Given the three-digit odd number is of the form 7ab and is divisible by 15, we want to find the highest two-digit number "ab" that satisfies these conditions.
Key points:
- The number is odd, so the last digit bbb must be odd.
- Divisible by 15 means divisible by both 3 and 5.
- Divisible by 5 means the last digit bbb must be 0 or 5.
- Since the number is odd, bbb cannot be 0, so b=5b=5b=5.
- Divisible by 3 means the sum of digits is divisible by 3.
Step-by-step:
- The number is 7ab=700+10a+b7ab=700+10a+b7ab=700+10a+b.
- Since b=5b=5b=5, the number is 7a57a57a5.
- Sum of digits: 7+a+5=12+a7+a+5=12+a7+a+5=12+a.
- For divisibility by 3, 12+a12+a12+a must be divisible by 3.
- The highest two-digit number "ab" means aaa and bbb form the two digits after 7.
- b=5b=5b=5, so we want the highest aaa such that 12+a12+a12+a is divisible by 3.
- 121212 is divisible by 3, so aaa must be divisible by 3.
- The highest digit aaa (0–9) divisible by 3 is 9.
Therefore:
- a=9a=9a=9
- b=5b=5b=5
- Highest two-digit number ab=95ab=95ab=95
The number 795795795 is odd, divisible by 5, and the sum 7+9+5=217+9+5=217+9+5=21 is divisible by 3, so 795795795 is divisible by 15. Answer: The highest two-digit number ababab is 95.
