A rogue, and a scallywag

The 3 digit integer may be 121, 221, 321, 421, 521, 621, 721, 821, 921. And the number is 9

And the 2 digit integer may be 20, 21, 22, 23, 24, 25, 26, 27, 28, 29. Total 10 integer.

So, there are $10+9= 19$ different integer could Bradan be thinking of

infirst case in first position we can put 1 to 9 means nine ways and in second part we can use 0 to 9 means ten ways as there or so 10+9=19

An integer with the last two digits known and the first one unknown can have $9$ possible first digits (and not $10$ because $0$ cannot be the first digit of a number. An integer with the first digit known and the last digit unknown can have $10$ possible ending digits. $9+10=\boxed{19}$ .

Three digit numbers having 21 as last two digits are 121,221,321,421,521,621,721,821,921 so these are 9 3-digit numbers bradan could be thinking of.2-digit no. having 2 as first digit are from 20 to 29 so these are 10 numbers.In total there are 9+10=19 numbers bradan could possibly think of.

Since the first two digits of the three digit number is fixed, there are only 9 possibilities from 1 to 9 as 0 is not applicable. for the two digit number, since it is the ones digit, there are ten possibilities from 0 to 9. In this case, 10+9=19

its simple for 3 digits = 9 1 1; because two last digit must 21 for 2 digits=10, (from 0 -9)

Clearly, if the number has $3$ digits, there are $9$ choices for the first digit, and the other two digits are fixed. If the number has two digits, the first digit is fixed, and there are $10$ choices for the second digit.

Thus, there are a total of $9 + 10 = \boxed {19}$ integers Bradan could be thinking of.

So, we have the three-digit number, of which 2 digits are given to us, and the two-digit number, of which 1 digit is given to us. So, there is only one digit in each number that can change. We cannot use 0 as the missing digit for the first number. So our choices for those missing digits are immediately limited to 1, 2, 3, 4, 5, 6, 7, 8, and 9 for the first number, and 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0 for the second number. Count all those and you get 19, which is the answer

Now first possibility goes like this : a21 is a three digit number.....so a must be an integer from 1 to 9....so there we have 9 possibilities.(a cannot be equal to 0 or else it would be a two digit number.) Then the second possibility is : 2b is a two digit number.... but here b must be an integer from 0 to 9.....so there are 10 possibilities. Hence total possibilities should be 10+9=19 possibilities.

There are 9 three-digit numbers ending with 21, and 10 two-digit numbers that has a tens digit of 2. 9+10=19. Therefore, the answer is 19.

Seja $x21$ o número de três algarismos. Os possíveis valores que "x" pode assumir são: {1, 2, 3, ..., 8, 9}. Portanto, 9 possibilidades! Veja:

$\\ 9 - 1 + 1 = \\ 9$

Já o número de dois dígitos é dado por $2y$ , então, os possíveis algarismos de "y" são: {0, 1, 2,..., 8, 9}.

Por isso,

$\\ 9 - 0 + 1 = \\ 10$

Segue que,

$9 + 10 = \\ \boxed{19}$