Certainly! Let's break down the questions and solve them step-by-step.
### Question 1: Sum of the digits of the greatest four-digit even square number
1. Find the greatest four-digit even square number:
- The greatest four-digit number is 9999.
- The largest integer whose square is less than or equal to 9999 is 99, because [tex]\(100^2 = 10000\)[/tex] is a five-digit number.
- However, we need an even square number. The largest even integer less than or equal to 99 is 98.
- Therefore, [tex]\(98^2 = 9801\)[/tex].
2. Sum of the digits of 9801:
- Split the number 9801 into its digits: 9, 8, 0, and 1.
- Sum these digits: [tex]\(9 + 8 + 0 + 1 = 18\)[/tex].
So, the sum of the digits of the greatest four-digit even square number is 18.
Therefore, the correct option is:
[tex]\[ \boxed{2 \text{\) 18}} \][/tex]
### Question 2: If the sum of the two consecutive odd numbers is [tex]\(P\)[/tex], what is the smaller number?
1. Expression for consecutive odd numbers:
- Let the smaller number be [tex]\(x\)[/tex].
- Then the next consecutive odd number would be [tex]\(x + 2\)[/tex] (since odd numbers differ by 2).
2. Sum of the numbers:
- The sum of these two numbers is given as [tex]\(P\)[/tex].
- Therefore, [tex]\(x + (x + 2) = P\)[/tex].
- Simplify the equation: [tex]\(2x + 2 = P\)[/tex].
3. Solve for [tex]\(x\)[/tex]:
- Rearrange the equation to solve for [tex]\(x\)[/tex]:
[tex]\[ 2x + 2 = P \][/tex]
[tex]\[ 2x = P - 2 \][/tex]
[tex]\[ x = \frac{P - 2}{2} \][/tex]
[tex]\[ x = \frac{P}{2} - 1 \][/tex]
So, the smaller of the two consecutive odd numbers is [tex]\(x = \frac{P}{2} - 1\)[/tex].
Therefore, the correct option is:
[tex]\[ \boxed{3 \text{\) \frac{P}{2} - 1}} \][/tex]
