Difference between revisions of "2015 AMC 10A Problems/Problem 18"

m (Problem 18)
Line 1: Line 1:
==Problem 18==
+
==Problem==
 
Hexadecimal (base-16) numbers are written using numeric digits <math>0</math> through <math>9</math> as well as the letters <math>A</math> through <math>F</math> to represent <math>10</math> through <math>15</math>. Among the first <math>1000</math> positive integers, there are <math>n</math> whose hexadecimal representation contains only numeric digits. What is the sum of the digits of <math>n</math>?
 
Hexadecimal (base-16) numbers are written using numeric digits <math>0</math> through <math>9</math> as well as the letters <math>A</math> through <math>F</math> to represent <math>10</math> through <math>15</math>. Among the first <math>1000</math> positive integers, there are <math>n</math> whose hexadecimal representation contains only numeric digits. What is the sum of the digits of <math>n</math>?
  

Revision as of 16:38, 14 February 2021

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

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
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