Difference between revisions of "2019 AMC 10B Problems/Problem 6"
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− | Rewrite <math>(n+1)! + (n+2)! = 440</math> as <math>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.</math> Factoring out the <math>n!</math> we get <math>n!(n + 1 + (n+1)(n+2)) = 440.</math> Expand this to get <math>n!(n^2 + 4n + 3) = 440.</math> Factor this to get <math>(n + 1)(n + 3) = 440.</math> If we take the prime factorization of <math>440</math> we see that it is <math>2^3 * 5 * 11.</math> Intuitively, we can find that <math>n + 1 = 20</math> and <math>n + 3 = 22.</math> Therefore, <math>n = 19.</math> Since the problem asks for the sum of the didgits of <math>n</math>, we finally calculate <math>1 + 9 = 10</math> and get answer choice <math>\boxed{\textbf{(C) }10}</math>. | + | Rewrite <math>(n+1)! + (n+2)! = 440 \cdot n!</math> as <math>(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.</math> Factoring out the <math>n!</math> we get <math>n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.</math> Expand this to get <math>n!(n^2 + 4n + 3) = 440 \cdot n!.</math> Factor this and divide by <math>n!</math> to get <math>(n + 1)(n + 3) = 440.</math> If we take the prime factorization of <math>440</math> we see that it is <math>2^3 * 5 * 11.</math> Intuitively, we can find that <math>n + 1 = 20</math> and <math>n + 3 = 22.</math> Therefore, <math>n = 19.</math> Since the problem asks for the sum of the didgits of <math>n</math>, we finally calculate <math>1 + 9 = 10</math> and get answer choice <math>\boxed{\textbf{(C) }10}</math>. |
− | == Video Solution == | + | ~pnacham |
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+ | ==Video Solution== | ||
+ | https://youtu.be/7p_ESPPF2es | ||
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+ | ~Education, the Study of Everything | ||
+ | |||
+ | == Video Solution== | ||
https://youtu.be/ba6w1OhXqOQ?t=1956 | https://youtu.be/ba6w1OhXqOQ?t=1956 | ||
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==Video Solution== | ==Video Solution== | ||
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~IceMatrix | ~IceMatrix | ||
+ | ==Video Solution== | ||
https://youtu.be/6YFN_hwotUk | https://youtu.be/6YFN_hwotUk | ||
Latest revision as of 10:24, 24 June 2023
- The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.
Contents
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
Solution 5
Rewrite as
Factoring out the
we get
Expand this to get
Factor this and divide by
to get
If we take the prime factorization of
we see that it is
Intuitively, we can find that
and
Therefore,
Since the problem asks for the sum of the didgits of
, we finally calculate
and get answer choice
.
~pnacham
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1956
Video Solution
~IceMatrix
Video Solution
~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.