Difference between revisions of "2018 AMC 10A Problems/Problem 18"
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This looks like balanced ternary, in which all the integers with absolute values less than <math>\frac{3^n}{2}</math> are represented in <math>n</math> digits. There are 8 digits. Plugging in 8 into the formula gives a maximum bound of <math>|x|=3280.5</math>, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are <math>3280+1=\boxed{3281}</math> integers or <math>\boxed{D}</math>. | This looks like balanced ternary, in which all the integers with absolute values less than <math>\frac{3^n}{2}</math> are represented in <math>n</math> digits. There are 8 digits. Plugging in 8 into the formula gives a maximum bound of <math>|x|=3280.5</math>, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are <math>3280+1=\boxed{3281}</math> integers or <math>\boxed{D}</math>. | ||
+ | |||
+ | <math>QED\blacksquare</math> |
Revision as of 17:09, 8 February 2018
How many nonnegative integers can be written in the form where for ?
Solution
This looks like balanced ternary, in which all the integers with absolute values less than are represented in digits. There are 8 digits. Plugging in 8 into the formula gives a maximum bound of , which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are integers or .