Math Olympiad Problem, you should know this trick! 

I don't know if you know but this video is being blatantly copied, here's the link: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Um53h6yq_6o.html If you look at the comments here there are various ways in which you can solve the problem but in the specific video they have the same script/format with how you systematically solved the problem, even down to the explanation of the constant "e" and how they used 1/6 instead of 1/49. Even the title is copied.

I divided both sides by base 49 to the power of 50. That yielded base 50/49 to the power of 50 compared with 49. Algebra shows that the base 50/49 requires a power greater than 193 to exceed 49, so base 50 to the power of 50 is less than base 49 to the power of 51.

I multiply every thing by zero. No more problem. Back to dog videos.

Divide both by 50⁵⁰: 1 vs. (49/50)⁵⁰ × 49 = (0.98)⁵⁰ × 49 Take natural logs of both and rearrange slightly: 0 vs. 50•ln(1 - 0.02) + ln49 To a very good approximation, ln(1 - 0.02) ≈ -0.02, so this is to a good approximation, 0 vs. 50(-0.02) + 2ln7 or 0 vs. -1.0 + 2ln7 The rhs is easily positive, so the circle encloses a "

The math is interesting to be shown but I really feel this more of a logic test as with the base numbers being so close just being multiplied that 1 extra time is a massive jump over the other. So it had to be the bigger of the two.

If you go in a calculus approach, you can consider the function f(a) = (50-a)^(50+a) = e^{(50+a)ln(50-a)} You can take the derivative, which is f’(a) = (50-a)^(50+a) [(50+a)/(50-a) + ln(50-a)] One can see that f(a) monotonously increases at least in the range such that (50+a)/(50-a)>0 and (50-a)>1 => f monotonously increases for -50 < a f(0) < f(1) => 50^50 < 49^51 Which is what we wanted. And it is nice to see one can immediately get 51^49 < 49^51 And the like exercises.

I saw a lot of people doing complicated solutions but i did it in a simpler way, don't know if it would work in all cases. I first tested this same case with smaller numbers that you can actually calculate: 4^4 and 3^5, in this case 4^4 > 3^5 and the diference between them is 13. Then i tested it again with 5^5 and 4^6, and in this case 5^5 < 4^6, and the differencre between is -971. With that we can see the difference will just keep getting bigger with higher values, so from that point on the number with the highest exponent will be greater, so 49^51 > 50^50.

See, I just used a calculator. But 50^50 and 49^51 are so large they would probably just give an error, so instead i took _the log of both,_ and 51log49 is larger than 50log50.

Divide both by 49*50^50. Then LHS=1/49, the RHS =(49/50)^50=(1-1/50)^50 which is very much 1/e. e=2.718

I did that with derivatives, that was tough! You should just take derivative from x^(100-x) then solve the transcendental equation, the solution is easily guessed: it is somewhere near 24 or 25. On infinity the derivative is negative, so the function is lesser for 50^50 as 50 > 49 and both 50 and 49 are greater than 25.

I choose 49^(51) because it is usually the bigger exponent who would give the bigger value.

Take the log of each side, know the log power rule making 50log(50) vs 51log49 and we all know that log grows sublinearly, but monotonically increasing, therefore 49^51 is easily bigger.

50 to the fiftieth power is approximately 8.8817E84. 49 to the fifty-first power is approximately 1.5848E86.

I think this is just for knowing how numbers work. You can just do 10^3 and 11^2. It is 1000 for 10^3 and 121 for 11^2

Appreciate ur work... WELLLLL DONEEE

6:57 Why should the limit of (1 + 1/n)^n, as n goes to infinity, is smaller than 3 imply that (1 + 1/n)^n is smaller than 3, for ALL n?

Great video and great question and great solution to problem

I just used binomial theroem at (1+(1/49))^50 *(1/49)≈ (1+(50/49))*(1/49)=99/49^2 so yeah maybe not the best but it did the job basically it states that (1+x)^n ≈ 1 + nx

Since the question was which side is greater, without requiring proof, I intuitively knew that the side with the higher power is greater for anything greater than 3^2 vs 2^3

I'd like to know: if you take two integers x^x and (x-1)^(x+1), is the second one always larger? 4^4 = 256 and 3^5 = 243. Then 5^5 = 3125 but 4^6 = 4096. Now the second is larger. Does that relationship continue always now, as x gets even larger?

Nice approach, but you could easily take “log” from each side and the answer would appear much sooner

You could also have done it using logarithms Taking log ( base 49) of both sides and the approximating value of log(base 49 ) (50) and getting the answer

I think i found an easier way: you subsitite x with 50, you get on one side x^x for 50^50 and for 49^51 you get (x-1)^(x+1) and then using binomial formulas and etc you get: 50^50 < 50^51 - 2498 (as 2489 is much smaller than 50^50 x 50) Tell me what you think!

