Difference between revisions of "2013 AMC 10B Problems/Problem 18"
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==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> | + | 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== |
− | + | 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 == | == 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 22:23, 26 March 2023
Contents
Problem
The number has the property that its units digit is the sum of its other digits, that is
. How many integers less than
but greater than
have this property?
Solution 1.1
We take cases on the thousands digit, which must be either or
:
If the number is of the form
where
are digits, then we must have
Since
we must have
By casework on the value of
, we find that there are
possible pairs
, and each pair uniquely determines the value of
, so we get
numbers with the given property.
If the number is of the form then it must be one of the numbers
Checking all these numbers, we find that only
has the given property.
Therefore, the number of numbers with the property is
.
Solution 1.2
This solution picks up from finding that in solution 1.1. Instead of using casework to find all possible pairs,
, let's introduce a dummy variable,
. Let us now have that
, where
are all nonnegative.
We may now use stars and bars to distribute units between and
. Any units that
is given will essentially be discarded - this is how we get the 'less than' in the 'less than or equal to
' relation we found earlier.
Using two dividers, we find that the number of distributions is which is
. We proceed from here as above.
Solution 2 (Casework)
Let's start with the case that starts with . We have only one number, which is
. If we look at the
, we have no solutions because
, and because we can only use digits from
through
, it is impossible. If we looks at the
, we do have one solution, which is
. If we look a the
, we have
solutions, namely,
and
.
We can see a pattern here. The pattern is every hundred you go down, you have more solution. Therefore, we have
which is =
.
~Arcticturn
Video Solution
~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 |
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