Final answer:
The greatest multiple of 36 that uses each of the digits 1 through 9 exactly once is 9,657,812,4, which was found using divisibility rules for 4 and 9 and trial and error.
Step-by-step explanation:
The question asks to find the greatest multiple of 36 that uses each of the digits 1 through 9 exactly once. First, we should note that a multiple of 36 must be divisible by both 4 and 9, since 36 = 4 × 9. A number is divisible by 4 if its last two digits form a number that is divisible by 4. For divisibility by 9, the sum of its digits must be divisible by 9.
When we consider the digits 1 through 9, their sum is 45 (1+2+3+4+5+6+7+8+9), which is divisible by 9. So any ordering of these digits will be divisible by 9. To satisfy the divisibility rule of 4, we need the last two digits of our number to form a number divisible by 4. Among the digits 1 through 9, 12, 16, 24, etc., we can find several pairings that meet this criteria.
By trial and error and keeping in mind the divisibility rules, the greatest multiple of 36 using all the digits 1 through 9 without repeating any is 9,657,812,4. This number satisfies both the rules for divisibility by 4 (last two digits are 24, which is divisible by 4) and by 9 (the sum of the digits is 45, which is divisible by 9).
The method we used involves understanding of divisibility rules, trial and error, and systematization by starting from the largest digit and going down to find the largest possible number that meets the criteria.