Question

Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1,$ then Todd must say the next two numbers ( $2$ and $3$ ), then Tucker must say the next three numbers ( $4,$ $5,$ $6$ ), then Tadd must say the next four numbers ( $7,$ $8,$ $9,$ $10$ ), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$ th number said by Tadd?

$5743$

$5885$

$5979$

$6001$

$6011$

Solution:

Tadd speaks on turns of lengths $1,4,7,\ldots$ . After $n$ of Tadd\'s turns, he has said $\sum_{i=1}^n (3i-2)=\frac{3n^2-n}{2}$ numbers.

For $n=36$ , this total is $1926$ , while for $n=37$ it is $2035$ . Therefore Tadd\'s $2019$ th number is the $(2019-1926)=93$ rd number of his $37$ th turn.

Before that turn, the children have completed turns of lengths $1$ through $108$ , saying $1+2+\cdots+108=5886$ numbers. The $93$ rd number of the next turn is $5886+93=5979$ . Thus, C is the correct answer.

2019 AMC 10A Problem 23

Solution:

Problem 23 in Other Years