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 "2013 AMC 10B Problems/Problem 18"

(→‎See also: Removed spam.)
m (→‎Solution 2 (Casework))
 
(12 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
The number <math>2013</math> has the property that its units digit is the sum of its other digits, that is <math>2+0+1=3</math>. How many integers less than <math>2013</math> but greater than <math>1000</math> share this property?
+
The number <math>2013</math> has the property that its units digit is the sum of its other digits, that is <math>2+0+1=3</math>. How many integers less than <math>2013</math> but greater than <math>1000</math> have this property?
  
 
<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58 </math>
 
<math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58 </math>
  
−
==Solution==
+
==Solution 1.1==
−
First, note that the only integer <math>2000\le x \le 2013</math> is <math>2002</math>.
+
We take cases on the thousands digit, which must be either <math>1</math> or <math>2</math>:
−
Now let's look at all numbers <math>x</math> where <math>1000<x<2000</math>
+
If the number is of the form <math>\overline{1bcd},</math> where <math>b, c, d</math> are digits, then we must have <math>d = 1 + b + c.</math> Since <math>d \le 9,</math> we must have <math>b + c \le 9 - 1 = 8.</math> By casework on the value of <math>b</math>, we find that there are <math>1 + 2 + \dots + 9 = 45</math> possible pairs <math>(b, c)</math>, and each pair uniquely determines the value of <math>d</math>, so we get <math>45</math> numbers with the given property.
−
Let the hundreds digit be <math>0</math>. Then, the tens and units digit can be <math>01, 12, 23, \hdots, 89</math>, which is <math>9</math> possibilities. We notice as the hundreds digit goes up by one, the number of possibilities goes down by one.
+
 
−
Thus, the number of integers is <math>1+2+3+4+5+6+7+8+9+1=\boxed{\textbf{(D)} 46}</math>
+
If the number is of the form <math>\overline{2bcd},</math> then it must be one of the numbers <math>2000, 2001, \dots, 2012.</math> Checking all these numbers, we find that only <math>2002</math> has the given property.
 +
Therefore, the number of numbers with the property is <math>45 + 1 = \boxed{\textbf{(D)}46}</math> .
 +
 
 +
==Solution 1.2==
 +
This solution picks up from finding that <math>b + c \le 8</math> in solution 1.1. Instead of using casework to find all possible pairs, <math>(b, c)</math>, let's introduce a dummy variable, <math>z</math>. Let us now have that <math>b + c + z = 8</math>, where <math>b, c, z</math> are all nonnegative.
 +
 
 +
We may now use stars and bars to distribute units between <math>b, c</math> and <math>z</math>. Any units that <math>z</math> is given will essentially be discarded - this is how we get the 'less than' in the 'less than or equal to <math>8</math>' relation we found earlier.
 +
 
 +
Using two dividers, we find that the number of distributions is <math>\binom{10}{2},</math> which is <math>45</math>. We proceed from here as above.
 +
 
 +
==Solution 2 (Casework)==
 +
Let's start with the case that starts with <math>200</math>. We have only one number, which is <math>2002</math>. If we look at the <math>1900s</math>, we have no solutions because <math>1+9 = 10</math>, and because we can only use digits from <math>1</math> through <math>9</math>, it is impossible. If we looks at the <math>1800s</math>, we do have one solution, which is <math>1809</math>. If we look a the <math>1700s</math>, we have <math>2</math> solutions, namely, <math>1708</math> and <math>1719</math>.
 +
 
 +
We can see a pattern here. The pattern is every hundred you go down, you have <math>1</math> more solution. Therefore, we have <math>1+0+1+2+3+4+5+6+7+8+9</math> which is = <math>\boxed{\textbf{(D)}46}</math> .
 +
 
 +
~Arcticturn
 +
 
 +
==Video Solution==
 +
https://youtu.be/2jNuQEfo1Rc
 +
 
 +
~savannahsolver
 +
 
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2013|ab=B|num-b=17|num-a=19}}
 
{{AMC10 box|year=2013|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:23, 26 March 2023

Problem

The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3$. How many integers less than $2013$ but greater than $1000$ have this property?

$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58$

Solution 1.1

We take cases on the thousands digit, which must be either $1$ or $2$: If the number is of the form $\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \le 9,$ we must have $b + c \le 9 - 1 = 8.$ By casework on the value of $b$, we find that there are $1 + 2 + \dots + 9 = 45$ possible pairs $(b, c)$, and each pair uniquely determines the value of $d$, so we get $45$ numbers with the given property.

If the number is of the form $\overline{2bcd},$ then it must be one of the numbers $2000, 2001, \dots, 2012.$ Checking all these numbers, we find that only $2002$ has the given property. Therefore, the number of numbers with the property is $45 + 1 = \boxed{\textbf{(D)}46}$ .

Solution 1.2

This solution picks up from finding that $b + c \le 8$ in solution 1.1. Instead of using casework to find all possible pairs, $(b, c)$, let's introduce a dummy variable, $z$. Let us now have that $b + c + z = 8$, where $b, c, z$ are all nonnegative.

We may now use stars and bars to distribute units between $b, c$ and $z$. Any units that $z$ is given will essentially be discarded - this is how we get the 'less than' in the 'less than or equal to $8$' relation we found earlier.

Using two dividers, we find that the number of distributions is $\binom{10}{2},$ which is $45$. We proceed from here as above.

Solution 2 (Casework)

Let's start with the case that starts with $200$. We have only one number, which is $2002$. If we look at the $1900s$, we have no solutions because $1+9 = 10$, and because we can only use digits from $1$ through $9$, it is impossible. If we looks at the $1800s$, we do have one solution, which is $1809$. If we look a the $1700s$, we have $2$ solutions, namely, $1708$ and $1719$.

We can see a pattern here. The pattern is every hundred you go down, you have $1$ more solution. Therefore, we have $1+0+1+2+3+4+5+6+7+8+9$ which is = $\boxed{\textbf{(D)}46}$ .

~Arcticturn

Video Solution

https://youtu.be/2jNuQEfo1Rc

~savannahsolver

See also

2013 AMC 10B (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=2013_AMC_10B_Problems/Problem_18&oldid=191415"