Question

A sequence of natural numbers is constructed by listing the first $4,$ then skipping one, listing the next $5,$ skipping $2,$ listing $6,$ skipping $3,$ and on the $n$ th iteration, listing $n+3$ and skipping $n.$ The sequence begins $1,2,3,4,6,7,8,9,10,13.$ What is the $500,000$ th number in the sequence?

$996,\!506$

$996,\!507$

$996,\!508$

$996,\!509$

$996,\!510$

Answer: A

Concepts:arithmetic sequence summation triangular number

Difficulty rating: 1790

Solution:

After $n$ full iterations, the number of listed terms is $4+5+\cdots+(n+3)=\frac{n(n+7)}2$ .

We need the largest $n$ with $\frac{n(n+7)}2<500000$ . Since $996\cdot1003=998988$ , after $996$ iterations there are $499494$ listed numbers.

The first number listed in iteration $997$ is one more than the total of all listed and skipped numbers so far, namely $996^2+4\cdot996+1=996001$ .

The $500000$ th listed number is the $500000-499494=506$ th number of this next block, so it is $996001+505=996506$ .

Thus, A is the correct answer.

2014 AMC 10A Problem 24

Answer: A

Difficulty rating: 1790

Solution:

Problem 24 in Other Years