# 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?

## Solution

### Solution 1 (Slow)

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 (Medium)

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

### Solution 3 (Fast)

We have the following: $n \equiv 0 \pmod{3}$ and $n \equiv 23\pmod{100}$. Then $n = 3a = 100b+23$ for some integers $a$ and $b$. Taking mod 3 gives: $0 \equiv b+2 \pmod 3 \implies b\equiv 1\pmod 3$ so $b=3c+1$ for some integer $c$. But $N = 100b+23 = 100(3c+1)+23$ so $n = 300c+123$. Bounding this gives us: $999 < 300c+123 < 10000$ so $876 < 300c < 9877$. Dividing by $300$ gives $2.92 < c < 32.9222$ so $3\le c \le 32$. This gives $32-3+1 = \boxed{\textbf{(B)} \ 30 }$

### Solution 4 (Extremely Fast)

The number is in the form $xy23$

Notice that the number is divisible by $3$ if the sum of the digits of $xy23$ is divisible by $3$

Our first number that is divisible by $3$ is $1023$, next is $1323$.. Notice $xy$ goes from $10\implies 97$

Hence, there are $\frac{97-10}{3}+1=30$ distinct four digit numbers.

### Solution 5 (Even Faster than Extremely Fast)

Following the form $xy23$ in Solution 4, notice that $x+y \equiv 1 \pmod 3$ to satisfy our condition. Choose a value of $x$ from 1 to 9. For $x=1, 4, 7$, there are exactly 4 values of $y: 0, 3, 6, 9$. For the remaining 6 digits, there are 3 choices for y. So our answer is $(3)(4)+(6)(3)=30$.

### Solution 6 (Even Faster than Even Faster than Extremely Fast)

There are $9 \cdot 10 = 90$ possible values for the first two digits. One-third of them yield a multiple of $3$, so the answer is $\frac{90}{3} = \boxed{\textbf{(B)}\ 30}$

### Solution 7 (Even Faster than Even Faster than Even Faster than Extremely Fast)

Note that the answer is equal to $\boxed{\textbf{(B)}\ 30}$ We will leave the steps to find this answer as an exercise for the reader to determine.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 