The number of integers greater than 5000 and divisible by 5 that can be formed with the digits 1, 3, 5, 7, 8, 9 where no digit is repeated is

We need to find integers that are > 5000, divisible by 5, using only digits 1, 3, 5, 7, 8, 9 without repetition.

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First, for a number to be divisible by 5, its last digit must be either 0 or 5.

Since 0 is not in our set of available digits, the only option is to use 5 as the last digit.

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Since we need numbers greater than 5000, they must have at least 4 digits. The maximum possible length is 6 digits (as we have 6 available digits).

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4-digit numbers:

• Last digit must be 5
  • First digit must be $\geq 6$ to ensure the number is > 5000 (We can't use 5 again)
  • From our available digits, only 7, 8, 9 are $\geq 6$
  • We have 3 choices for the first digit (7, 8, 9)
  • We have 4 choices for the second digit (all except first and 5)
  • We have 3 choices for the third digit (all except first, second, and 5)

Number of 4-digit numbers = $3 \times 4 \times 3 \times 1 = 36$

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5-digit numbers:

• Last digit must be 5
  • First digit can be any of the 5 digits except 5
  • Remaining positions filled with remaining digits

Number of 5-digit numbers = $5 \times 4 \times 3 \times 2 \times 1 = 120$

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6-digit numbers:

• Last digit must be 5
  • First digit can be any of the 5 digits except 5
  • Remaining positions filled with remaining digits

Number of 6-digit numbers = $5 \times 4 \times 3 \times 2 \times 1 \times 1 = 120$

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Total number of valid integers = $36 + 120 + 120 = 276$

Therefore, there are 276 integers greater than 5000 and divisible by 5 that can be formed with the given digits without repetition.