Difference between revisions of "2020 AMC 12B Problems/Problem 6"
MRENTHUSIASM (talk | contribs) m |
(→Video Solution by Omega Learn) |
||
(16 intermediate revisions by 3 users not shown) | |||
Line 4: | Line 4: | ||
<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: | |
− | We can now divide out a common factor of <math>n!</math> from each term of | + | <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> | |
− | <cmath>(n+2)(n+1)-(n+1)</cmath> | ||
− | |||
− | Factoring out <math>(n+1),</math> we get <cmath>(n+1 | ||
− | |||
which proves that the answer is <math>\boxed{\textbf{(D) } \text{a perfect square}}.</math> | which proves that the answer is <math>\boxed{\textbf{(D) } \text{a perfect square}}.</math> | ||
Line 27: | Line 23: | ||
This leaves <math>\boxed{\textbf{(D) } \text{a perfect square}}</math> as the only answer choice left. | 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 most likely | + | 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) | ||
+ | |||
+ | ~MRENTHUSIASM (Edits in Logic) | ||
+ | |||
+ | ~Countmath1 (Minor Edits in Formatting) | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
+ | |||
+ | https://youtu.be/1_JHg-1kboE | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
− | |||
− | |||
− | |||
== Video Solution == | == Video Solution == | ||
https://youtu.be/ba6w1OhXqOQ?t=2234 | https://youtu.be/ba6w1OhXqOQ?t=2234 | ||
− | |||
− | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/6ujfjGLzVoE | https://youtu.be/6ujfjGLzVoE | ||
− | |||
− | |||
==See Also== | ==See Also== |
Latest revision as of 15:02, 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 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.