Difference between revisions of "2015 AMC 10A Problems/Problem 18"
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| − | Notice that 1000 is 3E8 in hexadecimal. Thus, there are 399 valid <math>n</math>, corresponding to those 399 positive integers less than 1000 with hexadecimal representation less than 1000. (Notice that 399 < 3E8 in hexadecimal.) Our answer is <math>3 + 9 + 9 = 21</math> <math>\textbf{(E) }</math>. | + | Notice that 1000 is 3E8 in hexadecimal. The first digit could be 0, 1, 2, or 3 and the second two could be any digit 0-9. 4 • 10 • 10 = 400. However, this includes 3E9, so subtract one. Thus, there are 399 valid <math>n</math>, corresponding to those 399 positive integers less than 1000 with hexadecimal representation less than 1000. (Notice that 399 < 3E8 in hexadecimal.) Our answer is <math>3 + 9 + 9 = 21</math> <math>\textbf{(E) }</math>. |
Revision as of 19:03, 4 February 2015
Problem 18
Hexadecimal (base-16) numbers are written using numeric digits 
through 
as well as the letters 
through 
to represent 
through 
. Among the first 
positive integers, there are 
whose hexadecimal representation contains only numeric digits. What is the sum of the digits of ?
Solution
Notice that 1000 is 3E8 in hexadecimal. The first digit could be 0, 1, 2, or 3 and the second two could be any digit 0-9. 4 • 10 • 10 = 400. However, this includes 3E9, so subtract one. Thus, there are 399 valid , corresponding to those 399 positive integers less than 1000 with hexadecimal representation less than 1000. (Notice that 399 < 3E8 in hexadecimal.) Our answer is 
.
