AMC 10 Daily Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 24 Hard

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 2023^{\text{rd}} number said by Tadd?

  • A.

    5743

  • B.

    5885

  • C.

    5983

  • 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 2023 \leqslant\frac{k(3k-1)}{2}. The value of k will tell us in which round Tadd says his 2023^{\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 (2023-1926)=97^{\text{th}} 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 \frac{108\times(108+1)}{2}=5886. Thus the answer is 5886+97=\boxed{(\text{C})5983}.