# A Beautiful Problem

For some $m\in\mathbb N$, there are $2^m$ positive real numbers written on a board product of which is $1$. At each step, we can choose two numbers $a,b$ and replace both by $a+b$. Prove that after $2^{m-1}m$ steps the sum of all numbers on board will be at least $4^m$.

Note by Jubayer Nirjhor
6 years, 2 months ago

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How does this work? I'm confused by the #WishfulThinking tag. Is the question correct?

The sum of all the numbers is invariant and equal to $2^m$ and can never be $4^m$

. Also, the number of numbers on the boards decreases by one every step. So after $2^m -1$ steps, there should only be $1$ number left on the board, after which no moves are possible. So you can't have $2^{m-1}*m$ moves since $2^{m-1}*m > 2^{m} -1$ for $m \geq 2$

- 6 years, 2 months ago

We replace "both" of $a,b$ by $a+b$ so $(a,b)$ changes to $(a+b,a+b)$. Thus the number of terms remains same.

- 6 years, 2 months ago

Ah. Okay. The question is still ambiguously worded though. A better wording would be " we replace each of $a,b$ with $(a+b)$"

Anyways, the problem seems easy. Just apply the "step" to the two largest numbers which has a sum of $\geq 2$. Each "step" doubles the sum of the numbers. Initial sum = $a+b$, final sum $= 2(a+b)$. Applying it $2^{m-1}m$ times, the sum of those two numbers is now $\geq 2^{2^{m-1}m}$. which is obviously greater than $2^{2m}$

- 6 years, 2 months ago

Not so fast! If initial terms are $a_1,...,a_{2^m}$ then initial sum is $a_1+\cdots+a_{2^m}$, while the sum after applying the step is $2a_1+2a_2+a_3+\cdots+a_{2^m}$. So the sum is not doubled.

- 6 years, 2 months ago

Sorry. I wasn't clear. I meant the sum of the largest two numbers when I said the sum will be doubled.

My solution was to just ignore the other numbers. Take only the two largest numbers. Their sum has to $\geq 2$. Now, one step doubles the sum of these two numbers. Applying the step $2^{m-1}*m$ times we see that the sum of these two numbers is now $\geq 2^{2^{m-1}m}$. The sum of the numbers we ignored is $\geq 0$. so doesn't matter. Clearly, the sum of all the numbers is now $2^{2^{m-1}m}$ which is clearly greater than $2^{2m}$.

- 6 years, 2 months ago

You don't decide which numbers to choose. The statement holds for any order.

- 6 years, 2 months ago

Any order? No it doesn't. Take the case where $m = 2$. Take $x_1, x_2, 2-x_1,2-x_2$. Sum of these $2^2$ numbers is $2^2$

Apply the step 4 times on $x_1, x_2$. Then the four numbers are $4(x_1+x_2),4(x_1+x_2), 2-x_1, 2-x_2$.

Sum of these numbers is $4 + 7(x_1+x_2)$. If we take $x_1,x_2$ arbitrarily small, it will not be greater than $16$.

- 6 years, 2 months ago

Sorry. A change has been made. The initial product of the numbers is $1$. No restriction on the sum.

- 6 years, 2 months ago