Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
Notice that 1000 is 3E8 in hexadecimal. The first digit could be 0, 1, 2, or 3 and the second two could be any digit 0-9. 4 • 10 • 10 = 400. However, this includes 000, so subtract one. Thus, there are 399 valid <math>n</math>, corresponding to those 399 positive integers less than 1000 with hexadecimal representation less than 1000. (Notice that 399 < 3E8 in hexadecimal.) Our answer is <math>3 + 9 + 9 = 21</math> <math>\textbf{(E) }</math>.
+
Notice that <math>1000</math> is <math>3E8</math> in hexadecimal. 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 < 3E8</math> in hexadecimal.) Finally, our answer is <math>3 + 9 + 9 = \boxed{\textbf{(E) } 21}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2015|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:26, 5 February 2015

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 $399 < 3E8$ in hexadecimal.) Finally, our answer is $3 + 9 + 9 = \boxed{\textbf{(E) } 21}$

See Also

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_18&oldid=67634"