Difference between revisions of "2014 AMC 10A Problems/Problem 24"

m (See Also)
(One intermediate revision by one other user not shown)
Line 21: Line 21:
 
<math>1,2,3,4,5,...,500,000</math>
 
<math>1,2,3,4,5,...,500,000</math>
  
All we need to do is count how many numbers are skipped, <math>n</math>, and "push" (add on to) <math>500,000</math> however many numbers are skipped.
+
All we need to do is count how many numbers are skipped, <math>n</math>, and "push" (add on to) <math>500,000</math> according to however many numbers are skipped.
  
 
Clearly, <math>\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}</math>. This means that the number of skipped number "blocks" in the sequence is <math>999-3=996</math> because we started counting from 4.
 
Clearly, <math>\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}</math>. This means that the number of skipped number "blocks" in the sequence is <math>999-3=996</math> because we started counting from 4.
Line 27: Line 27:
 
Therefore <math>n=\frac{996(997)}{2}=496,506</math>, and the answer is <math>496,506+500000=\boxed{\textbf{(A)}996,506}</math>.
 
Therefore <math>n=\frac{996(997)}{2}=496,506</math>, and the answer is <math>496,506+500000=\boxed{\textbf{(A)}996,506}</math>.
  
==Video Solution==
+
 
 +
=== Video Solution by Richard Rusczyk ===
 +
 
 
https://www.youtube.com/watch?v=KfGtE4G6tBo&t=427s
 
https://www.youtube.com/watch?v=KfGtE4G6tBo&t=427s
 +
 +
~ dolphin7
  
 
==See Also==
 
==See Also==

Revision as of 13:28, 2 January 2021

Problem

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?

$\textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510$

Solution 1

If we list the rows by iterations, then we get

$1,2,3,4$

$6,7,8,9,10$

$13,14,15,16,17,18$ etc.

so that the $500,000$th number is the $506$th number on the $997$th row because $4+5+6+7......+999 = 499,494$. The last number of the $996$th row (when including the numbers skipped) is $499,494 + (1+2+3+4.....+996)= 996,000$, (we add the $1-996$ because of the numbers we skip) so our answer is $996,000 + 506 = \boxed{\textbf{(A)}996,506}$.

Solution 2

Let's start with natural numbers, with no skips in between.

$1,2,3,4,5,...,500,000$

All we need to do is count how many numbers are skipped, $n$, and "push" (add on to) $500,000$ according to however many numbers are skipped.

Clearly, $\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}$. This means that the number of skipped number "blocks" in the sequence is $999-3=996$ because we started counting from 4.

Therefore $n=\frac{996(997)}{2}=496,506$, and the answer is $496,506+500000=\boxed{\textbf{(A)}996,506}$.


Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=KfGtE4G6tBo&t=427s

~ dolphin7

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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. AMC logo.png