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.  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>
 
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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Observe that the number <math>n</math> first appears as the <math>n</math>th triangular number said. It follows that the <math>55</math>th number is <math>10</math> (<math>55</math> is the <math>10</math>th triangular number). Thus, the <math>53</math>rd number is <math>10 - 2 = 8</math>, which is answer choice \boxed{\textbf{(E)}}$.
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- Mathavi
  
 
==Video Solution==
 
==Video Solution==

Revision as of 15:10, 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 $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$, so Blair follows by saying $``1, 2"$. Jo then says $``1, 2, 3"$, and so on. What is the $53^{\text{rd}}$ number said?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

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", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because $53-45=8$. Since we are starting from 1 every turn, the 53rd number said will be $\boxed{\textbf{(E) }8}$.

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 $1 + 2 + 3 + 4 ...$ every time we finish a "turn" we notice the sum of these would be the largest number $\frac{n(n+1)}{2}$ under 53, we can easily see that if we double this it's $n^2 + n \simeq 106$, and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, $\frac{9(9+1)}{2} = 45$ and $53 - 45 = \boxed{\textbf{(E) }8}$

Solution 3

Observe that the number $n$ first appears as the $n$th triangular number said. It follows that the $55$th number is $10$ ($55$ is the $10$th triangular number). Thus, the $53$rd number is $10 - 2 = 8$, which is answer choice \boxed{\textbf{(E)}}$.

- Mathavi

Video Solution

https://youtu.be/a-3CAo4CoWc

~no one

See also

2013 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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