To find the number of positive integers with exactly four decimal digits that satisfy the given conditions, we can break down the problem into three parts and solve each individually.
(a) The number should be divisible by both 5 and 7.
To find numbers that are divisible by 5, we know that the last digit must be either 0 or 5 since any number ending with 0 or 5 is divisible by 5. Therefore, there are 2 choices for the last digit.
To find numbers that are divisible by 7, we cannot use a simple criteria based on a single digit. Instead, we can use the divisibility rule for 7, which states that a number is divisible by 7 if and only if the difference between twice the unit digit and the remaining number is divisible by 7.
So, we can start by writing down a chart with potential last digits (0 and 5) and then checking which numbers will satisfy the divisibility rule for 7:
For 0 as the last digit: We can have 10 choices for the first digit (0-9), 10 choices for the second digit, and 9 choices for the third digit (as it cannot be the same as the first digit). This gives us a total of 10 * 10 * 9 = 900 possibilities.
For 5 as the last digit: We can have 9 choices for the first digit (1-9, since 0 will make it a 3-digit number), 10 choices for the second digit (0-9), and 8 choices for the third digit (as it cannot be the same as the first or second digit). This gives us a total of 9 * 10 * 8 = 720 possibilities.
Now, to find the numbers that are divisible by both 5 and 7, we need to find the common multiples of 900 and 720 (since the last three digits can be arranged in any order). We can find the least common multiple (LCM) of 900 and 720, which is 3,600. Since the last three digits can be arranged in any order, we have 6 permutations for each number. So, the total number of positive integers satisfying condition (a) is 3,600/6 = 600.
(b) The number should have distinct digits.
To ensure that the digits are distinct, we can choose the digits from the remaining pool after selecting the last digit for condition (a).
For 0 as the last digit: We have 9 choices for the first digit (1-9, since 0 will make it a 3-digit number), 9 choices for the second digit (as it cannot be the same as the first digit), and 8 choices for the third digit (as it cannot be the same as the first or second digit). This gives us a total of 9 * 9 * 8 = 648 possibilities.
For 5 as the last digit: We have 9 choices for the first digit (1-9, since 0 will make it a 3-digit number), 9 choices for the second digit (as it cannot be the same as the first digit), and 7 choices for the third digit (as it cannot be the same as the first or second digit). This gives us a total of 9 * 9 * 7 = 567 possibilities.
Again, we can find the common multiples of 648 and 567 (last three digits arranged in any order), which is 3,888. Dividing by 6 (number of permutations) gives us 648.
(c) The number should not be divisible by either 5 or 7.
To find the numbers that are not divisible by 5 or 7, we need to find the total numbers minus the numbers found in (a) and (b).
Total numbers between 1000 and 9999 (inclusive): 9999 - 1000 + 1 = 9000.
Numbers satisfying (a): 600.
Numbers satisfying (b): 648.
Therefore, the number of positive integers satisfying condition (c) is 9000 - 600 - 648 = 7752.
In conclusion, there are 600 positive integers with exactly four decimal digits that are divisible by both 5 and 7, have distinct digits, and are not divisible by either 5 or 7.