Let's approach this problem systematically.
We are given the digits 9, 6, and 8, and need to form the largest 2-digit number using any two of these digits, such that the number is a multiple of 4.
### Steps to Solve:
1. Understanding Multiples of 4:
A number is a multiple of 4 if the number formed by its last two digits is divisible by 4.
2. Form Possible 2-Digit Numbers:
- Using 9 and 6: We can form the numbers 96 and 69.
- Using 9 and 8: We can form the numbers 98 and 89.
- Using 6 and 8: We can form the numbers 68 and 86.
3. Check Divisibility by 4:
- 96: Checking divisibility by 4, [tex]\( 96 \div 4 = 24 \)[/tex]. So, 96 is a multiple of 4.
- 69: Checking divisibility by 4, [tex]\( 69 \div 4 = 17.25 \)[/tex]. So, 69 is not a multiple of 4.
- 98: Checking divisibility by 4, [tex]\( 98 \div 4 = 24.5 \)[/tex]. So, 98 is not a multiple of 4.
- 89: Checking divisibility by 4, [tex]\( 89 \div 4 = 22.25 \)[/tex]. So, 89 is not a multiple of 4.
- 68: Checking divisibility by 4, [tex]\( 68 \div 4 = 17 \)[/tex]. So, 68 is a multiple of 4.
- 86: Checking divisibility by 4, [tex]\( 86 \div 4 = 21.5 \)[/tex]. So, 86 is not a multiple of 4.
4. Identify the Largest Multiple of 4:
From the valid results:
- 96
- 68
The largest number that meets the criteria is 96.
So, the largest 2-digit number that can be formed using the digits 9, 6, and 8 and is a multiple of 4 is 96. The possible 2-digit numbers that are multiples of 4 are 96 and 68.
