be the term-inator

Probability Level 4

We have a series where the sum of any 7 consecutive terms is negative and the sum of any 11 consecutive terms is positive. What is the maximum number of terms in this series?

The answer is 16.

2 solutions

Changming Xu
Dec 19, 2013

If there were 17 terms we see that there are 7 groups of 11 terms and 11 groups of 7 terms. If we sum the 7 groups of 11 terms, since each group has a positive sum, the total would be positive. However, if we sum the 11 groups of 7 terms, since each group has a negative sum, the total would be negative. This is a contradiction. This contradiction does not occur with 16 or less terms thus the maximum number of terms is 16.

An example of 16 terms is necessary to complete the argument.

George G - 7 years, 5 months ago

{5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13, 5, 5}, I think this works?

Changming Xu - 7 years, 5 months ago

almost, I think the signs are reversed.

George G - 7 years, 5 months ago

@George G – Okay fixed

Changming Xu - 7 years, 5 months ago

It's also important to be clear that 17 is the first number such that each term gets added the same number of times. With 16, per say, the sums of the 7-sets and 11-sets don't have to be the same, because some terms get counted more than others. The key concept here is double counting the sum.

Riley Pinkerton - 7 years, 5 months ago

Sorry for spamming your solution, this is for the Brilliant staff. This problem is very well-known since it was included in IMO 1977 and featured in Arthur Engel's book Problem-Solving Strategies. I think that problems that are very popular such as this one shouldn't be featured in Brilliant since it's very easy to just look it up or cheat in different ways.

José Marín Guzmán - 7 years, 5 months ago

Exactly, it's IMO 1977 .

Jon Haussmann - 7 years, 5 months ago

"If we sum the 7 groups of 11 terms, since each group has a positive sum, the total would be positive. "

I don't follow..

Akshaj Kadaveru - 7 years, 5 months ago

The problem states that every group of 11 terms has a positive sum so summing 7 groups of 11 terms would also have a positive sum. Does that make more sense?

Changming Xu - 7 years, 5 months ago

I recommend you to start reading problem-solving strategies

Matthew Fan - 7 years, 5 months ago

Thanks for the feedback!

Changming Xu - 7 years, 5 months ago

Adit Mohan
Feb 17, 2014

first i showed it is not possible to have 17 terms as follows.
if we make an array :.
a1 a2 a3 a4... a7.
a2 a3 a4 a5 a8.
. . a11 a12 ..... a17.
if we take sums by horizontal rows we get the sum of all the terms in the array is positive(since sum of 7 consecutive terms is positive) similarly if we sum by columns we get the sum of array negative. this is contradictory so 17 terms are not possible.
next, notice that 16 terms are possible for ex.
8,8,-21,8,8,8,-21,8,8,-21,8,8,8,-21,8,8

be the term-inator

The answer is 16.

2 solutions

0 pending reports