compare $m=50^{50}$ with $n=49^{51}$

Question

A multiple choice question:

If $m=50^{50}$ and $n=49^{51}$, then

(A) $m>n$

(B) $m<n$

(C) $m=n$

(D) The given information is not enough

My attempt:

Since ordinary calculators can not evaluate large numbers as $m$ and $n$, then we can use a trick, which is taking the logarithm of both $m$ and $n$ to the same base, lets use $\ln$ (log to the base $e$).

$50\ln(50)$ VS $51\ln(49)$

$195.60$ VS $198.48$

Hence $49^{51}$ is greater.

So, B must be the correct choice.

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This question was asked in a national exam for high school students.

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However:

• Calculators are not allowed.

• Log tables are not provided.

• Students may not have any knowledge about logarithms and their properties.

• Students should have basic knowledge about exponents like $(a/b)^k=a^k/b^k$, $a^j \times a^k = a^{(j+k)}$, and some other basics.

• The average time to solve a question in this exam is 75 seconds.

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How can we answer this question?

Answer 1

Hint: Use the inequality $$\left(1-\frac{1}{x}\right)^x>\frac{1}{x-1}$$

Answer 2

We have $$\frac{n}{m} = \frac{49^{51}}{50^{50}} = 49 \cdot \left(\frac{49}{50}\right)^{50} = 49 \cdot \left(1-\frac{1}{50}\right)^{50} \approx \frac{49}{e} > 1$$ Even if you dont know that for large $n$ $$ \left(1-\frac{1}{n}\right)^{n} \approx \frac{1}{e}$$ as long as you're able to tell that $$\left(1-\frac{1}{50}\right)^{50} > \frac{1}{49}$$ you're fine. You can get that for example from Bernoulli's inequality. For $x>-1$: $$ (1+x)^n \ge 1+xn$$ so $$ \left(1-\frac{1}{50}\right)^{50} = \left(\left(1-\frac{1}{50}\right)^{25} \right)^2 \ge \left(1-\frac{1}{50}\cdot 25\right)^2 = \left(\frac12\right)^2 = \frac14 > \frac{1}{49}$$

Answer 3

Use Bernoulli's inequality with $x=-\frac{1}{50}$ and $r=48$:

$$\frac{49^{51}}{50^{50}}=\frac{49^3}{50^2}\cdot\frac{49^{48}}{50^{48}}=\frac{49^3}{50^2}\left(1-\frac{1}{50}\right)^{48}\ge\frac{49^3}{50^2}\left(1-\frac{48}{50}\right)=\frac{2\cdot49^3}{50^3}>1$$

Answer 4

In this answer, it is shown that $\left(1+\frac1{n-1}\right)^n$ is decreasing. That means that its reciprocal $\left(1-\frac1n\right)^n$ is increasing. Thus, for $n\ge2$, we have $$ \left(1-\frac1n\right)^n\ge\frac14\tag1 $$ Therefore, $$ \begin{align} \frac{49^{51}}{50^{50}} &=49\left(1-\frac1{50}\right)^{50}\\ &\ge\frac{49}4\tag2 \end{align} $$

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Another Proof of $\boldsymbol{(1)}$

Using Theorem $1$ from this answer with $m=2$, we get $$ \begin{align} \left(1-\frac1n\right)^n &\ge1-\frac{n}{n}+\frac{n(n-1)}{2n^2}-\frac{n(n-1)(n-2)}{6n^3}\\ &=\frac{n^2-1}{3n^2}\tag3 \end{align} $$ For $n\ge2$, $(3)$ gives $(1)$.

Answer 5

Knowing the following fairly common limit is very useful:

$$ \lim_{x \rightarrow \infty}\left(1+\frac{1}{x} \right)^x=e \approx 3$$

$$50^{50}<49^{51}$$ $$\left(\frac{50}{49}\right)^{50}<49$$ $$\left(1+\frac{1}{49}\right)^{50}<49$$ $$\left(1+\frac{1}{49}\right)^{50} \approx e \approx 3$$ $$3<49$$ Therefore we conclude that $50^{50}<49^{51}$

Answer 6

Note that $(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+\dots+x^n$

$50^{50}$ and $49^{51}$

$50^{50}$ and $49^{50}\cdot49$

$(\frac{50}{49})^{50}$ and $49$

$(1+\frac{1}{49})^{50}$ and $49$

$1+50(\frac{1}{49})+\underset{\text{Negligible terms}}{\underbrace{\frac{50\times49}{2}(\frac{1}{49})^2+\frac{50\times49\times48}{6}(\frac{1}{49})^3+\dots+(\frac{1}{49})^{50}}}$ and $49$

$1+\frac{50}{49}$ and $49$

Clearly, $1+\frac{50}{49}<49$. Hence $50^{50}<49^{51}$. Thus, B is the correct choice.

Answer 7

Why do you just compute the two numbers? The following code is in Julia.

m = BigInt(50)^50
n = BigInt(49)^51
sign(m-n)

The output is -1 which means $m-n < 0$.