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Difference between revisions of "2001 AMC 10 Problems/Problem 12"

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== Problem ==
 
== Problem ==
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Suppose that <math> n </math> is the product of three consecutive integers and that <math> n </math> is divisible by <math> 7 </math>. Which of the following is not necessarily a divisor of <math> n </math>?
 
Suppose that <math> n </math> is the product of three consecutive integers and that <math> n </math> is divisible by <math> 7 </math>. Which of the following is not necessarily a divisor of <math> n </math>?
  
 
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42 </math>
 
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42 </math>
  
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== Solution ==
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== Solutions ==
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=== Solution 1 ===
  
 
Whenever <math> n </math> is the product of three consecutive integers, <math> n </math> is divisible by <math> 3! </math>, meaning it is divisible by <math> 6 </math>.
 
Whenever <math> n </math> is the product of three consecutive integers, <math> n </math> is divisible by <math> 3! </math>, meaning it is divisible by <math> 6 </math>.
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In our answer choices, the one that is not a factor of <math> 42 </math> is <math> \boxed{\textbf{(D)}\ 28} </math>.
 
In our answer choices, the one that is not a factor of <math> 42 </math> is <math> \boxed{\textbf{(D)}\ 28} </math>.
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=== Solution 2 ===
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We can look for counterexamples. For example, letting <math>n = 13 \cdot 14 \cdot 15</math>, we see that <math>n</math> is not divisible by 28, so <math>\boxed{\textbf{(D) }28}</math> is our answer.
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===Solution 3(elimination)===
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No matter what 3 integers you choose, one of them has to be even, so since <math>14 = 7 \cdot 2</math>, and it has 7 and 2 as a divisor, answer B is out. Now, if it wasn't divisible by 3, it could be A or C(<math>21 = 7 \cdot 3</math>,and <math>6 = 2 \cdot 3</math>)m so it must be divisible by 3. Therefore, it is either D or E. Since we eliminated 6, if it was E, it would be not divisible by 6(<math>42 = 7 \cdot 6</math>), but it is not, so the answer is <math> \boxed{\textbf{(D)}\ 28} </math>.
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~idk12345678
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==Video Solution by Daily Dose of Math==
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https://youtu.be/Ce2zWT2A0sU?si=FxjhpB2Tq0vHcl9B
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~Thesmartgreekmathdude
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2001|num-b=11|num-a=13}}
 
{{AMC10 box|year=2001|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 20:42, 15 July 2024

Problem

Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7$. Which of the following is not necessarily a divisor of $n$?

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42$

Solutions

Solution 1

Whenever $n$ is the product of three consecutive integers, $n$ is divisible by $3!$, meaning it is divisible by $6$.

It also mentions that it is divisible by $7$, so the number is definitely divisible by all the factors of $42$.

In our answer choices, the one that is not a factor of $42$ is $\boxed{\textbf{(D)}\ 28}$.

Solution 2

We can look for counterexamples. For example, letting $n = 13 \cdot 14 \cdot 15$, we see that $n$ is not divisible by 28, so $\boxed{\textbf{(D) }28}$ is our answer.

Solution 3(elimination)

No matter what 3 integers you choose, one of them has to be even, so since $14 = 7 \cdot 2$, and it has 7 and 2 as a divisor, answer B is out. Now, if it wasn't divisible by 3, it could be A or C($21 = 7 \cdot 3$,and $6 = 2 \cdot 3$)m so it must be divisible by 3. Therefore, it is either D or E. Since we eliminated 6, if it was E, it would be not divisible by 6($42 = 7 \cdot 6$), but it is not, so the answer is $\boxed{\textbf{(D)}\ 28}$.

~idk12345678

Video Solution by Daily Dose of Math

https://youtu.be/Ce2zWT2A0sU?si=FxjhpB2Tq0vHcl9B

~Thesmartgreekmathdude

See Also

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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