# 2015 AMC 10A Problems/Problem 18

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## Problem

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

## Solution 2 (Casework)

First, we set a bound by writing $1000$ in base- $16$. $1000_{10}=3E8_{16}$. Therefore, we are considering numbers with a maximum of $3$ digits, and a maximum of $3$ in the $256$ths-place (the first place in a $3$-digit number).

Case $1$: $1$-digit numbers: There are evidently $9$ numbers that fit this category.

Case $2$: $2$-digit numbers: There are $9\cdot10=90$ numbers that fit this category.

Case $3$: $3$-digit numbers: There are $3\cdot10\cdot10=300$ numbers that fit this category

Adding these up, we get $9+90+300=399$ numbers. $3 + 9 + 9 = \boxed{\textbf{(E) } 21}$ ~sosiaops

## Solution 3 (Quick)

We can quickly see that $400$ in hexadecimal = $0+0+256*4$ = 1024. If we go down to 399 in hexadecimal, we have $9+9*16+3*256$ which is $921$, which is obviously less than 1000. Therefore, the answer is $3+9+9$ = $\boxed{\textbf{(E) } 21}$

~Arcticturn

## Video Solutions

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