Question

Jo and Blair take turns counting from $1$ to one more than the last number said by the other person.

Jo starts by saying"1", so Blair follows by saying "1, 2". Jo then says "1, 2, 3", and so on.

What is the $53^{rd}$ number said?

$2$

$3$

$5$

$6$

$8$

Solution:

The sequence $1,2 \cdots n$ is said first after the $n^{th}$ triangular number $T_n,$ which is: $T_n = \dfrac{n(n+1)}2.$

(The triangular number $T_n$ is defined as being the sum of the numbers $1$ to $n.$ As for why it has this formula, there's a bunch of ways you can prove it using induction or arithmetic sequences. We'll leave that to you, though!)

Moving on, writing out some triangular numbers, we notice that $T_9 = 45$ is the closest triangular number less than $53.$ This means that the sequence $1,2 \cdots 9$ is first said after $T_9 = \dfrac{9\cdot 10}2 = 45$ numbers.

Therefore, the $46^{th}$ number starts this new sequence at $1,$ and counting forwards, we can see that the $53rd$ number is $7$ more than this, with a value of $8.$

Thus, the correct answer is E .

2013 AMC 10B Problem 13

Solution:

Problem 13 in Other Years