"Powerful" inequality

Algebra Level 2

$\Large \color{#3D99F6}{n}^{\color{#D61F06}{200}} < \color{#20A900}{5}^{\color{#624F41}{300}}$

Find the largest integer $n$ that satisfies the above inequality.

The answer is 11.

12 solutions

Joshua Acusta
May 8, 2015

$(n^2)^{100} < (5^3)^{100} \\ n^2 < 5^3 \\ n^2 < 125$

The largest perfect square number less than 125 is 121 and its square root is 11. $\square$

this is the fastest way to do

Neil Yabut - 6 years, 1 month ago

Probably the most easy and productive method...did the same way.. Already upvoted..!! Joshua Acusta

Rahul Singh - 6 years ago

I did that and then I messed it up by rounding up

William Barela - 5 years, 12 months ago

You cannot cancel out the 100th power without knowing the nature of n. If n is negative, it may change the inequality sign.

Rishik Jain - 5 years, 12 months ago

You can, since n^2 is positive, the inequality sign will not change.

Owen Leong - 5 years, 8 months ago

Same way, its very easy

Sulung Agus - 5 years ago

Same way, probably the most simple.

Z G - 2 years, 4 months ago

That’s smart

a byatt - 9 months, 3 weeks ago

Curtis Clement
May 9, 2015

By dividing each exponent by 200: $\ n < 5^{\frac{300}{200}} = 5^{3/2}$ $\ = 5 \sqrt{5} = \sqrt{125}$ $11 < \sqrt{125} < 12$ $\therefore\ Max (n) = 11$

Exactly the same way i solved

Ayush Sharma - 6 years, 1 month ago

Ventus 蒼
May 6, 2015

n^200 < (sqrt(5^3))^200
200 > 1
so we know n < sqrt(125)
125 <144 means n < sqrt(144) = 12
max(int n) = 11

It shouldn't be computer science. @Nihar Mahajan

A Former Brilliant Member - 6 years, 1 month ago

I changed it to number theory again.

Nihar Mahajan - 6 years, 1 month ago

Challenge Student note : Nicely done.Welcome to brilliant.

Nihar Mahajan - 6 years, 1 month ago

Lol Challenge master Second note : Welcome, To brilliant. And please learn Latex.

A Former Brilliant Member - 6 years, 1 month ago

whats latex?

Jun Arro Estrella - 6 years, 1 month ago

$\LaTeX$ is the typesetting you use to write math here on Brilliant. You can read more about it here .

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – oh thank you Pras..:)

Jun Arro Estrella - 6 years, 1 month ago

Steven Zheng
May 6, 2015

We can use logarithms. $300 \ln{5} =200\ln{n}$ . Solving for $n$ , we get $n=2.414...$ . Evaluating $\left\lfloor { e }^{ 2.414 } \right\rfloor$ yields 11.

Moderator note:

You made it unnecessarily difficult. You could have take the 100th root between the two numbers and the answer would be immediate.

Even using logarithms, you don't need to restrict yourself to the natural logarithm.

The real valued function $f(x,b)=\log_bx$ is strictly increasing $\forall~b\in(1,+\infty)$ . Hence, we can simply take logarithm of base $5$ on both sides:

$n^{200}\lt 5^{300}\implies 200\log_5n\lt 300\implies \log_5n\lt \frac{3}{2}$

The real valued function $g(x,b)=b^x$ is also strictly increasing $\forall~b\in(1,+\infty)$ . Hence, we have,

$\log_5n\lt \frac{3}{2}\implies n\lt 5^{3/2}=\sqrt{125}\in [11,12)\\ \implies n\leq 11~\textrm{for integer }n\implies \max(n)=11~\textrm{for integer }n$

Prasun Biswas - 6 years, 1 month ago

Fabio Bittar
May 22, 2015

This is my first attempt at writing a solution, so bear with me.

${ n }^{ 200 }\quad <\quad { 5 }^{ 300 }\quad ;\quad n\quad is\quad a\quad positive\quad integer\\ \\ 300\log { 5 } \quad >\quad 200\log { n } \quad \Longrightarrow \quad 1.5\log { 5 } \quad >\quad \log { n } \\ \log { 5^{ 1.5 } } \quad >\quad \log { n } \quad \longrightarrow \quad \log { 11.18...\quad >\quad \log { n } } \\ \\ n\quad =\quad 11$

Swayamsiddha Mohapatra
Jun 13, 2015

To find the largest integer closest but less than $5^{300}$ let us first equate the two and find the exact solution for which $n^{200}$ = $5^{300}$

Hence

n= $5^{300/200}$

n= $5^{1.5}$

n= $\sqrt{125}$

n=11.180339887

So the closest whole number value of n which is less than 11.18033 is 11. and hence n=11 would be the largest integer to satisfy the condition

Moderator note:

This is a very common mistake made, which is to write it as an equality instead of an inequality. You have to be careful that the steps which you take are valid as an inequality, and not just as an equality.

For example, in solving $x^2 < 1$ , we cannot simply use $x^2 = 1 \Rightarrow x = -1, 1$ and pretend that the answer is $x < -1, 1$ .

Andrew Effendy
Nov 29, 2015

5 log x=300
log x/log 5=300
log x=300 log 5
n log x=200 log x/log n=200 log x=200 log n

300 log 5 = 200 log n 3/2 = 5 log n n = 5^3/2 n=11,...

so, the integer must be 11.

A Former Brilliant Member
May 14, 2015

n^200 < 5^300

Raising both sides to the power of 1^100 gives:

n^2 < 5^3

n^2 <125

Since 11^2 = 121 and 12^2 = 144, the largest value of n^2 is greater than 11 but less than 12.

So the largest integer n which satisfies the inequality is n=11.

Pratyush Pushkar
May 13, 2015

taking log both side 200logn<300log5 , logn<(3/2)log5 , n<5^1.5 , n=11

Saunak Bhattacharjee
May 12, 2015

n^200 > 5^300, implies that n^2 > 5^3=125. Now the largest perfect square number which comes before 125 ,is 121=11^2. hence, n=11.

Shrinath Jani
May 8, 2015

n<(5)^(300/200) n<5^1.5 n<11.somthng so 11 is largest number

Anurag Malakar
May 8, 2015

n<5^(300/200)

n<5^(3/2)

So, n<5*root(5)<12

The largest integer is, therefore 11

"Powerful" inequality

The answer is 11.

12 solutions

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