Difference between revisions of "2019 AMC 10A Problems/Problem 23"
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==Solution== | ==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 & 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} | ||
+ | |||
+ | </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== | ==See Also== |
Revision as of 20:01, 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 , then Todd must say the next two numbers (
and
), then Tucker must say the next three numbers (
,
,
), then Tadd must say the next four numbers (
,
,
,
), 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
is reached. What is the
th number said by Tadd?
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.
Notice that at the end of each round, the last number said is the triangular number where
is the round number.
Tadd says number in round 1,
numbers in round 2,
numbers in round 3, and in general
numbers in round n. At the end of round n, the number of numbers Tadd has said so far is
or
. We want the smallest positive integer
such that
. The value of
will tell us which round Tadd says his 2019th number. Through guess and check,
.
Using our formula , Tadd says
numbers in the first 36 rounds, meaning we are looking for the
number Tadd says in the 37th round.
We found that the last number said at the very end of the round is
triangular number. In particular, for
, the
triangular number is
. Finally,
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.