2013 AMC 12B Problems/Problem 7

The following problem is from both the 2013 AMC 12B #7 and 2013 AMC 10B #13, so both problems redirect to this page.

Problem

Jo and Blair take turns counting from to one more than the last number said by the other person. Jo starts by saying , so Blair follows by saying . Jo then says , and so on. What is the number said?

Solution

We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because . Since we are starting from 1 every turn, the 53rd number said will be .

Faster Solution

We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. We notice that the number of numbers is every time we finish a "turn" we notice the sum of these would be the largest number under 53, we can easily see that if we double this it's , and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, and

Solution 3

Observe that the number first appears as the th triangular number said. It follows that the th number is ( is the th triangular number). Thus, the rd number is , which is answer choice \boxed{\textbf{(E)}}$.

- Mathavi

Video Solution

https://youtu.be/a-3CAo4CoWc

~no one

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.