Well IIRC, if the number of digits on the base numbers were in the same range (like 10 with 11 being 2-digited numbers), even if one of the base numbers were bigger, the bigger exponent will always have bigger value no matter which one has it. So by that logic, because the difference between 50 and 49 was small enough, I can assume that the value of 49^51 would be at least 10 times larger than 50^50.

You can demo the 2an method as well ? Really like you channel btw 😊

I solved it with the properties of logs. Taking the log of both sides simplifies to 50 log 50 and 51 log 49. Log 50 and log 49 are almost the same due to log based anything big, squishes changes hard. Therefore, 50 and 51 are compared alone.

I really knew this answer in a second with just pure logical thinking man

I think the proof using binomial expansion of 50^50 (== (49+1)^50) is simpler. Just my opinion

Never-ending, I was not thinking straight. I got it now.

This is a great video!

I just used a calculator to observe that 5^5 is less than 4^6. I therefore generalized that for any positive number n greater than 2, n^n < (n-1)^(n+1).

If you have any interesting & splendid questions to provide, just comment! I will choose some of them to make Shorts ❤

I know more about my weak spot in Mathematics in this video, I shall train more about this skill.

Extract root 50 both sides which yields 50 ? 49x49^(1/50), and knowing that numbers < 1 rised to any positive power online be 1 at the infinite, so 49x0,... will be < 50.

really nice the way you do it.

Every time you are increasing the power like 10^6 and 10^7, it becomes a huge difference because it would have more zeros at the end, so the number with the larger power would likely to be larger than smaller one. Unless it more complicated, I could not say that 2^3 is larger than 4^2, so it varies on cases, but since the digits are close, I think you can tell.

Method 1: We must know this Fundamental Result that in the Sequence, given by [1+(1/n)]^n or [(n+1)/n)]^n, that is (2/1)^1, (3/2)^2, (4/3)^3, (5/4)^4, … the successive terms are larger and finally converge to some largest value which is greater than 2 and less than 3. We can obtain these results just from the Binomial Expansion of [1+(1/n)]^n and rearrangement of the terms, without going in to any concept of the number e. In fact, this observation contributed most to the beginning of the History of ‘e’, which, due to its unique properties, is used as the base for Natural Logarithm. Similarly, in the Sequence of the inverted terms, namely (1/2)^1, (2/3)^2, (3/4)^3, (4/5)^4, … the successive terms are smaller and finally converge to some smallest value which is lesser than 1/2 and greater than 1/3. Now, coming to the Question, Let R = (49^51)/(50^50) Or, R = [(49×49)(49^49)]/[(50)(50^49)] Or, R = (49×49/50)(49/50)^49 We know that (49/50)^49 is larger than 1/3. Therefore, R is larger than (49×49/50)/3, which is 49×49/150. Or, R is greater than 1. Hence 49^51 > 50^50 Method 2: Consider the Function f(x) = (50-x)^(50+x). Using Calculus [Take log and differentiate], for this Question, in particular, we observe that at x=0 to x=1, f(x) is increasing. So, f(0) < f(1); Or 50^50 < 49^51. In general, we can show that f(x) is increasing at x = .., -2, -1, 0, 1, 2, etc. until x nears 26 or so. Therefore, … < 52^48 < 51^49 < 50^50 < 49^51 < 48^52 < … Related Practice Questions: Q1: Which is Larger: 24^76 or 25^75 Q2: Which is Larger: 23^77 or 24^76 Q3: Prove by simple Algebra that the successive terms in the Sequence (2/1)^1, (3/2)^2, (4/3)^3, (5/4)^4, … are larger.

In our jee preparation we can solve these type of problems orally . You can use binomial to quickly get answer

WOW, You are really clever with this one. Hats off to you!

When x is large, (x-1)^(x+1) > x^x. Let x be a positive integer. We want to compare x^x and (x-1)^(x+1). We address here the case when x is large (>>1). To do so, we take the log: log ((x-1)^(x+1)) = (x+1) log(x-1). x being large, log(x-1) = log(x(1-1/x)) = log(x) + log(1-1/x) \approx log(x) - 1/x because log(1+u) \approx u when u is small. So, (x+1) log(x-1) \approx (x+1) (log(x) - 1/x) = x log(x) + log(x) - (x+1)/x We conclude that when x is large, (x+1) log(x-1) - x log(x) \approx log(x) - (x+1)/x And obviously, log(x) - (x+1)/x is also dominated by log(x) which not only is positive, but diverges! So, when x is large, (x-1)^(x+1) > x^x. Remains the question of 'large'. Well, instead of developing the log(1-1/x) around zero, you can keep its exact value and study the difference, (x+1) log(x-1) - x log(x) = log(x) + (x+1) log(1-1/x). Starting from x=5, you can show that this difference is monotonic, increasing, so large is not so large, it's x=5.😉

The exponents are 50 and 51 so by binomial theorem there will be 51 and 52 terms so 49^51>50^50 5 second question

x^x>(x-1)^(x+1) as long as x>4.141041525..., the value being easily obtained from Newton's iterative formula for non-linear equations.

