Difference between revisions of "2019 AMC 10B Problems/Problem 6"
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Dividing both sides by <math>n!</math> as before gives <math>(n+1)+(n+1)(n+2)=440</math>. Now factor out <math>(n+1)</math>, giving <math>(n+1)(n+3)=440</math>. By considering the prime factorization of <math>440</math>, a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>n=19</math>, so the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>. | Dividing both sides by <math>n!</math> as before gives <math>(n+1)+(n+1)(n+2)=440</math>. Now factor out <math>(n+1)</math>, giving <math>(n+1)(n+3)=440</math>. By considering the prime factorization of <math>440</math>, a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>n=19</math>, so the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>. | ||
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+ | ===Solution 4=== | ||
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+ | Since <math>(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!</math>, the result can be factored into <math>(n+1)(n+3)n!=440 \cdot n!</math> and divided by <math>n!</math> on both sides to get <math>(n+1)(n+3)=440</math>. From there, it is easier to complete the square with the quadratic <math>(n+1)(n+3) = n^2 + 4n + 3</math>, so <math>n^2+4n+4=441 \Rightarrow (n+2)^2=441</math>. Solving for <math>n</math> results in <math>n=19,-23</math>, and since <math>n>0</math>, <math>n=19</math> and the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>. | ||
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+ | ~Randomlygenerated | ||
== Video Solution == | == Video Solution == |
Revision as of 23:22, 24 October 2021
- The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.
Contents
[hide]Problem
There is a positive integer such that
. What is the sum of the digits of
?
Solution
Solution 1
Solving by the quadratic formula, (since clearly
). The answer is therefore
.
Solution 2
Dividing both sides by gives
Since
is non-negative,
. The answer is
.
Solution 3
Dividing both sides by as before gives
. Now factor out
, giving
. By considering the prime factorization of
, a bit of experimentation gives us
and
, so
, so the answer is
.
Solution 4
Since , the result can be factored into
and divided by
on both sides to get
. From there, it is easier to complete the square with the quadratic
, so
. Solving for
results in
, and since
,
and the answer is
.
~Randomlygenerated
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1956
~ pi_is_3.14
Video Solution
~IceMatrix
~savannahsolver
See Also
2019 AMC 10B (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 10 Problems and Solutions |
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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.