Difference between revisions of "2021 AMC 10B Problems/Problem 13"
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Latest revision as of 08:51, 2 March 2021
Contents
Problem
Let be a positive integer and be a digit such that the value of the numeral in base equals , and the value of the numeral in base equals the value of the numeral in base six. What is
Solution
We can start by setting up an equation to convert base to base 10. To convert this to base 10, it would be Because it is equal to 263, we can set this equation to 263. Finally, subtract from both sides to get .
We can also set up equations to convert base and base 6 to base 10. The equation to covert base to base 10 is The equation to convert base 6 to base 10 is
Simplify so it becomes Setting the above equations equal to each other, we have
Subtracting 4 from both sides gets
We can then use equations to solve for . Set equal to and solve to find that .
Plug back into the equation . Subtract 261 from both sides to get your final equation of Solve using the quadratic formula to find that the solutions are 9 and -10. Because the base must be positive,
Adding 2 to 9 gets
-Zeusthemoose (edited for readability)
Solution 2
is greater than when both are interpreted in base 10, so is less than . Some trial and error gives . in base 9 is , so the answer is .
-SmileKat32
Video Solution by OmegaLearn (Bases and System of Equations)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/L1iW94Ue3eI?t=880
~IceMatrix
Video Solution by Interstigation
~Interstigation
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.