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

Math Olympiad Problem

[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.

[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.

[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 .

[mda]

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

[mingwangzhong]

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.

[TheSoteriologist]

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

[nasabdul]

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

[rchatte]

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

[ckshene]

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.

[vladpetre]

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.

[Skaahn]

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

[MsLeober]

Нужно показать что 50^50<49^49х49^2. После логарифмирования
50 ln 50 - 49 ln 49<2 ln 49. По теореме о среднем
левая часть неравенства равна f'(c)(50-49), где
f(x)=x ln x, 49<c<50, f'(c)=ln c +1<ln 50+1=ln 50 e<ln150 .
С другой стороны 2 ln 49=ln 49^2=ln 2451.
Неравенство доказано.

[aakashanantharaman]

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

[Ben-pwqe]

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 😊

[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.

[tonybantu]

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.

[hznerle]

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 .

[tassiedevil]

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.

[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.

[stem-orgayafelizardoiiiy.]

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