Problem of the Week Problem E and Solution Wipe Away 3

Problem

Tyra writes consecutive positive integers on a whiteboard starting with the integer $$1$$. However, when she writes a number that is a multiple of $$9$$, or contains the digit $$9$$, Juliana immediately erases it. If they continue this for a long time, what is the 400th number that Juliana will erase?

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, $$214\,578=9 \times 23\,842.$$

Solution

We first consider the integers between $$1$$ and $$999$$, inclusive. Since $$999 = 111 \times 9$$, there are $$111$$ multiples of $$9$$ between $$1$$ and $$999$$.

Now letâ€™s figure out how many of the integers from $$1$$ to $$999$$ contain the digit $$9$$. The integers from $$1$$ to $$99$$ 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 $$99$$ that contain the digit $$9$$. Since there are $$19$$ integers from $$1$$ to $$99$$ 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 $$999$$, every integer contains the digit $$9$$. Thus, there are $$100$$ numbers that contain the digit $$9$$. Thus, in total, $$171+100 = 271$$ of the integers from $$1$$ to $$999$$ contain the digit $$9$$.

However, some of the integers that contain the digit $$9$$ are also multiples of $$9$$, so were counted twice. 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 $$999$$ that contain the digit $$9$$, and are also multiples of $$9$$.

Thus, there are $$1+2+28=31$$ integers from $$1$$ to $$999$$ that contain the digit $$9$$, and are also multiples of $$9$$. It follows that Juliana erases $$111+271-31=351$$ of the numbers from $$1$$ to $$999$$ from the whiteboard. Since we are looking for the 400th number that Juliana erases, we need to keep going.

Next, we consider the integers between $$1000$$ and $$1099$$, inclusive. Since $$1099=(122 \times 9)+1$$, there are $$122$$ multiples of $$9$$ between $$1$$ and $$1099$$. Since there are $$111$$ multiples of $$9$$ between $$1$$ and $$999$$, it follows that there are $$122-111=11$$ multiples of $$9$$ between $$1000$$ and $$1099$$. The integers between $$1000$$ and $$1099$$ that contain the digit $$9$$ are $$1009,~1019, \ldots, 1079,~ 1089$$ as well as $$1090,~ 1091, \ldots, 1097~ , 1098,~ 1099$$. Thus, there are $$19$$ integers from $$1000$$ to $$1099$$ that contain the digit $$9$$. Of these, the only integers that are also multiples of $$9$$ are $$1089$$ and $$1098$$. Thus, Juliana erases $$11+19-2=28$$ of the numbers from $$1000$$ to $$1099$$ from the whiteboard. In total, she has now erased $$351+28=379$$ numbers.

Next, we consider the integers between $$1100$$ and $$1189$$, inclusive. Since $$1189=(132 \times 9) +1$$, there are $$132$$ multiples of $$9$$ between $$1$$ and $$1189$$. Since there are $$122$$ multiples of $$9$$ between $$1$$ and $$1099$$, it follows that there are $$132-122=10$$ multiples of $$9$$ between $$1100$$ and $$1189$$. The integers between $$1100$$ and $$1189$$ that contain the digit $$9$$ are $$1109,~1119,\dots,1179,~1189$$. Thus there are $$9$$ integers from $$1100$$ to $$1189$$ that contain the digit $$9$$. The only one of these that is also a multiple of $$9$$ is $$1179$$. Thus, Juliana erases $$10+9-1=18$$ of the numbers from $$1100$$ to $$1189$$ from the whiteboard. In total, she has now erased $$379+18=397$$ numbers.

The next three numbers that Juliana will erase are $$1190,~1191,$$ and $$1192$$. Thus, the 400th number that Juliana erases is $$1192$$.