Difference between revisions of "2013 AMC 12B Problems/Problem 7"
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We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", <math>1+2+3+4+5+6+7+8+9=45</math> numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because <math>53-45=8</math>. Since we are starting from 1 every turn, the 53rd number said will be <math>\boxed{\textbf{(E) }8}</math>. | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", <math>1+2+3+4+5+6+7+8+9=45</math> numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because <math>53-45=8</math>. Since we are starting from 1 every turn, the 53rd number said will be <math>\boxed{\textbf{(E) }8}</math>. | ||
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| + | ==Faster Solution== | ||
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| + | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. We notice that the number of numbers is <math>1 + 2 + 3 + 4 ...</math> every time we finish a "turn" we notice the sum of these would be the largest number <math>\frac{n(n+1)}{2}</math> under 53, we can easily see that if we double this it's <math>n^2 + n \simeq 106</math>, and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, <math>\frac{9(9+1)}{2} = 45</math> and <math>53 - 45 = \boxed{\textbf{(E) }8}</math> | ||
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| + | ==Solution 3== | ||
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| + | Let <math>T(n)</math> denote the <math>n</math>th triangle number. Then, observe that the <math>T(n)</math>th number said is <math>n</math>. It follows that the <math>55</math>th number is <math>10</math> (as <math>55 = T(10)</math>). Thus, the <math>53</math>rd number is <math>10 - 2 = 8</math>, which is answer choice <math>\boxed{\textbf{(E)}}</math>. | ||
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| + | ~Mathavi | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/a-3CAo4CoWc | https://youtu.be/a-3CAo4CoWc | ||
| − | ~ | + | ~no one |
== See also == | == See also == | ||
Latest revision as of 15:18, 16 October 2022
- The following problem is from both the 2013 AMC 12B #7 and 2013 AMC 10B #13, so both problems redirect to this page.
Problem
Jo and Blair take turns counting from 
to one more than the last number said by the other person. Jo starts by saying 
, so Blair follows by saying 
. Jo then says 
, and so on. What is the 
number said?
Solution
We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", 
numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because 
. Since we are starting from 1 every turn, the 53rd number said will be 
.
Faster Solution
We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. We notice that the number of numbers is 
every time we finish a "turn" we notice the sum of these would be the largest number 
under 53, we can easily see that if we double this it's 
, and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, 
and 
Solution 3
Let 
denote the 
th triangle number. Then, observe that the th number said is . It follows that the 
th number is 
(as 
). Thus, the 
rd number is 
, which is answer choice 
.
~Mathavi
Video Solution
~no one
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
