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Difference between revisions of "2012 AMC 8 Problems/Problem 15"
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<math> \textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}61\text{ and }65\qquad\textbf{(E)}\hspace{.05in}66\text{ and }99 </math>
 
<math> \textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}61\text{ and }65\qquad\textbf{(E)}\hspace{.05in}66\text{ and }99 </math>
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==Video Solution==
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https://youtu.be/rQUwNC0gqdg?t=172
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https://www.youtube.com/watch?v=Vfsb4nwvopU  ~David
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https://youtu.be/hOnw5UtBSqI ~savannahsolver
  
 
==Solution==
 
==Solution==
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To find the answer to this problem, we need to find the least common multiple of <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math> and add <math>2</math> to the result. The least common multiple of the four number is <math>60</math>, and adding <math>2</math>, we find the number we want is <math>62</math>. Now we need to find the range which contains <math>62</math>. The only such range and our final answer is <math> \boxed{\textbf{(D)}\ 61\text{ and }65} </math>.
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To find the answer to this problem, we need to find the least common multiple of <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math> and add <math>2</math> to the result. To calculate the least common multiple, we need to find the prime factorization for each of them. <math>3 = 3^{1}</math>, <math>4 = 2^{2}</math>, <math>5 = 5^{1}</math>, and <math>6 = </math>2^{1}*{3^{1}}<cmath>. So the least common multiple of the four numbers is </cmath>2^{2}*{3^{1}*{5^{1}}}<math> = 60</math>, and by adding <math>2</math>, we find that that such number is <math>62</math>. Now we need to find the only given range that contains <math>62</math>. The only such range is answer <math>\textbf{(D)}</math>, and so our final answer is <math> \boxed{\textbf{(D)}\ 61\text{ and }65} </math>.
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~NXC
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=14|num-a=16}}
 
{{AMC8 box|year=2012|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 15:32, 16 November 2024

Problem

The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?

$\textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}61\text{ and }65\qquad\textbf{(E)}\hspace{.05in}66\text{ and }99$

Video Solution

https://youtu.be/rQUwNC0gqdg?t=172

https://www.youtube.com/watch?v=Vfsb4nwvopU ~David

https://youtu.be/hOnw5UtBSqI ~savannahsolver

Solution

To find the answer to this problem, we need to find the least common multiple of $3$, $4$, $5$, $6$ and add $2$ to the result. To calculate the least common multiple, we need to find the prime factorization for each of them. $3 = 3^{1}$, $4 = 2^{2}$, $5 = 5^{1}$, and $6 =$2^{1}*{3^{1}}\[. So the least common multiple of the four numbers is\]2^{2}*{3^{1}*{5^{1}}}$= 60$, and by adding $2$, we find that that such number is $62$. Now we need to find the only given range that contains $62$. The only such range is answer $\textbf{(D)}$, and so our final answer is $\boxed{\textbf{(D)}\ 61\text{ and }65}$.

~NXC

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AJHSME/AMC 8 Problems and Solutions

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