The question in the uploaded image asks:

"How many numbers are there between 100 and 1000 such that every digit is either 2 or 9?"

Solution:

1. A number between 100 and 1000 is a 3-digit number.

   • The first digit must be either 2 or 9 (to ensure it remains a 3-digit number).
   • The second digit can be either 2 or 9.
   • The third digit can also be either 2 or 9.

2. Since there are two choices (2 or 9) for each of the three digits, the total number of numbers is:

   $2 \times 2 \times 2 = 8$

The Numbers:

The possible numbers formed by digits {2, 9} are:

• 222, 229, 292, 299, 922, 929, 992, 999.

So, there are 8 numbers in total.

Would you like a further breakdown or have any questions about the solution? 😊

Related Questions:

1. How many numbers between 10 and 100 have every digit as 2 or 9?
2. How many 4-digit numbers can be formed using only the digits 2, 5, and 9?
3. If we add another digit option (e.g., 3), how many numbers between 100 and 1000 can we form?
4. What is the sum of all numbers between 100 and 1000, where every digit is 2 or 9?
5. How many numbers between 100 and 1000 have at least one digit as 9?

Tip:

When solving problems like this, always identify how many digits the number has and analyze constraints digit by digit.