Three digit numbers

There are 900 3 digit integers, and $9^3$ contain no zero. Subtraction gives the answer of 171.

19 numbers that satisfy it per hundred .9 hundreds. $19 \times 9$ =171

Tendo Números de 3 Dígitos Começamos Pelo 100 , Assim Temos :

100,101,102.....110\times9 = 90 , Multiplicamos por 9 , Porque Vai do 100 ao 999 Agora Olharemos o Números com a Unidade 0 , Temos = 110,120,130.....190\times98 =81

Somaremos o Resultado das Possibilidades das Dezena com os da Unidade , Assim Temos = 90+81= 171

It is easier to approach this problem by firstly, finding out the number of three digit positive integers without 0 as any of its digits, then, subtracting that from the total number of three digit positive integers.

To find the number of three digit positive integers without 0 as any of its digits, consider that each of the three digits can be an integer from 1 to 9 inclusive, so number of integers without 0 as any of its digits $= 9^3 = 729$ .

The total number of three digit positive integers is just the number of integers from 100 to 999 inclusive, hence total number of integers $= 999-100+1 = 900$ .

Hence, number of three digit positive integers with 0 as at least one of its digits $= 900-729 = \fbox{171}$ .

In 3 digit number we can consider 3 positions to set a digit.

In 1st position we can set digits 1,2,3,....,9(9 digits).

Then we have 2 possibility.

Case 1: If we set 0 in 2nd position then

we can set 0,1,2,...10(10 digits) in 3rd position.

Then we will get 9 * 1 * 10=90 numbers.

Case 2: If we set 1,2,3,...,9(9 digits) in 2nd position,

then we must set 0 at 3rd position.

Then we will get 9 * 9 * 1=81 numbers.

So, ultimately we get 90+81=171 numbers.

Imagine that there are three positions which can be filled with digits. The hundreds' digit cannot be 0, but it can be any of the other 9 digits. The tens and units positions can be a zero or any of the other nine digits. To have at least one zero, we can consider the scenarios that the units' digit must be 0 and the scenario that the tens' digit must be zero. If the units' digit must be zero, the hundreds' digit can be any of the 9 digits (1-9) and the tens' digit can be any of the 10 digits, therefore, there are 9x10=90 ways for this number to exist. In the second case, the tens' digit must be 0. The hundreds' digit can be any of the 9 digits (1-9), but to avoid repetition from the previous case, the units' digit can only go from 1-9. Therefore, there are 9x9=81 ways for this number to exist. Lastly, we must add the total possible ways for this integer to exist 81+90=171 .

If second digit must be 0: Then we can choose 1-9 for the first digit and third digit, so the possible solution is 9x9 = 81 If third digit must be 0: Then we can choose 1-9 for the first digit and second digit, so the possible solution is 9x9 = 81 If second and third digit must be 0: Then we can choose 1-9 for the first digit, so the possible solution is 9

Because they are three different conditions, so the solution is: 81+81+9 = 171

We can divide the numbers that have atleast one $0$ into three groups:

1. Numbers of the form $\overline{ab0}$
2. Numbers of the form $\overline{a0b}$
3. Numbers of the form $\overline{a00}$

where $a, b \neq 0$ .

Since in the first two groups $a, b$ can hold any number from $1$ to $9$ , there are $9\times 9 = 81$ possible numbers for each group. In group three, $a$ can only have $9$ possible values. Therefore, $9$ numbers in group three. Summing it all up we get $81 + 81 + 9 = 171$

0 cannot be at the hundreds place.

Case 1:Let 0 be at the tens place $\Rightarrow$ number of ways = $9 \times 1 \times 9 = 81$ .

Case 2: let's take that zero can be at tens or ones place $\Rightarrow$ number of ways = $9 \times 10 \times 1 = 90$

Case 2 includes the numbers which have 0 at their ones place only.Total numbers = $81+90 = 171$