# 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 .