How many times will the digit 8 be written when listing the integers from 1 to 1000?

Answer

Hint:
Here, we need to find the number of times the digit 3 comes in 1 to 1000. First, we will use cases to find the number of integers that have the digit 3. Then, we will use the cases to find the number of times 3 comes in those integers. Then we will add them to get the required answer.

Complete Step by Step Solution:
The numbers 1 to 1000 can be divided as follows:
1 to 9 are one digit numbers
10 to 99 are two digit numbers
100 to 999 are three digit numbers
1000 is a four digit number
First, we will find the number of integers having the digit 3 while listing the one digit numbers 1 to 9.
The only number that has the digit 3 in the integers from 1 to 9 is 3.
Therefore, the number of times the digit 3 comes in 1 to 9 is 1.
Next, we will find the number of integers having the digit 3 while listing the two digit numbers 10 to 99.
Let us use three cases:
Case 1: Both the digits are 3.
The only two digit number that has 3 in both places is 33.
Therefore, the number of two digit numbers having the digit 3 in both places is 1.
Here, the digit 3 comes twice.
Therefore, the number of times 3 appears is 2.
Case 2: The digit at ten’s place is 3.
If the digit at ten’s place is 3, then the digit in the unit’s place can be any number other than 3 (if the number at unit’s place is also taken as 3, it will include 33. This will result in the counting the number 33 twice, both in case 1 and case 2).
The remaining digits are 0, 1, 2, 4, 5, 6, 7, 8, 9.
Therefore, the digit at ten’s place is 3 (1 way), and the digit at unit’s place can be any of the remaining 9 numbers (9 ways).
The number of two digit numbers having 3 at ten’s place is the product of the number of ways to select the digit at ten’s place and the digit at unit’s place.
The number of two digit numbers having 3 at ten’s place is \[1 \times 9 = 9\] numbers.
Each of these numbers has only one 3 at the ten’s place.
Therefore, the number of times 3 appears in these numbers is 9.
Case 3: The digit at unit’s place is 3.
If the digit at unit’s place is 3, then the digit in the ten’s place can be any number other than 0 or 3 (if the number at ten’s place is 0, then the number is not a two digit number).
The remaining digits are 1, 2, 4, 5, 6, 7, 8, 9.
Therefore, the digit at unit’s place is 3 (1 way), and the digit at ten’s place can be any of the remaining 8 numbers (8 ways).
The number of two digit numbers having 3 at unit’s place is \[1 \times 8 = 8\] numbers.
Each of these numbers has only one 3 at the unit’s place.
Therefore, the number of times 3 appears in these numbers is 8.
Thus, using the 3 cases, we get the number of the digit 3 comes in the numbers 10 to 99 as \[2 + 9 + 8 = 19\].
Next, we will find the number of integers having the digit 3 while listing the three digit numbers 100 to 999.
Let us use seven cases:
Case 1: All the three digits are 3.
The only three digit number that has 3 in all places is 333.
Therefore, the number of three digit numbers having the digit 3 in all places is 1.
Here, the digit 3 comes thrice.
Therefore, the number of times 3 appears is 3.
Case 2: The digit at hundred’s place is 3.
If the digit at hundred’s place is 3, then the digit in the ten’s and unit’s place can be any number other than 3 (if the number at unit’s place or ten’s place is also taken as 3, it will include numbers from other cases also and will lead to incorrect answer).
The remaining digits are 0, 1, 2, 4, 5, 6, 7, 8, 9.
Therefore, the digit at hundred’s place is 3 (1 way), the digits at ten’s place and unit’s place can be any of the remaining 9 numbers (9 ways each).
The number of three digit numbers having 3 at hundred’s place is the product of the number of ways to select the digit at hundred’s place, ten’s place and unit’s place.
The number of three digit numbers having 3 at hundred’s place is \[1 \times 9 \times 9 = 81\] numbers.
Each of these numbers has only one 3 at the hundred’s place.
Therefore, the number of times 3 appears in these numbers is 81.
