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

Revision as of 21:23, 18 July 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 Solutions

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

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

@@ Line 5: / Line 5: @@
 ==Solution==
-Notice that <math>1000</math> is <math>3E8</math> when converted to hexadecimal(<math>3 \cdot 16^2 + 14 \cdot 16^1 + 8 \cdot 16^0</math>). We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.
+Notice that <math>1000</math> is <math>3E8</math> when converted to hexadecimal (<math>3 \cdot 16^2 + 14 \cdot 16^1 + 8 \cdot 16^0</math>). We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.
 The first digit could be <math>0,</math> <math>1,</math> <math>2,</math> or <math>3,</math> and the second two could be any digit <math>0 - 9</math>, giving <math>4 \cdot 10 \cdot 10 = 400</math> combinations. However, this includes <math>000,</math> so this number must be diminished by <math>1.</math>  Therefore, there are <math>399</math> valid <math>n</math> corresponding to those <math>399</math> positive integers less than <math>1000</math> that consist of only numeric digits.  (Notice that <math>399</math> is the least hexadecimal number using only decimal digits before <math>3E8</math>.)  Therefore, our answer is <math>3 + 9 + 9 = \boxed{\textbf{(E) } 21}</math>

2015 AMC 10A (Problems • Answer Key • Resources)
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

Page

Toolbox

Search