Math Olympiad Question | Which Is Larger? | You should learn this method

1.005^200 and 2, which one is larger? A fantastic math problem. A common method to this kind of questions.

Remind: Please give a mention If you want to use this question & the explaining steps to make videos.

[DavidMFChapman]

Most proofs in the comments here are needlessly complicated. The binomial expansion of (1+x)^n is 1+nx + additional positive terms. The first two terms add to 2 in this instance (n=200, x=.005) so the additional terms make the LHS larger than 2.

Note added: in no way am I criticizing the original video. There’s all different ways to solve problems. However I do find some of the methods discussed in the comments to be needlessly complicated. I shared my own thoughts on it and many agreed. I don’t feel I was being nasty about it.

[AlexBrogan]

0.005 is half a percent, and if you increase something by half a percent 200 times it will more than double. Intuitive way I solved the problem.

[WookieRookie]

As it was mentioned, 1.005^200 is (1 + 1/200)^200, so it is a number from the sequence that defines the e number. That sequence is monotone, it increases, and its second number is (1 + 1/2)^2 = 2.25, which is larger than 2. So 200th number is surely even larger than 2.

[GDyoutube]

There is a faster and “stupider” way, but which I think was meant to be the preferred solution: 1.005^2 is 1.010025, which means that at every multiplication we add 0.005 and a “tiny” bit. Since we are doing this 200 times we can just do 200 x 0.005 and we add 1 so we are already at 2. But we have a lot of “tiny” bits we didn’t add, so we can be sure the answer is above 2. I think this could be the preferred solution as it would justify why the question involves precisely 1.005 and 200 :)

[JB_-wtby]

Be careful, the part at 6:35 is not always true ! It only works with positive numbers. Take a= -1, b=3, c=2, here b>c (3>2) yet a/b > a/c (-1/3 > -1/2)

[antonalexandrov]

If it didn't need to be proven then one can easily use the rule of 72 that can be used to approximate how many times an interest rate has to applied to double the initial value. This is very often used to quickly check how many years it would take to double your initial capital. Since the "interest" here is 0.5% 72/0.5 = 144. So 1.005^144 is approximately equal to 2, which would mean the 200th power will be a lot larger (somewhere around e).

[StRichLee]

I think it can just use Bernoulli's Inequality (1+x)^n>=1+nx for x>=-1 is real number and n>=1 is integer.
When x isn't equal to 0 and n isn't equal to 1, it can be (1+x)^n>1+nx, so x=0.005 n=200, (1+0.005 )^200>1+200*0.005=2 Q.E.D

[kenhaley]

Here's what I did:
Using binomial expansion: (a+b) ^ n = a^n + n ( a^n-1)(b) + ...etc (there are n+1 terms but in this case we only need to look at the first 2)
So, (1 + .005) ^ 200 = 1^200 + 200(1^199)(.005) + ... (the other 199 terms are much smaller but all are positive)
which simplifies to 1 + 200(1)(.005) + ....(199 smaller positive numbers).
which further simplifies to 1 + 1 + ....(199 smaller positive numbers) = 2 + .... (199 smaller positive numbers).
Thus 1.005^200 must be larger than 2.

[sunnydelight]

I just thought of it as 1.005 represents a 0.5% increase. If that increase was additive, then after 200 increases, the value would be 200% (2). But here it is compounded, which would always be greater.

[chegra]

This is how I would do it.
f(x) = 1.005^x is strictly increasing. Thus, we need to find the value of x when f(x) is 2. If x is less than 200 then 1.005^200 is greater than 2 otherwise it is not.
1. 1.005^x = 2
2. log(1.005^x) = log(2)
3. xlog(1.005) = log(2)
4. x = log(2)/log(1.005)
5. x =138.976
Since the value for x that causes f(x) equal to 2 is less than 200 and f(x) is strictly increasing then 1.005^200 > 2

[josemario]

Using a taylor's expansion of ln(x) around x=1, it is easy to see that ln(1.005) is very close to 0.005, so ln(1.005^200)=200*ln(1.005) is approximately 1. From this we can conclude that 1.005^200 is very close to euler's number "e" (since ln(e)=1), which is around 2.71, hence 1.005^200 is approximately 2.71.

