A Beautiful Problem

For some mNm\in\mathbb N, there are 2m2^m positive real numbers written on a board product of which is 11. At each step, we can choose two numbers a,ba,b and replace both by a+ba+b. Prove that after 2m1m2^{m-1}m steps the sum of all numbers on board will be at least 4m4^m.

Note by Jubayer Nirjhor
4 years, 5 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 2m 2^m and can never be 4m 4^m

. Also, the number of numbers on the boards decreases by one every step. So after 2m1 2^m -1 steps, there should only be 1 1 number left on the board, after which no moves are possible. So you can't have 2m1m 2^{m-1}*m moves since 2m1m>2m1 2^{m-1}*m > 2^{m} -1 for m2 m \geq 2

Siddhartha Srivastava - 4 years, 5 months ago

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

Jubayer Nirjhor - 4 years, 5 months ago

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

Anyways, the problem seems easy. Just apply the "step" to the two largest numbers which has a sum of 2 \geq 2 . Each "step" doubles the sum of the numbers. Initial sum = a+b a+b , final sum =2(a+b) = 2(a+b) . Applying it 2m1m 2^{m-1}m times, the sum of those two numbers is now 22m1m \geq 2^{2^{m-1}m} . which is obviously greater than 22m 2^{2m}

Siddhartha Srivastava - 4 years, 5 months ago

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@Siddhartha Srivastava Not so fast! If initial terms are a1,...,a2ma_1,...,a_{2^m} then initial sum is a1++a2ma_1+\cdots+a_{2^m}, while the sum after applying the step is 2a1+2a2+a3++a2m2a_1+2a_2+a_3+\cdots+a_{2^m}. So the sum is not doubled.

Jubayer Nirjhor - 4 years, 5 months ago

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@Jubayer Nirjhor 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 2 \geq 2 . Now, one step doubles the sum of these two numbers. Applying the step 2m1m 2^{m-1}*m times we see that the sum of these two numbers is now 22m1m \geq 2^{2^{m-1}m} . The sum of the numbers we ignored is 0 \geq 0 . so doesn't matter. Clearly, the sum of all the numbers is now 22m1m 2^{2^{m-1}m} which is clearly greater than 22m 2^{2m} .

Siddhartha Srivastava - 4 years, 5 months ago

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

Jubayer Nirjhor - 4 years, 5 months ago

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@Jubayer Nirjhor Any order? No it doesn't. Take the case where m=2 m = 2 . Take x1,x2,2x1,2x2 x_1, x_2, 2-x_1,2-x_2 . Sum of these 22 2^2 numbers is 22 2^2

Apply the step 4 times on x1,x2 x_1, x_2 . Then the four numbers are 4(x1+x2),4(x1+x2),2x1,2x2 4(x_1+x_2),4(x_1+x_2), 2-x_1, 2-x_2 .

Sum of these numbers is 4+7(x1+x2) 4 + 7(x_1+x_2) . If we take x1,x2 x_1,x_2 arbitrarily small, it will not be greater than 16 16 .

Siddhartha Srivastava - 4 years, 5 months ago

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@Siddhartha Srivastava Sorry. A change has been made. The initial product of the numbers is 11. No restriction on the sum.

Jubayer Nirjhor - 4 years, 5 months ago

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