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: https://youtu.be/Um53h6yq_6o 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.

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.

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 "<" "less than"

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.

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.

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 <49 Thus, given [0,1] is in (-50,49) => 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.

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<<49 therefore LHS is smaller.

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.

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

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

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

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

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.

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

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