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

Latest revision as of 23:36, 28 March 2025

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

Solution 4

Let $n = a \cdot (a + 1) \cdot (a + 2)$ . Minimizing $a$ so that 7 is a factor, we see that when $a = 5$ , $n = 5 \cdot 6 \cdot 7$ . Taking the prime factorization of this, we see that you cannot make 28 from $2 \cdot 3 \cdot 5 \cdot 7$ . Therefore, the answer is $\boxed{\textbf{(D)}\ 28}$ .

~ mohizoid

Video Solution by Daily Dose of Math

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

~Thesmartgreekmathdude

@@ Line 22: / Line 22: @@
 ~idk12345678
+=== Solution 4 ===
+Let <math>n = a \cdot (a + 1) \cdot (a + 2)</math>. Minimizing <math>a</math> so that 7 is a factor, we see that when <math>a = 5</math>, <math>n = 5 \cdot 6 \cdot 7</math>. Taking the prime factorization of this, we see that you cannot make 28 from <math>2 \cdot 3 \cdot 5 \cdot 7</math>. Therefore, the answer is <math>\boxed{\textbf{(D)}\ 28}</math>.
+~ mohizoid
+==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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