[ianbarquette]

1.005²⁰⁰= (1+0.005)²⁰⁰

The *Bernoulli Aproximation:*
*_(1+a)^x ≈ 1+ax, for a ≪ 1_*

Using it:
(1+0.005)²⁰⁰ ≈ 1+0.005*200 = 2

Now, let's see if the *aproximation* is *greater* or *lower* than the real one:
(1+0.005)² - 1+0.005*2 = 1, 010025 - 1, 01 = 0, 000025.

As the *difference is positive: Real > Aproximation*

Thus, *Real > 2* and so *1.005²⁰⁰ > 2*

I give the proof that *Bernoulli's Aproximation* is always smaller than the real one for you 😉

[daledot]

I did it this way on a slide rule. ln(1.005) is about 0.00495 (LL1 to D). Then the product of that and 200 is 0.99 (standard CD multiply). Get rid of the ln by exponentiation, e raised to 0.99 is 2.68 (D to LL2) just a bit under e. e>2.

[g_lise]

Square and multiply with force to the lower bound of 1.005^2 being 2.771. Since 200 has a low amount of ones in it's binary form (11001000), it makes it even easier. Start with 1 in your result and 1.005 as your working number. If the last digit in binary of the exponent is 1, multiply result by your working number. Square the working number and round it DOWN to 3-4 decimals, the more you do, the closer the lower bound will be. Delete the rightmost digit in the binary exponent. Repeat until you delete everything from your exponent. You just did a multiplication of 2 4-digit numbers exactly 10 times and proven, that 1.005^2 > 2.771 and that's certainly greater than 2. That's what I thought of when I saw this first time, nonetheless, the video shows a much more elegant solution :D

[usiek]

If you are familiar with Euler's number (e), you instantly know, without any calculations, that the first number must be greater than 2 since it is fairly close to e.

[chimkinovania]

1.005 = 1 + 1/200
1.005^2 = (1 + 1/200)(1 + 1/200)
1 + 2/200 + x where x is a small positive integer (we can ignore x because everything related to x will be positive)
1.005^3 = (1 + 1/200)(1 + 1/200)(1 + 1/200)
(1 + 1/200)(1 + 2/200)
1 + 3/200 + some other positive number x
Thus (1 + 1/200)^y = 1 + y/200 + some positive x
y is 200 in this case so
1 + 200/200 plus some positive x
2 + x
Again, x is positive and thus
1.005^200 > 2

Edit: Fixed typo and wording

[enderforces]

other than the application of the e limit that you can use, you can also just calculate log2(1.005) if you're given a calculator, or you can also think about how 1.005 x 1.005 behaves. 1.005*1.005 equals to something higher than 1.01, which i don't care about (easy to calculate, i just care about it being a higher number). So we can apply a minoration and multiply 1.01x1.01 and do the same thing for 3 times. Same reasoning for multiplication by 5 done 2 times (200=2^3*5^2). With the minoration we should get to 2. This means that we know for sure that the 200th root of 2 is a smaller number than 1.005, elevate both members to the 200th power and we get the disequation we were trying to figure out earlier.

[SmartWorkingSmartWorker]

Very cool method. I guess using the binomial theorem also works.

[goldfing]

This is quite easy to answer, because it deals with interest calculation we know from daily life. We distinguish between linear and exponential return. Usually, banks do linear return when the time span is less than a year, but exponential (compound interest) over several years. The factor 1.005 means that the capital is multiplied by 1.005 after one period, that means that 0.5 percent interest rate is applied. 0.5 percent is 1/200.

Now if linear return is applied, it takes 200 periods to double the initial capital (e.g. 1000 dollars initial capital, interest is 5 dollars per period, after 200 periods you have gained 200 times 5 dollars = 1000 dollars, resulting in 2000 dollars overall capital).

However, if exponential return is applied, the capital of course is growing faster (due to the interest of interest effect), so you will have more than 2000 dollars after the same 200 time periods. And this is exactly the result of 1.005^200, using the exponential function with base 1.005 and exponent 200. So you will more than double your capital within these 200 time periods.

In summary, 1.005^200 > 2.

[alessandroblanco]

Multiply both terms for 1000^200, on the left you get 1005^200 while on the right 2*(1000^200) and written like this it is pretty clear that the left one is bigger, even though i have proved nothing like this