For some m∈N, there are 2m 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 2m−1m steps the sum of all numbers on board will be at least 4m.
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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 and can never be 4m
. Also, the number of numbers on the boards decreases by one every step. So after 2m−1 steps, there should only be 1 number left on the board, after which no moves are possible. So you can't have 2m−1∗m moves since 2m−1∗m>2m−1 for m≥2
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 ≥2. Each "step" doubles the sum of the numbers. Initial sum = a+b, final sum =2(a+b). Applying it 2m−1m times, the sum of those two numbers is now ≥22m−1m. which is obviously greater than 22m
@Siddhartha Srivastava
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Not so fast! If initial terms are a1,...,a2m then initial sum is a1+⋯+a2m, while the sum after applying the step is 2a1+2a2+a3+⋯+a2m. So the sum is not doubled.
@Jubayer Nirjhor
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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. Now, one step doubles the sum of these two numbers. Applying the step 2m−1∗m times we see that the sum of these two numbers is now ≥22m−1m. The sum of the numbers we ignored is ≥0. so doesn't matter. Clearly, the sum of all the numbers is now 22m−1m which is clearly greater than 22m.
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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 and can never be 4m
. Also, the number of numbers on the boards decreases by one every step. So after 2m−1 steps, there should only be 1 number left on the board, after which no moves are possible. So you can't have 2m−1∗m moves since 2m−1∗m>2m−1 for m≥2
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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.
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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 ≥2. Each "step" doubles the sum of the numbers. Initial sum = a+b, final sum =2(a+b). Applying it 2m−1m times, the sum of those two numbers is now ≥22m−1m. which is obviously greater than 22m
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a1,...,a2m then initial sum is a1+⋯+a2m, while the sum after applying the step is 2a1+2a2+a3+⋯+a2m. So the sum is not doubled.
Not so fast! If initial terms areLog in to reply
My solution was to just ignore the other numbers. Take only the two largest numbers. Their sum has to ≥2. Now, one step doubles the sum of these two numbers. Applying the step 2m−1∗m times we see that the sum of these two numbers is now ≥22m−1m. The sum of the numbers we ignored is ≥0. so doesn't matter. Clearly, the sum of all the numbers is now 22m−1m which is clearly greater than 22m.
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m=2. Take x1,x2,2−x1,2−x2. Sum of these 22 numbers is 22
Any order? No it doesn't. Take the case whereApply the step 4 times on x1,x2. Then the four numbers are 4(x1+x2),4(x1+x2),2−x1,2−x2.
Sum of these numbers is 4+7(x1+x2). If we take x1,x2 arbitrarily small, it will not be greater than 16.
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1. No restriction on the sum.
Sorry. A change has been made. The initial product of the numbers is