2015 AMC 10A Problems/Problem 18

Problem 18

Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$. Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$?

$\textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21$

Solution

Notice that $1000$ is $3E8$ in hexadecimal. We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.

The first digit could be $0,$ $1,$ $2,$ or $3,$ and the second two could be any digit $0 - 9$, giving $4 \cdot 10 \cdot 10 = 400$ combinations. However, this includes $000,$ so this number must be diminished by $1.$ Therefore, there are $399$ valid $n$ corresponding to those $399$ positive integers less than $1000$ that consist of only numeric digits. (Notice that $399$ is the least hexadecimal number using only decimal digits before $3E8$.) Therefore, our answer is $3 + 9 + 9 = \boxed{\textbf{(E) } 21}$

See Also

Video Solution: https://www.youtube.com/watch?v=2DVSkWu_H1g

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS