Difference between revisions of "2021 AMC 10B Problems/Problem 13"
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Revision as of 23:42, 11 February 2021
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 3
+2+d. Because it is equal to 263, we can set this equation to 263. Finally, subtract from both sides to get 3+2 = 263-.
We can also set up equations to convert base and base 6 to base 10. The equation to covert base to base 10 is 3+2+4. The equation to convert base 6 to base 10 is 
+
+6+1.
Simplify ++6+1 so it becomes 6+253. Setting the above equations equal to each other, we have 3+2n+4 = 6d+253. Subtracting 4 from both sides gets 3+2n = 6d+249.
We can then use 3+2 = 263- and 3+2 = 6+249 to solve for . Set 263- equal to 6+249 and solve to find that =2.
Plug =2 back into the equation 3+2 = 263-. Subtract 261 from both sides to get your final equation of 3+2-261 = 0. Solve using the quadratic formula to find that the solutions are 9 and -10. Because the base must be positive, =9.
Adding 2 to 9 gets 
-Zeusthemoose
| 2021 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
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| All AMC 10 Problems and Solutions | ||
