Difference between revisions of "2019 AMC 10A Problems/Problem 23"

Revision as of 19:01, 9 February 2019

Problem

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?

$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$

Solution

A round will be defined as one complete rotation through each of the three children.

We should create a table to keep track of what numbers each child says for each round.

$\begin{tabular}{||c c c c||} \hline Round & Tadd & Todd & Tucker \\ [0.5ex] \hline\hline 1 & 1 & 2-3 & 4-6 \\ \hline 2 & 7-10 & 11-15 & 16-21 \\ \hline 3 & 22-28 & 29-36 & 37-45 \\ \hline 4 & 46-55 & 56-66 & 67-78 \\ [1ex] \hline \end{tabular}$

Notice that at the end of each round, the last number said is the $3n^{\text{th}}$ triangular number where $n$ is the round number.

Tadd says $1$ number in round 1, $4$ numbers in round 2, $7$ numbers in round 3, and in general $3n - 2$ numbers in round n. At the end of round n, the number of numbers Tadd has said so far is $1 + 4 + 7 + ... + (3n - 2)$ or $\frac{n(3n-1)}{2}$ . We want the smallest positive integer $k$ such that $2019 \leq \frac{k(3k-1)}{2}$ . The value of $k$ will tell us which round Tadd says his 2019th number. Through guess and check, $k = 37$ .

Using our formula $\frac{n(3n-1)}{2}$ , Tadd says $1926$ numbers in the first 36 rounds, meaning we are looking for the $93^{\text{rd}}$ $(2019 - 1926)$ number Tadd says in the 37th round.

We found that the last number said at the very end of the $n^{\text{th}}$ round is $3n^{\text{th}}$ triangular number. In particular, for $n = 36$ , the $108^{\text{th}}$ triangular number is $5886$ . Finally, $5886 + 93 = \boxed{\textbf{(C) }5979}$

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "2019 AMC 10A Problems/Problem 23"

Revision as of 19:01, 9 February 2019

Problem

Solution

See Also

@@ Line 11: / Line 11: @@
 ==Solution==
+A round will be defined as one complete rotation through each of the three children.
+We should create a table to keep track of what numbers each child says for each round.
+<math>
+ \begin{tabular}{||c c c c||}
+ \hline
+ Round & Tadd & Todd & Tucker \\ [0.5ex]
+ \hline\hline
+& 1 & 2-3 & 4-6 \\
+ \hline
+& 7-10 & 11-15 & 16-21 \\
+ \hline
+& 22-28 & 29-36 & 37-45 \\
+ \hline
+& 46-55 & 56-66 & 67-78 \\ [1ex]
+ \hline
+\end{tabular}
+</math>
+Notice that at the end of each round, the last number said is the <math>3n^{\text{th}}</math> triangular number where <math>n</math> is the round number.
+Tadd says <math>1</math> number in round 1, <math>4</math> numbers in round 2, <math>7</math> numbers in round 3, and in general <math>3n - 2</math> numbers in round n. At the end of round n, the number of numbers Tadd has said so far is <math>1 + 4 + 7 + ... + (3n - 2)</math> or <math>\frac{n(3n-1)}{2}</math>. We want the smallest positive integer <math>k</math> such that <math>2019 \leq \frac{k(3k-1)}{2}</math>. The value of <math>k</math> will tell us which round Tadd says his 2019th number. Through guess and check, <math>k = 37</math>.
+Using our formula <math>\frac{n(3n-1)}{2}</math>, Tadd says <math>1926</math> numbers in the first 36 rounds, meaning we are looking for the <math>93^{\text{rd}}</math> <math>(2019 - 1926)</math> number Tadd says in the 37th round.
+We found that the last number said at the very end of the <math>n^{\text{th}}</math> round is <math>3n^{\text{th}}</math> triangular number. In particular, for <math>n = 36</math>, the <math>108^{\text{th}}</math> triangular number is <math>5886</math>. Finally, <math>5886 + 93 = \boxed{\textbf{(C) }5979}</math>
 ==See Also==