Difference between revisions of "2020 AMC 12A Problems/Problem 10"
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Adding the digits together, we have <math>2+5+6=\boxed{\textbf{E) }13}</math> ~quacker88 | Adding the digits together, we have <math>2+5+6=\boxed{\textbf{E) }13}</math> ~quacker88 | ||
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+ | ==See Also== | ||
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+ | {{AMC12 box|year=2020|ab=A|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Revision as of 11:47, 1 February 2020
Problem
There is a unique positive integer such thatWhat is the sum of the digits of
Solution
Any logarithm in the form .
so
becomes
Using property of addition, we can expand the parentheses into
Expanding the RHS and simplifying the logs without variables, we have
Subtracting from both sides and adding to both sides gives us
Multiplying by , raising the logs to exponents of base to get rid of the logs and simplifying gives us
Adding the digits together, we have ~quacker88
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 |
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