Difference between revisions of "2020 AMC 12B Problems/Problem 6"
Countmath1 (talk | contribs) (→Solution 2) |
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<cmath>(n+2)(n+1)-(n+1)</cmath> | <cmath>(n+2)(n+1)-(n+1)</cmath> | ||
− | Factoring out <math>(n+1)</math> | + | Factoring out <math>(n+1),</math> we get <cmath>(n+1)(n+2-1) = (n+1)^2</cmath> |
− | 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> |
==Solution 2== | ==Solution 2== | ||
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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 added to encourage the selection of choice <math>\textbf{(B)}</math> | + | 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 added to encourage the selection of choice <math>\textbf{(B)}.</math> |
~DBlack2021 (Solution Writing) | ~DBlack2021 (Solution Writing) |
Revision as of 11:18, 18 October 2021
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 this expression:
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 most likely added to encourage the selection of choice
~DBlack2021 (Solution Writing)
~Countmath1 (Minor edits in formatting)
~MRENTHUSIASM (Edits in Logic)
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=2234
~ pi_is_3.14
Video Solution
~IceMatrix
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.