Difference between revisions of "2014 AMC 10A Problems/Problem 24"
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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 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>. | + | 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>. |
+ | |||
+ | ===Note=== | ||
+ | One may also note we simply need to add the number of skipped numbers to <math>500,000</math> to get our answer. The number of skipped numbers is <math>\frac{996\cdot 997}{2}</math> which has a units digit of <math>6</math>. Looking at the answer choices, it becomes apparent that the answer is <math>\boxed{\textbf{(A) }996,506}</math> | ||
+ | |||
+ | ~firebolt360 | ||
==Solution 2== | ==Solution 2== | ||
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<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. | ||
− | 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>. |
+ | |||
+ | ==Solution 3 (AOPS Video Transcript)== | ||
+ | |||
+ | First, we group the numbers together in the following way: <math>{1, 2, 3, 4, (5)}; {6, 7, 8, 9, 10, (11, 12)}; ...</math> We quickly realize that the number of terms in the curly braces follow a pattern: <math>5, 7, 9, 11, ... , n</math> (where <math>n</math> is the <math>n^\text{th}</math> block. Now, we can tell that the last number in a curly brace will be the number of terms in the set added to the number of terms in all the previous sets. Luckily for us, odd numbers are easy to add. If we pretend that there was a <math>1, 3</math> at the beginning, then the sum of all of the numbers before and including <math>n</math> is <math>(n+2)^2</math>. However, we have to subtract <math>1+3</math> which results in <math>n^2+4n</math>. The amount of numbers in the parenthesis are the <math>n^\text{th}</math> triangular number or <math>\frac{n(n+1)}{2}</math>. Next, we want to find the greatest <math>n</math>, where <math>(n^2+4n) - \frac{n(n+1)}{2}<500000</math>. Simplifying, we get <math>n^2+7n<1000000</math>. We realize that <math>n=1000</math> results in a number just <math>7000</math> greater than our target. Next, we square <math>999</math>: <math>(1000 - 1)^2 = 1000000 - 2001</math>. As we decrease <math>n</math> by <math>1</math>, we decrease the result of the equation by approximately <math>2000</math>. In order to decrease by at least <math>7000</math>, we have to decrease <math>4</math> times leading to <math>n=1000-4=996</math>. We plug it in to <math>n^2+4n</math> getting <math>996\cdot1000=996000</math>. This is the last number in the <math>996^\text{th}</math> set. The number of terms used is <math>\frac{n\cdot(n+7)}{2}=\frac{996\cdot1003}{2}=498\cdot1003=498000+1494=499494</math>. We need to add <math>500000-499494=506</math> terms to get an answer of <math>\boxed{\textbf{(A) }996,506}</math>. | ||
+ | |||
+ | ~MathFun1000 | ||
+ | |||
+ | ==Solution 4(Cheap)== | ||
+ | We look at each option starting from <math>A</math>. Clearly <math>500000 + n(n+1)/2 = 996506</math> where <math>n</math> is the number of the skipped numbers. So we have <math>n^2 + n - 987012 = 0</math>. This factors as <math>(n+997)(n-996)</math>. Since <math>n</math> is an integer the answer is <math>A</math>. | ||
+ | |||
+ | ~coolmath_2018 | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://www.youtube.com/watch?v=KfGtE4G6tBo | ||
+ | |||
+ | ~ dolphin7 | ||
==See Also== | ==See Also== |
Latest revision as of 21:36, 30 October 2021
Contents
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 .
Note
One may also note we simply need to add the number of skipped numbers to to get our answer. The number of skipped numbers is which has a units digit of . Looking at the answer choices, it becomes apparent that the answer is
~firebolt360
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 .
Solution 3 (AOPS Video Transcript)
First, we group the numbers together in the following way: We quickly realize that the number of terms in the curly braces follow a pattern: (where is the block. Now, we can tell that the last number in a curly brace will be the number of terms in the set added to the number of terms in all the previous sets. Luckily for us, odd numbers are easy to add. If we pretend that there was a at the beginning, then the sum of all of the numbers before and including is . However, we have to subtract which results in . The amount of numbers in the parenthesis are the triangular number or . Next, we want to find the greatest , where . Simplifying, we get . We realize that results in a number just greater than our target. Next, we square : . As we decrease by , we decrease the result of the equation by approximately . In order to decrease by at least , we have to decrease times leading to . We plug it in to getting . This is the last number in the set. The number of terms used is . We need to add terms to get an answer of .
~MathFun1000
Solution 4(Cheap)
We look at each option starting from . Clearly where is the number of the skipped numbers. So we have . This factors as . Since is an integer the answer is .
~coolmath_2018
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=KfGtE4G6tBo
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.