2019 AMC 10A Problem 23

Problem 23 of 25HarderAlgebra

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?

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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.

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