2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 23 Easy

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 10000 is reached. What is the 2019th number said by Tadd? (2019 AMC 10A Problem, Question#23)

  • A.

    5743

  • B.

    5885

  • C.

    5979

  • D.

    6001

  • E.

    6011

Answer:C

Define a round as one complete rotation through each of the three children.

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

    Round            Tadd            Todd            Tucker
       1                   1              2\sim3           4\sim6
       2              7\sim10      11\sim15      16\sim21
       3            22\sim28      29\sim36      37\sim45
       4            46\sim55      56\sim66      67\sim78

Notice that at the end of the n , the last number said is the 3n^{\text{th}} triangular 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+\cdots+(3n-2)=\frac{n(3n-1)}{2}, by the arithmetic series sum formula. We therefore want the smallest positive integer k such that 2019 \leqslant\frac{k(3k-1)}{2}. The value of k will tell us in which round Tadd says his 2019^{\text{th}} number. Through guess and check (or by actually solving the quadratic inequality), k=37.

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

We found that the last number said at the very end of the n^{\text{th}} round is the 3n^{\text{th}} triangular number.

For n=36, the 108^{\text{th}} triangular number is 5886. Thus the answer is 5886+96=\boxed{(\text{C})5979}.

Firstly, as in Solution 1, we list how many words Tadd says, Todd says, and Tucker says in each round.

Tadd: 1, 4, 7, 10, 13\cdots

Todd: 2, 5, 8, 11, 14\cdots

Tucker: 3, 6, 9, 12, 15\cdots

We can find a general formula for the number of numbers each of the kids say after the nth round. For Tadd, we can either use the arithmetic series sum formula (like in Solution 1) or standard summation results to get \sum _{i=1}^{n}3n-2=-2n+3\sum _{i=1}^{n}n=-2n+\dfrac{3n(n+1)}{2}=\dfrac{3n^2-2}{2}

.Now, to find the number of rotations Tadd and his siblings go through before Tadd says his 2019th word, we know the inequality \frac{3n^2-n}{2}<2019 must be satisfied, and testing numbers gives the maximum integer value of n as 36.

The next main insight, in order to simplify the computation process, is to notice that the 2019th number Tadd says is simply the number of numbers Todd and Tucker say plus the 2019 Tadd says, which will be the answer since Tadd goes first.

Carrying out the calculation thus becomes quite simple:

\left(\sum _{i=1}^{36}{3n}+\sum _{i=1}^{36}{3n}-1 \right)+2019=\left(\sum _{i=1}^{36}{6n}-1 \right)+2019=\left(5+11+17\cdots+255\right)+2019=\dfrac{36(260)}{2}+2019

At this point, we can note that the last digit of the answer is 9, which gives \boxed{(\text{C})5979}.