2003 AMC 10B Problems/Problem 25

Problem

How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?

$\textbf{(A) }27\qquad\textbf{(B) }30\qquad\textbf{(C) }33\qquad\textbf{(D) }81\qquad\textbf{(E) }90$

Solution

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 $23$ , the sum of the digits is $2+3 = 5$ (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.

$5+1 = 6$ ,

$5+4 = 9$ , and so on.

However since the largest four-digit number ending with $23$ is $9923$ , the maximum sum is

$5+18 = 23$ .

Using that process we can fairly quickly compile a list of the sum of the first two digits of the number.

$\{1, 4, 7, 10, 13, 16\}$

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 $xy$ in separate cases.

$I. x+y = 1, \{10\} = 1$ $II. x+y = 4, \{13, 22, 31, 40\} = 4$ $III. x+y = 7, \{16, 25, 34, 43, 52, 61, 70\} = 7$ $IV. x+y = 10, \{19, 28, 37, 46, 55, 64, 73, 82, 91\} = 9$ $V. x+y = 13, \{49, 58, 67, 76, 85, 94\} = 6$ $VI. x+y = 16, \{79, 88, 97\} = 3$

And finally, we add the number of elements in each set.

$1+4+7+9+6+3 = \boxed{\mathrm{(B)}\ 30}$

2002 AMC 10A (Problems • Answer Key • Resources)
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2003 AMC 10B Problems/Problem 25

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