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"

(→‎Solution)
m (→‎Reasoning for the 400 Combinations)
 
(12 intermediate revisions by 8 users not shown)
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>?
  
 
<math> \textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21 </math>
 
<math> \textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21 </math>
  
−
==Solution==
+
==Solution 1==
−
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>
 
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>
  
−
===Reasoning for the 400 Combinations===
+
==Solution 2 (Casework)==
−
This is because we don't want more than <math>9</math> for each digit since that would lead to a letter digit(A, B, C, D, E, or F.) Thus we have <math>3 \cdot 16^2 + 9 \cdot 16^1 + 9 \cdot 16^0.</math> Then continue as follows.
+
First, we set a bound by writing <math>1000</math> in base-<math>16</math>. <math>1000_{10}=3E8_{16}</math>. Therefore, we are considering numbers with a maximum of <math>3</math> digits, and a maximum of <math>3</math> in the <math>256</math>ths-place (the first place in a <math>3</math>-digit number).
  
−
~Clarification by mathboy282
+
Case <math>1</math>: <math>1</math>-digit numbers:
 +
There are evidently <math>9</math> numbers that fit this category.
 +
 
 +
Case <math>2</math>: <math>2</math>-digit numbers:
 +
There are <math>9\cdot10=90</math> numbers that fit this category.
 +
 
 +
Case <math>3</math>: <math>3</math>-digit numbers:
 +
There are <math>3\cdot10\cdot10=300</math> numbers that fit this category
 +
 
 +
Adding these up, we get <math>9+90+300=399</math> numbers. <math>3 + 9 + 9 = \boxed{\textbf{(E) } 21}</math> ~sosiaops
 +
 
 +
==Solution 3 ==
 +
We can quickly see that <math>400</math> in hexadecimal = <math>0+0+16^2*4</math> = 1024. If we go down to 399 in hexadecimal, we have <math>9+9*16+3*256</math> which is <math>921</math>, which is obviously less than 1000. Therefore, the answer is <math>3+9+9</math> = <math>\boxed{\textbf{(E) } 21}</math>
 +
 
 +
~Arcticturn
 +
 
 +
== Video Solutions ==
 +
https://youtu.be/ZhAZ1oPe5Ds?t=4596
 +
 
 +
https://www.youtube.com/watch?v=2DVSkWu_H1g
 +
 
 +
https://youtu.be/jkxWTsfbAjQ
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
−
Video Solution:
 
−
https://www.youtube.com/watch?v=2DVSkWu_H1g
 
  
 
{{AMC10 box|year=2015|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2015|ab=A|num-b=17|num-a=19}}

Latest revision as of 08:45, 10 July 2024

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 1

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 $3 + 9 + 9 = \boxed{\textbf{(E) } 21}$

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

We can quickly see that $400$ in hexadecimal = $0+0+16^2*4$ = 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

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

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

https://youtu.be/jkxWTsfbAjQ

~savannahsolver

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=222003"