Difference between revisions of "2014 AMC 10A Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | A sequence of natural numbers is constructed by listing the first <math>4</math>, then skipping one, listing the next <math>5</math>, skipping <math>2</math>, listing <math>6</math>, skipping <math>3</math>, and, on the <math>n</math>th iteration, listing <math>n+3</math> and skipping <math>n</math>. The sequence begins <math>1,2,3,4,6,7,8,9,10,13</math>. What is the <math>500,000</math>th number in the sequence? | + | A sequence of natural numbers is constructed by listing the first <math>4</math>, then skipping one, listing the next <math>5</math>, skipping <math>2</math>, listing <math>6</math>, skipping <math>3</math>, and, on the <math>n</math>th iteration, listing <math>n+3</math> and skipping <math>n</math>. The sequence begins <math>1,2,3,4,6,7,8,9,10,13</math>. What is the <math>500,\!000</math>th number in the sequence? |
− | <math> \textbf{(A)}\ 996,506\qquad\textbf{(B)}\ | + | <math> \textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510 </math> |
− | + | ==Solution 1== | |
− | ==Solution== | ||
If we list the rows by iterations, then we get | If we list the rows by iterations, then we get | ||
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<math>13,14,15,16,17,18</math> etc. | <math>13,14,15,16,17,18</math> etc. | ||
− | so that the <math>500,000</math>th number is the <math>506</math>th number on the <math>997</math>th row | + | so that the <math>500,000</math>th number is the <math>506</math>th number on the <math>997</math>th row because <math>4+5+6+7......+999 = 499,494</math>. The last number of the <math>996</math>th row (when including the numbers skipped) is <math>499,494 + (1+2+3+4.....+996)= 996,000</math>, (we add the <math>1-996</math> because of the numbers we skip) so our answer is <math>996,000 + 506 = \boxed{\textbf{(A)}996,506}</math>. |
+ | |||
+ | ==Solution 2== | ||
+ | Let's start with natural numbers, with no skips in between. | ||
+ | |||
+ | <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> 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. | ||
+ | |||
+ | 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 by Richard Rusczyk === | ||
+ | |||
+ | https://www.youtube.com/watch?v=KfGtE4G6tBo&t=427s | ||
+ | |||
+ | ~ dolphin7 | ||
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2014|ab=A|num-b=23|num-a=25}} | {{AMC10 box|year=2014|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 13:28, 2 January 2021
Problem
A sequence of natural numbers is constructed by listing the first , then skipping one, listing the next , skipping , listing , skipping , and, on the th iteration, listing and skipping . The sequence begins . What is the th number in the sequence?
Solution 1
If we list the rows by iterations, then we get
etc.
so that the th number is the th number on the th row because . The last number of the th row (when including the numbers skipped) is , (we add the because of the numbers we skip) so our answer is .
Solution 2
Let's start with natural numbers, with no skips in between.
All we need to do is count how many numbers are skipped, , and "push" (add on to) according to however many numbers are skipped.
Clearly, . This means that the number of skipped number "blocks" in the sequence is because we started counting from 4.
Therefore , and the answer is .
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=KfGtE4G6tBo&t=427s
~ dolphin7
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
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 |
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