The highest possible two-digit number ab is 95.

Key idea

  • A number is divisible by 15 if it is divisible by 3 and 5.
  • The three-digit number is 7ab7ab7ab, and it is odd , so the units digit bbb must be 5 (it cannot be 0, or the number would be even).
  • So the number is 7a57a57a5.

Divisibility conditions

  1. Divisible by 5
    • Last digit is 5 → already satisfied.
  1. Divisible by 3
    • Sum of digits must be divisible by 3.
 * Sum of digits: 7+a+5=12+a7+a+5=12+a7+a+5=12+a.
 * So 12+a12+a12+a must be a multiple of 3.
 * Since 12 is already a multiple of 3, any digit a=0,1,2,…,9a=0,1,2,\dots,9a=0,1,2,…,9 makes 12+a12+a12+a still divisible by 3 exactly when aaa itself is a multiple of 3 (i.e., a=0,3,6,9a=0,3,6,9a=0,3,6,9).

So the valid numbers are:

  • 705,735,765,795705,735,765,795705,735,765,795.

Their corresponding two-digit parts ababab are:

  • 05, 35, 65, 95.

Among these, the highest two-digit number ab is 95.

Therefore, for the three-digit odd number 7ab7ab7ab to be divisible by 15, the maximum possible value of ab is 95.

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