2021 AMC 10B Problems/Problem 13
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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