Difference between revisions of "2013 AMC 10B Problems/Problem 18"

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

@@ Line 1: / Line 1: @@
-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?
+==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> have this property?
 <math> \textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58 </math>
+==Solution 1.1==
+We take cases on the thousands digit, which must be either <math>1</math> or <math>2</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.
+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 ==
+{{AMC10 box|year=2013|ab=B|num-b=17|num-a=19}}
+{{MAA Notice}}

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

Page

Toolbox

Search