Problem D and Solution

Wipe Away 2

Ajay writes the positive integers from \(1\) to \(1000\) on a whiteboard. Jamilah then erases all the numbers that are multiples of \(9\). Magdalena then erases all the remaining numbers that contain the digit \(9\). How many numbers are left on the whiteboard?

Note: In solving this problem, it may be helpful to use the fact that a number is divisible by \(9\) exactly when the sum of its digits is divisible by \(9\). For example, the number \(214\,578\) is divisible by \(9\) since \(2+1+4+5+7+8 = 27\), which is divisible by \(9\). In fact, \(214\,578=9 \times 23\,842.\)

We first calculate the number of integers that Jamilah erases, which is the number of multiples of \(9\) between \(1\) and \(1000\). Since \(1000 = (111 \times 9) + 1\), there are \(111\) multiples of \(9\) between \(1\) and \(1000\). Thus, Jamilah erases \(111\) numbers from the whiteboard.

Now letâ€™s figure out how many of the integers from \(1\) to \(1000\) contain the digit \(9\). The integers from \(1\) to \(100\) that contain the digit \(9\) are \(9,19, \ldots, 79, 89\) as well as \(90, 91, \ldots, 97 , 98, 99\). Thus, there are \(19\) positive integers from \(1\) to \(100\) that contain the digit \(9\). Since there are \(19\) integers from \(1\) to \(100\) that contain the digit \(9\), it follows that there are \(19 \times 9 = 171\) integers from \(1\) to \(899\) that contain the digit \(9\).

Between \(900\) and \(1000\), there are \(100\) integers that contain the digit \(9\), namely, every number except for \(1000\). Thus, in total, \(171+100 = 271\) of the integers from \(1\) to \(1000\) contain the digit \(9\).

However, some of the integers that contain the digit \(9\) are also multiples of \(9\), so were erased by Jamilah. To determine how many of these such numbers there are, we use the fact that a number is divisible by \(9\) exactly when the sum of its digits is divisible by \(9\).

The only one-digit number that contains the digit \(9\) and is also a multiple of \(9\) is \(9\) itself.

The only two-digit numbers that contain the digit \(9\) and are also multiples of \(9\) are \(90\) and \(99\).

To find the three-digit numbers that contain the digit \(9\) and are also multiples of \(9\), we will look at their digit sum.

**Case 1:**Three digit-numbers with a digit sum of \(9\):

The only possibility is \(900\). Thus, there is \(1\) number.**Case 2:**Three digit-numbers with a digit sum of \(18\):If two of the digits are \(9\), then the other digit must be \(0\). The only possibilities are \(909\) and \(990\). Thus, there are \(2\) numbers.

If only one of the digits is \(9\), then the other two digits must add to \(9\). The possible digits are \(9\), \(4\), \(5\), or \(9\), \(3\), \(6\), or \(9\), \(2\), \(7\), or \(9\), \(8\), \(1\). For each of these sets of digits, there are \(3\) choices for the hundreds digit. Once the hundreds digit is chosen, there are \(2\) choices for the tens digit, and then the remaining digit must be the ones digit. Thus, there are \(3 \times 2=6\) possible three-digit numbers for each set of digits. Since there are \(4\) sets of digits, then there are \(4 \times 6 = 24\) possible numbers.

**Case 3:**Three digit-numbers with a digit sum of \(27\):

The only possibility is \(999\). Thus, there is \(1\) number.

Therefore, there are \(1+2+24+1=28\) three-digit numbers from \(1\) to \(1000\) that contain the digit \(9\), and are also multiples of \(9\).

Thus, there are \(1+2+28=31\) numbers that contain the digit \(9\), but were erased by Jamilah. It follows that Magdalena erases \(271-31=240\) numbers from the whiteboard.

Hence, the number of numbers left on the whiteboard is \(1000 - 111 - 240 = 649\).