It seems that you've uploaded an image containing a set of math problems. Let me go through them and provide solutions:

Problem 7:

"Using the digits 1, 3, 5, 8, how many 2-digit numbers divisible by 9 can be formed?"

To form a two-digit number divisible by 9, the sum of its digits must be divisible by 9. The possible sums of digits from the numbers 1, 3, 5, 8 are:

• 1 + 3 = 4
• 1 + 5 = 6
• 1 + 8 = 9 (divisible by 9)
• 3 + 5 = 8
• 3 + 8 = 11
• 5 + 8 = 13

Only the sum of 1 and 8 results in a number divisible by 9. The two numbers that can be formed are 18 and 81, so the correct answer is:

Answer: C) 3

Problem 8:

"AB and BA are two-digit natural numbers. If A + B = 17, find the sum of AB + BA."

Let’s define the numbers:

• AB = 10A + B (since A is the tens digit and B is the ones digit)
• BA = 10B + A (since B is the tens digit and A is the ones digit)

Given that A + B = 17, we can substitute to find AB + BA:

$AB + BA = (10A + B) + (10B + A)$ $AB + BA = 11A + 11B = 11(A + B)$

Since A + B = 17:

$AB + BA = 11 \times 17 = 187$

Answer: A) 187

Problem 9:

"Which of the following statements is correct?"

• A) An odd number is divisible only by 1.
• B) All even numbers are even.
• C) All odd numbers are even.
• D) All even numbers are odd.

The correct statement is:

• A) An odd number is divisible only by 1 (is false).
• B) All even numbers are even (is true).

Answer: B) All even numbers are even.

Problem 10:

"52 people are divided into groups of six and seven people. How many groups can be formed?"

We need to divide 52 by 6 and 7:

• Dividing by 6: 52 ÷ 6 = 8 groups of 6 people (remainder 4).
• Dividing by 7: 52 ÷ 7 = 7 groups of 7 people (remainder 3).

Thus, the solution is dividing into 8 groups of 6 and 3 groups of 7, for a total of 20 groups.

Answer: B) 20

Let me know if you need further explanations or if you have any other questions!