Difference between revisions of "2003 AMC 10B Problems/Problem 25"
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Revision as of 11:11, 4 July 2013
Problem
How many distinct four-digit numbers are divisible by 
and have 
as their last two digits?
Solution
Solution 1
To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are , the sum of the digits is 
(therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.
,
, and so on.
However since the largest four-digit number ending with is 
, the maximum sum is
.
Using that process we can fairly quickly compile a list of the sum of the first two digits of the number.
![\[\{1, 4, 7, 10, 13, 16\}\]](http://latex.artofproblemsolving.com/6/4/d/64ddcf298f712b3a213eeb5a83e875865ccb8407.png)
Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers 
in separate cases.
![\[I. x+y = 1, \{10\} = 1\]](http://latex.artofproblemsolving.com/9/4/3/943a6fca640b2ef0375212e22d8e53b886155d61.png)
![\[II. x+y = 4, \{13, 22, 31, 40\} = 4\]](http://latex.artofproblemsolving.com/e/9/e/e9e1cddd5cb18b50d8e6a3fbf53287d9ebe3f467.png)
![\[III. x+y = 7, \{16, 25, 34, 43, 52, 61, 70\} = 7\]](http://latex.artofproblemsolving.com/c/5/3/c53383c2670ef646e03ed75d328e01a70c39bba7.png)
![\[IV. x+y = 10, \{19, 28, 37, 46, 55, 64, 73, 82, 91\} = 9\]](http://latex.artofproblemsolving.com/8/c/9/8c9642d529a637ed6678565dacbf2b8dd86221a9.png)
![\[V. x+y = 13, \{49, 58, 67, 76, 85, 94\} = 6\]](http://latex.artofproblemsolving.com/e/5/d/e5db4469f46fe2074b96fed15d09fef52f30de45.png)
![\[VI. x+y = 16, \{79, 88, 97\} = 3\]](http://latex.artofproblemsolving.com/a/f/a/afa411e8f5ae7e3ba4228cd38945cec13c3180c7.png)
And finally, we add the number of elements in each set.
![\[1+4+7+9+6+3 = \boxed{\textbf{(B)}\ 30}\]](http://latex.artofproblemsolving.com/9/1/0/91018c437f8d8e47a97363f6d79fa803645ae6e7.png)
Solution 2
A number divisible by has all its digits add to a multiple of 
The last two digits are 
and and add up to 
Therefore the first two digits must add up to 
digits (including 
) are 
are 
and are 
The following combinations are equivalent to 
:
![\[0\ (\text{mod}\ 3)+1\ (\text{mod}\ 3) \equiv 1\ (\text{mod}\ 3)+0\ (\text{mod}\ 3) \equiv 2\ (\text{mod}\ 3) +2\ (\text{mod}\ 3)\]](http://latex.artofproblemsolving.com/2/c/d/2cde0c1540ededeb77780da629a47fff63361977.png)
Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.
![\[3\cdot3 + 3\cdot4 + 3\cdot3 = 9 + 12 + 9 = \boxed{\textbf{(B)}\ 30}\]](http://latex.artofproblemsolving.com/0/a/5/0a57342d9ed7b8da88c30c113012d36f02e40212.png)
See also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last problem | |
| 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. 