A simpler step could be dividing the (50/49)^50 inside term, that will be 1. smth and overall approx to 1 now this 1/49 is obviously small term hence the final answer is that 50^50 is smaller than 49^51

I thought of other method that you write 49 as (50-1) and then apply binomial expansion on it and neglecting some terms. Then compare it

I started looking at (1+1/n)^n and soon figured out that it tends to e (the base of the natural logarithm) which implies that the second number is (very) approximately 50/e times larger.

Excellent math question and more excellent explanation thereon😊

My estimate would be 49^51 greater. Because for two numbers being only 1 unit appart, also being powered at least 50 times, wich is a lot, 51>50. But i do not know the verification yet.

make them to the nth power,(let n -> infinity ) take log , you will find right side is bigger than left side

Would it be possible to simply time bith sides by 50? This makes lefthand side equal to 50^51 and righthand (49*50)^51 which is obviously greater?

In fact I became interested in the question for which integers m>1 the expression m^m < (m-1)^(m+1) is valid, which is m>4. A nice one.

If you are just checking to see which is larger do you absolutely need to keep the exponents at 50 and 51 or can you "simplify" this to 50^2 and 49^3?

e ＜ 3, it is OK, however, how do you proof (1 + 1/n)^n is less than 3 when n equals to 3? I do not think it is obvious.

if you formulate the question x*x and (x-1)*x+1 and start to put small numbers in these formulas you would see except for one (4*4 and 3*5), (x-1) * (x+1) is always greater than x*x

generic solution x^x vs (x-1)^(x+1) transforming it to 1/(x-1) vs (1-1/x)^x then left part is 0 < 1/(x-1) < 1/4 for x>5 right part: 1/4 < (1-1/x)^x < e for x>2, thus for any x> 5 right part is always larger. Done

i knew the correct answer immediately just based on the principle that in large numbers, the exponent is always more influential than the base; for example although in the case of 2^3 < 3^2 the base matters more . . . this is an exception to the rule because the numbers are so small . . . the rule that "exponents have more influence than the base" applies for even slightly bigger numbers; for example 5^6 > 6^5. I have no proof to offer, but there is a general rule that holds when comparing numbers x^y vs (x-1)^(y+1) as long as both x and y are greater than 2 then (x-1)^(y+1) will always be greater than x^y

I thought of it in a simplar way. We have that 50^2 is less than 49^3, then 50^n is less than 49^(n+1), so 50^50 is less than 49^51.

Спасибо.

No need. A slightly bigger number multiplied certain times is way smaller than a slightly smaller multiplied one more time. Because the last added multiplacation is done to the total number reached before it while the increase in the multiplied number is an increment.

3:08 50 and 49 they have special relationship that is 50 is one more than 49. Very special relationship 😂

heyy idk if it defies anyone here, but 49⁵¹ can be written as (50-1)⁵¹ now as we know it is (50-1)*(50-1)...(50-1) 51 times, using basic algebra, the first term of this polynomial is 50⁵¹, last term is 1, now all the inbetween terms would be negative as power is odd, as 50(-1) power n times is 50 if its even, and here it is even only, as there are 50+1 terms(the term whose 50 we are multiplying wont be considered, obviously, in the same way the term 50 multiplied by 50 gives a positive value greater than 50, so the negative terms are def lesser than 50 which gives, 50⁵⁰-50+x, just signifies that x is any value lesser than +infinity and greater than 50, simply x belongs to (50,inf) no limits needed it shows that 49⁵¹ is greater

We can compare the two numbers by taking their ratio: (50^50) / (49^51) Simplifying this expression, we can rewrite it as: (50/49) ^ 50 Since 50/49 is greater than 1, raising it to the 50th power will give us a number that is greater than 1. Therefore, (50^50) / (49^51) is greater than 1, which means that 50^50 is greater than 49^51. Therefore, 50 to the power of 50 is larger.

