Positive eights, negative fives
A finite sequence of real numbers has the property that every block of 8 consecutive terms has a positive sum, and every block of 5 consecutive terms has a negative sum.
What is the greatest possible length of the sequence?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
To show that 12 terms are impossible, write the terms in a rectangular array and add it up in two ways.
Use the array whose entry in row , column is .
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s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. For 12 terms, a array of terms has rows that are 8-blocks (positive sums) and columns that are 5-blocks (negative sums). The same total cannot be both positive and negative. Then an example shows that 11 terms are possible.
Step 1: 12 is impossible. Suppose works. Put in row and column , for and ; indices go up to 12. Each row is 8 consecutive terms, so the total of the array is a sum of 5 positive numbers: positive. Each column is 5 consecutive terms, so the total is also a sum of 8 negative numbers: negative. Contradiction. So no sequence has 12 or more terms.
Step 2: 11 works. Take Every 5 consecutive terms contain two and three 1: sum . Every 8 consecutive terms contain three and five 1: sum .
So the greatest length is 11.
In general. With blocks of and terms, the maximum length is : here .