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

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 == (Easy to understand). Denote the four digit number by $ab23_{10}$ . Then it is well known the sum of the digits of an integer $N$ determine whether or not they are divisible by $3$ . Thus we want $a + b + 2 + 3 \equiv 0 \pmod 3$ .

$\underline{a = 1}: \implies$ $b \equiv 0 \pmod 3 \implies b = 0, 3, 6, 9$ . $\underline{a = 2}: \implies$ $b \equiv 2 \pmod 3 \implies b = 2, 5, 8$ . $\underline{a = 3}: \implies$ $b \equiv 1 \pmod 3 \implies b = 1, 4, 7$ .

Now we have that when $a = 4$ this is the same case as $a = 1$ as $4 \equiv 1 \pmod 3$ . Hence if we count the above number of solutions and multiply by $3$ (as $1 \leq a \leq 9$ ), we are home free. There are $4 + 3 + 3 = 10$ solutions. Multiplying this by $3$ yields $\boxed{30}$ . $\blacksquare$

2003 AMC 10B (Problems • Answer Key • Resources)
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