2015 AMC 10A Problems/Problem 18

Revision as of 22:32, 17 January 2021 by Pi is 3.14 (talk | contribs) (Solution)

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 $3E8$ when converted to hexadecimal($3 \cdot 16^2 + 14 \cdot 16^1 + 8 \cdot 16^0$). 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 $3 + 9 + 9 = \boxed{\textbf{(E) } 21}$

Reasoning for the 400 Combinations

This is because we don't want more than $9$ for each digit since that would lead to a letter digit(A, B, C, D, E, or F.) Thus we have $3 \cdot 16^2 + 9 \cdot 16^1 + 9 \cdot 16^0.$ Then continue as follows.

~Clarification by mathboy282

Video Solution

https://youtu.be/ZhAZ1oPe5Ds?t=4596

~ pi_is_3.14

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
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