - that the error when taking power can be expressed as power times fractional error. But soon I realized this could not be true. Because you just keep multiplying too many times. Then I got the right idea in a moment. That is, to apply the idea of continuous growth. Using this I thought the first number must be really really close to 'e', and concluded that one must be larger. And then I proceeded to see the solution in this video. It is nothing short of ingenious. The math here is so beautiful.

[@santafehiker]

Seems unnecessarily complicated. A simple Taylor series expansiion gives 1.005^200 > 1 + 0.005*200

[@andrewz2854]

This question is beautiful because of its simultaneous simplicity and complexity.

[@saurabhshankar9545]

Easiest solution 1.005 ^200 = (1 + 1/200) ^ 200 which is of the form ( 1 + 1/n)^n and as n goes higher the value will approach number e which is greater than 2.

[@tahalateef]

The easiest explanation is perhaps through the binomial theorem. 1.005^200 = (1+0.005)^200 = 1^200 + 200x1^199x0.005 + subsequent terms = 1 + 1 + subsequent terms = 2 + subsequent terms (which will always be greater than 2).

[@himanshupaliwal2026]

you can write (1.005)^200 as (1 + 0.005)^200 which is similar to standard binomial expression (1+x)^n , whose first 2 terms are nC0 and nC1(x). For this expression it's 200C0 = 1 and 200C1(0.005) = 1 so we can say that only first 2 terms add up to 2 so it's is definitely greater than 2.

[@wp4297]

f(n)=(1+1/n)^n increasing (whose limit for n->infty goes to e), and f(1)=2. Thus for everything comes next, m>n, f(m)>2

[@EhsanZia-Academi]

Thank you for this video. Great idea you used to solve this problem.

[@ffff3c]

(1+0.10)^2 = 1.10^2 = 1.21 > 1.20 = (1 + 0.10 *2) Intuitively, increasing something by 10% twice in a row, is better than increasing by 20%=2*10% at once, because of compound interest. We can then easily extrapolate that for x>=0 (1+x)^n > 1+x*n Which answers the question here for x=0.005 and n=200

[@dylanfitzpatrick9423]

Compound interest is a beautiful thing

[@PickYourPoisson]

In a similar vein to how the problem is rewritten at the beginning, in math we know (1+1/x)^x (we'll label this f) approaches e and never has a negative derivative for the domain x > 0. At x = 2, we get a value of 2.25 (1.5^2) . Since f approaches approaches e and only does so from below, we can safely assume for the value x=200, that we'll get a number in-between 2.25 and e this being greater than 2

[@prabhakarakshay]

We can also take log, ln(1.005^200) = 200ln(1.005) Using Taylor series expansion, ln (1+0.005) = 0.005 + very small terms So, 200*0.005 =1 While ln 2 = 0.693, here also we can do ln(1+1) and estimate using Taylor series that it is less than 1. As 1> 0.693, hence 1.005^200>2 Taylor series: ln(1+x) = x-x^2/2 + x^3/3 -....

[@itachu.]

i know a method , cant tell exact value but, for x , y > 0 if x^n > y then x > y^1/n and 1.005 is greater than 2^1/200. (basically approximations :P

[@HotelPapa100]

Simpler aproach: (1+z)^n where z is very small is approximately 1+z x n for small n, this estimate is always a bit low, and the estimation gets worse for larger n (this follows from the binomial formula). So (1+1/200)^200 must be larger than 1 + 200/200.

[@karldavis7392]

Before putting it into a calculator, I guessed that 1.005^200 was larger because ln(2)=.693, and .693/.005 is 138, and 200>138, so it more than doubled. This is a perfect question for logarithms done in your head.

[@DANGJOS]

Without watching the video (other than the first 24 seconds before being forced to pause), 1.005^200 is definitely larger. .005 times 200 plus 1 gets you to 2 already, so the extra factors added to it would have to make it larger than 2

[@polantcrumbosh3848]

simple way i looked at it, we know that (1 + 1/n)^n is equal to e when n approaches infinity, if we substitute n with 1, the lowest positive integer, is obviously just (1+1)^1 which is 2, and we know as n increases it grows from 2 to e, so n = 200 has to obviously fall somewhere between and thus is larger than 2

[@yash1152]

best thing abt this vid is that is uses extremely minimal set of prerequisites... otherwise after 0:33 one can easily use binomial expansion & careful approximation

[@Parasmunt]

The Marvel of Math is how a way becomes clear when you simply describe it in a different way.

[@elcharara7926]

Or you can elevate a Log in an equality and you find that the fraction is strictly higher of lower than 1 using a logarithmic inequality

[@blank4502]

Think about it this way: the formula for e is (1+1/n)^n. Here, n = 200. Now, even if n = *1* , it evaluates to 2. Therefore, since every higher n gets closer to e, it must be greater than 2.

[@flame_orange]

Its really simple with Binomial Theorem!👀

[@mkostya]

The sequence (1+1/n)^n has limit of e when n-> infinity, and it is monotonically increasing. For n=1 we have 2^1 = 2 meaning all consecutive members of this sequence for n > 1 will be greater than 2.

[@lechat_dnipro]

cool proof visually, however, from binomial expansion formula for (1+1/200)^200, it's obvious that two first parts have given 2 already: 2 < 1 + 200*1/200 + ... "199 parts each >0". So, this question has just half a second solution...