Russian Math Olympiad Question

[DucLe-kghx]

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

[redfluxbluedawn]

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

[Bermatematika]

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

[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.

[Woodside]

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).

[legendexists]

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

[thomasgreene]

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.

[echthroi]

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

[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 .

[davidking]

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.

[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.

[jonsrecordcollection]

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.

[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.

[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.

[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

[manloeste]

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)

[DanBurgaud]

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

[hobocraft]

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

[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.

[zjz]

If you do banking, you know compound interest > simple interest. Now let interest be 0.5% and 200 periods. You got the answer.