2003 AMC 10B Problems/Problem 25

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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

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 $23$, 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.

\[1+4+7+9+6+3 = \boxed{\textbf{(B)}\ 30}\]

Solution 2

A number divisible by $3$ has all its digits add to a multiple of $3.$ The last two digits are $2$ and $3$ and add up to $5 \equiv 2\ (\text{mod}\ 3).$ Therefore the first two digits must add up to $1\ (\text{mod}\ 3).$ $4$ digits (including $0$) are $0\ (\text{mod}\ 3),$ $3$ are $1\ (\text{mod}\ 3),$ and $3$ are $2\ (\text{mod}\ 3).$ The following combinations are equivalent to $1\ (\text{mod}\ 3)$:

\[0\ (\text{mod}\ 3)+1\ (\text{mod}\ 3) \equiv 1\ (\text{mod}\ 3)+0\ (\text{mod}\ 3) \equiv 2\ (\text{mod}\ 3) +2\ (\text{mod}\ 3)\]

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.

\[3\cdot3 + 3\cdot4 + 3\cdot3 = 9 + 12 + 9 = \boxed{\textbf{(B)}\ 30}\]

See also

2003 AMC 10B (ProblemsAnswer KeyResources)
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