Case 3: The digit at ten’s place is 3.
If the digit at ten’s place is 3, then the digit in the hundred’s place can be any of the remaining numbers except 0 or 3, and the digit at the unit’s place can be any number other than 3.
The remaining digits are 0, 1, 2, 4, 5, 6, 7, 8, 9.
Therefore, the digit at ten’s place is 3 (1 way), the digits at hundred’s place can be any number except 0 or 3 (8 ways) and the digit at unit’s place can be any of the remaining 9 numbers (9 ways).
The number of three digit numbers having 3 at ten’s place is \[8 \times 1 \times 9 = 72\] numbers.
Each of these numbers has only one 3 at the ten’s place.
Therefore, the number of times 3 appears in these numbers is 72.
Case 4: The digit at unit’s place is 3.
If the digit at unit’s place is 3, then the digit in the hundred’s place can be any of the remaining numbers except 0 or 3, and the digit at the ten’s place can be any number other than 3.
The remaining digits are 0, 1, 2, 4, 5, 6, 7, 8, 9.
Therefore, the digit at unit’s place is 3 (1 way), the digits at hundred’s place can be any number except 0 or 3 (8 ways) and the digit at ten’s place can be any of the remaining 9 numbers (9 ways).
The number of three digit numbers having 3 at unit’s place is \[8 \times 9 \times 1 = 72\] numbers.
Each of these numbers has only one 3 at the unit’s place.
Therefore, the number of times 3 appears in these numbers is 72.
Case 5: The digit at hundred’s place and ten’s place is 3.
If the digit at hundred’s and ten’s place is 3, then the digit in the unit’s place can be any number other than 3.
The remaining digits are 0, 1, 2, 4, 5, 6, 7, 8, 9.
Therefore, the digit at hundred’s and ten’s place is 3 (1 way each), and the digits at unit’s place can be any of the remaining 9 numbers (9 ways each).
The number of three digit numbers having 3 at hundred’s and ten’s place is \[1 \times 1 \times 9 = 9\] numbers.
Each of these numbers has two 3, that is at the hundred’s and ten’s place.
Therefore, the number of times 3 appears in these 9 numbers is 18.
Case 6: The digit at hundred’s place and unit’s place is 3.
If the digit at hundred’s and unit’s place is 3, then the digit in the ten’s place can be any number other than 3.
The remaining digits are 0, 1, 2, 4, 5, 6, 7, 8, 9.
Therefore, the digit at hundred’s and unit’s place is 3 (1 way each), and the digits at ten’s place can be any of the remaining 9 numbers (9 ways each).
The number of three digit numbers having 3 at hundred’s and unit’s place is \[1 \times 9 \times 1 = 9\] numbers.
Each of these numbers has two 3, that is at the hundred’s and unit’s place.
Therefore, the number of times 3 appears in these 9 numbers is 18.
Case 7: The digit at ten’s place and unit’s place is 3.
If the digit at ten’s and unit’s place is 3, then the digit in the hundred’s place can be any number other than 0 or 3.
The remaining digits are 1, 2, 4, 5, 6, 7, 8, 9.
Therefore, the digit at ten’s and unit’s place is 3 (1 way each), and the digits at hundred’s place can be any of the remaining 8 numbers (8 ways each).
The number of three digit numbers having 3 at ten’s and unit’s place is \[8 \times 1 \times 1 = 8\] numbers.
Each of these numbers has two 3, that is at the ten’s and unit’s place.
Therefore, the number of times 3 appears in these 8 numbers is 16.
Thus, using the 7 cases, we get the number of the digit 3 comes in the numbers 100 to 999 as \[3 + 81 + 72 + 72 + 18 + 18 + 16 = 280\].
The four digit number 1000 has no digit 3.
Therefore, the number of times 3 comes in 1 to 1000 is the sum of the number of times 3 comes in 1 to 9, 10 to 99, and 100 to 999.
Thus, we get
Number of times 3 comes in integers 1 to 1000 \[ = 1 + 19 + 280 = 300\]

Therefore, the correct option is option (a).

Note:
A common mistake is to write the answer as 271 by adding the number of integers having the digit 3 in them. Some of the integers from 1 to 1000 have the digit 3 more than once. Therefore, the answer 271 is the number of integers, and not the number of times 3 appears in them. Hence, this is incorrect.

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