Difference between revisions of "2020 AMC 12B Problems/Problem 6"
MRENTHUSIASM (talk | contribs) (→Solution 2: The original solution violates the condition that n>=9. I am adding explanation why violating it is OK.) |
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<cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following? | <cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following? | ||
− | <math>\textbf{(A) } \text{a multiple of } | + | <math>\textbf{(A) } \text{a multiple of 4} \qquad \textbf{(B) } \text{a multiple of 10} \qquad \textbf{(C) } \text{a prime number} \qquad \textbf{(D) } \text{a perfect square} \qquad \textbf{(E) } \text{a perfect cube}</math> |
==Solution 1== | ==Solution 1== | ||
We first expand the expression: | We first expand the expression: | ||
− | <cmath>\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}</cmath> | + | <cmath>\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}.</cmath> |
+ | We can now divide out a common factor of <math>n!</math> from each term of the numerator: | ||
+ | <cmath>(n+2)(n+1)-(n+1).</cmath> | ||
+ | Factoring out <math>(n+1),</math> we get <cmath>[(n+2)-1](n+1) = (n+1)^2,</cmath> | ||
+ | which proves that the answer is <math>\boxed{\textbf{(D) } \text{a perfect square}}.</math> | ||
− | + | ==Solution 2== | |
+ | In the numerator, we factor out an <math>n!</math> to get <cmath>\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1).</cmath> Now, without loss of generality, test values of <math>n</math> until only one answer choice is left valid: | ||
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+ | * <math>n = 1 \implies (3)(2) - (2) = 4,</math> knocking out <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(E)}.</math> | ||
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+ | * <math>n = 2 \implies (4)(3) - (3) = 9,</math> knocking out <math>\textbf{(A)}.</math> | ||
− | < | + | This leaves <math>\boxed{\textbf{(D) } \text{a perfect square}}</math> as the only answer choice left. |
− | + | This solution does not consider the condition <math>n \geq 9.</math> The reason is that, with further testing it becomes clear that for all <math>n,</math> we get <cmath>(n+2)(n+1)-(n+1) = (n+1)^{2},</cmath> as proved in Solution 1. The condition <math>n \geq 9</math> was added most likely to encourage picking <math>\textbf{(B)}</math> and discourage substituting smaller values into <math>n.</math> | |
− | + | ~DBlack2021 (Solution) | |
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+ | ~MRENTHUSIASM (Edits in Logic) | ||
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+ | ~Countmath1 (Minor Edits in Formatting) | ||
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+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
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+ | https://youtu.be/1_JHg-1kboE | ||
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+ | ~Education, the Study of Everything | ||
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== Video Solution == | == Video Solution == | ||
https://youtu.be/ba6w1OhXqOQ?t=2234 | https://youtu.be/ba6w1OhXqOQ?t=2234 | ||
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/6ujfjGLzVoE | https://youtu.be/6ujfjGLzVoE | ||
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==See Also== | ==See Also== |
Latest revision as of 14:02, 8 June 2023
Contents
Problem
For all integers the value of is always which of the following?
Solution 1
We first expand the expression: We can now divide out a common factor of from each term of the numerator: Factoring out we get which proves that the answer is
Solution 2
In the numerator, we factor out an to get Now, without loss of generality, test values of until only one answer choice is left valid:
- knocking out and
- knocking out
This leaves as the only answer choice left.
This solution does not consider the condition The reason is that, with further testing it becomes clear that for all we get as proved in Solution 1. The condition was added most likely to encourage picking and discourage substituting smaller values into
~DBlack2021 (Solution)
~MRENTHUSIASM (Edits in Logic)
~Countmath1 (Minor Edits in Formatting)
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=2234
Video Solution
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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All AMC 12 Problems and Solutions |
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