Difference between revisions of "2001 AMC 10 Problems/Problem 12"
(→Solution 1) |
|||
(4 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | |||
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> | ||
− | === Solution 1=== | + | == Solutions == |
+ | === 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>. | ||
Line 13: | Line 13: | ||
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>. | ||
− | == Solution 2 == | + | === Solution 2 === |
− | 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 (D) is our answer. | + | 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. |
+ | |||
+ | ===Solution 3(elimination)=== | ||
+ | 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>. | ||
+ | |||
+ | ~idk12345678 | ||
+ | |||
+ | ==Video Solution by Daily Dose of Math== | ||
+ | |||
+ | https://youtu.be/Ce2zWT2A0sU?si=FxjhpB2Tq0vHcl9B | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
== See Also == | == See Also == |
Latest revision as of 20:42, 15 July 2024
Contents
Problem
Suppose that is the product of three consecutive integers and that is divisible by . Which of the following is not necessarily a divisor of ?
Solutions
Solution 1
Whenever is the product of three consecutive integers, is divisible by , meaning it is divisible by .
It also mentions that it is divisible by , so the number is definitely divisible by all the factors of .
In our answer choices, the one that is not a factor of is .
Solution 2
We can look for counterexamples. For example, letting , we see that is not divisible by 28, so is our answer.
Solution 3(elimination)
No matter what 3 integers you choose, one of them has to be even, so since , 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(,and )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(), but it is not, so the answer is .
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.