Clever Approximation #1

Algebra Level 4

Let there be a natural number N N that satisfies:

N = 1 2 ⋅ 3 4 ⋅ 5 6 ⋅ ⋯ ⋅ 99 100 N=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\text{ }\cdots\text{ }\cdot\frac{99}{100}

Add up all the numbers of the true statements from the list below.

1 . N > 1 100 2 . N > 1 16 4 . N > 1 10 8 . N > 1 8 16 . 1 2 ⋅ 3 2 ⋅ 5 2 ⋅ 7 2 ⋅ ⋯ ⋅ 9 9 2 < 100 ! 9 \begin{aligned} & \boxed{1}.\quad N>\frac{1}{100} \\ &\boxed{2}.\quad N>\frac{1}{16} \\ &\boxed{4}.\quad N>\frac{1}{10} \\ &\boxed{8}.\quad N>\frac{1}{8} \\ &\boxed{16}.\quad 1^2\cdot3^2\cdot5^2\cdot7^2\cdot\cdots\cdot99^2<\frac{100!}{9} \end{aligned}


The answer is 19.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Jun 15, 2017

Let P P and Q Q satisfy:

P = 2 3 ⋅ 4 5 ⋅ 6 7 ⋅ ⋯ ⋅ 96 97 ⋅ 98 99 ⋅ 100 101 Q = 2 3 ⋅ 4 5 ⋅ 6 7 ⋅ ⋯ ⋅ 96 97 ⋅ 98 99 \begin{aligned} &P=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\text{ }\cdots\text{ }\cdot\frac{96}{97}\cdot\frac{98}{99}\cdot\frac{100}{101} \\ &Q=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\text{ }\cdots\text{ }\cdot\frac{96}{97}\cdot\frac{98}{99} \end{aligned}

Then we can see, from the fact that 1 2 < 2 3 \dfrac{1}{2}<\dfrac{2}{3} , 3 4 < 4 5 \dfrac{3}{4}<\dfrac{4}{5} , ⋯ \cdots , 99 100 < 100 101 \dfrac{99}{100}<\dfrac{100}{101} ,

N < P N 2 < N P = 1 2 ⋅ 2 3 ⋅ 3 4 ⋅ ⋯ ⋅ 100 101 N 2 < 1 101 < 1 100 N < 1 10 \begin{aligned} &N<P \\ &N^2<NP=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\text{ }\cdots\text{ }\cdot\frac{100}{101} \\ &N^2<\frac{1}{101}<\frac{1}{100} \\ &N<\frac{1}{10} \end{aligned}

Therefore, we can rule out 4 \boxed{4} and 8 \boxed{8} .

Also, we can see, from the fact that 2 3 < 3 4 \dfrac{2}{3}<\dfrac{3}{4} , 4 5 < 5 6 \dfrac{4}{5}<\dfrac{5}{6} , ⋯ \cdots , 98 99 < 99 100 \dfrac{98}{99}<\dfrac{99}{100} ,

1 2 N > Q 1 2 N 2 > N Q = 1 2 ⋅ 2 3 ⋅ 3 4 ⋅ ⋯ ⋅ 99 100 N 2 > 1 200 > 1 225 N > 1 15 \begin{aligned} &\frac{1}{2}N>Q \\ &\frac{1}{2}N^2>NQ=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\text{ }\cdots\text{ }\cdot\frac{99}{100} \\ &N^2>\frac{1}{200}>\frac{1}{225} \\ &N>\frac{1}{15} \end{aligned}

Therefore, we can say that 1 \boxed{1} and 2 \boxed{2} are true.

Since N < 1 10 < 1 9 N<\dfrac{1}{10}<\dfrac{1}{9} ,

We can say that 100 ! ⋅ N < 100 ! 9 100!\cdot N<\dfrac{100!}{9} .

Therefore, 16 \boxed{16} is true.

So, the sum we're trying to get is 1 + 2 + 16 = 19 1+2+16=\boxed{19} .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I think a better way to ask this question, is: Which of the following options is true:
- N < 1 100 N < \frac{1}{100}
- 1 100 < N < 1 16 \frac{ 1}{100} < N < \frac{1}{16}
- ETC


Thoughts? Let me know if I should change the options.

Calvin Lin Staff - 3 years, 11 months ago

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Mmm yeah that'd be better. But you'll have to change the answer too!

Boi (보이) - 3 years, 11 months ago

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