I took the ln on both sides and solved it really easily. Admittedly a calculator is required. ln50^50 = ln49^51 50 ln 50 = 195 51 ln 49 = 198

все куда проще - находим производную функции (a-x)^(a+x) a=50 убеждаемся, что на больше нуля

(1+1/49)^50 is a number very small approximately 1. If it was raised by infinity it would be 1 im the limit. Using an approximation, the result should be 1,xxx (near 1) multiply by 1/49, which would be a number below 1. So, the result would be 50^50

Just an easy way, even no pen or paper needed: 3^3 > 2^4; and 4^4 > 3^5, but already next group 5^5 < 4^6 with a gap increasing on 6^6 < 5^7 and so on. Safe to interpolate 50^50 < 49^51.

take log of both the numbers and compare them

Or you could just know that bigger exponent than base is bigger than bigger base than exponent for the same numbers, as long as both numbers are greater than e. And this problem is just using a midpoint which would obviously be between them. Not a hard proof but it works. Yeah 2^3

Для чего введено "меньше чем 1/6"? Ведь и так видно, что 3*50 меньше чем 49*49, значит - дробь меньше единицы. А всё остальное - классно! Браво!

consider n^n < (n-1)^(n+1) for n >= 5, then prove it by induction is what I did. Your solution is definitely more elegant!

Pretty simple expnentional increases makes this easy to figure out without even doing any math

This video is basically a tutorial of how to turn a simple problem to a larger one to look cool

I did it in a very simple manner using basics of binomial... I wrote 50^50 as (49+1)^50 and expanded it a bit as.. (50c0.49^50 + 50c1.49^49 + 50c2.49^48+ - - - - + 50c50.1).... EQN. 1 Now i evaluated the numerically greatest term in the expansion using (n+1) /(1+|a/b|) - 1

Bro simple apply log on both sides and check by basic values

I thought this was rather obvious. The higher exponent is going to result in a much bigger number with the bases being so close in value.

Another way that I approached it employed the binomial theorem and root approximation: 50^50 ? 49^51 (49+1)^50 ? 49*49^50 (1+1/49)^50 ? 49 1+1/49 ? 49^(1/50) Now x^(1/a) ~ t+ (x-t)/a*t^(a-1) where t is a close estimate to the root letting t = 1, then 49^(1/50)= 1+ (49-1)/50*1^(50-1) = 1+48/50*1^49 = 1+48/50 Comparing now 1 + 1/49 ? 1 + 48/50, leaves 1/49 ? 48/50 so 50 ? 48*49 where ? must be "

You can just do 1+nlog(x) and find the number of digits

Just take log on both side we can easily bring power down

Can t we just take example like 50 ²

Honestly these are waaay more simple than they seem, just graph x^x=y and (x-1)^(x+1), then compare ‘em, they intersect at around 4.136 and never look back

you can otherwise just use the log, its done in less than 3 lines

49^51 is the larger number. If you know which operation contains more juice so to speak, then you have the ability to solve this problem.

one of the simplest way to solve this question by taking logarithm both side and solve it it easily showed 49^51 is greater than 50^50 its very simple problem if we didn't use your method my friend

The suggested solution is incorrect. The fact that (1+1/n)^n converges to e as n→∞ says nothing about how large (1+1/49)^49 might be. It only says something about the asymptotic behavior of the function that maps n to (1+1/n)^n. An easy and correct way is to use the inequality ln(1+1/n) < 1/n, derived from the Taylor expansion of the natural logarithm, to deduce ln( (1+1/49)^49 ) = 49·ln(1+1/49) < 49·1/49 = 1, which implies (1+1/49)^49 < e. To participants in a math olympiad this should be a trivial exercise.

Can't we do comparative analysis? For eg. 50^3 = 125,000 and 49^4 = 5,764,801, hence 50^3 < 49^4 so 50^50 should be lesser than 49^51

Write 49 as 50-1 and use binomial expansion

In physics all things work the same but they can have different scales 4⁶>5⁵ which means 49⁵¹>50⁵⁰

I made 50 to the power of 2, then, I made 49 to the power of 3. It worked fine but idk if this was a matter of luck or this really works

my quastion is why or from where he decided to make e < 3 like why 3?

50^50/(49^50)

Blud really said “changements”

The easiest approach to this question is to take the difference between the two given numbers.

Thank god there is subtitle. I couldnt understand right hand side due to the accent.

49 almost = 100/2. So roughly estimating whereas 49^51 will be close to a 102 digit number but 50^50 will be close to a 100 digit number.

I'm all praise for this Prof. I myself am prof of mathematics

Please ma'am solve this question...what is the remainder of 128^2023 divided by 126.... It's a Olympiad question. Please sir solve this

My guessing on a multiple choice guess would have done the same

I am like "wow great" So, whats the answer? 😂

Anyone who's interested in solving IMO problems doesn't need more than two minutes (from 0:54 to 2:59) to get that (50^50)/(49^51) = (50/49)^50 x 1/49.

7:33 when did 1/6 came from? Is there any reason to be specifically 1/6 or it's just to simplify the multiplication... (but then we could just use 1/3 and it would be 50/49 and LARGER than 1)

Intuitively, bigger power = bigger number. It'd be good to see a question where it's not the case.