How many numbers can be formed using digits $0, 1, 2..........9$ which is more than or equal to $6000$ and less than $7000$ and is divisible by $5$ whereas any number can be repeated as many times?

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Solution

Thousand Place must be filled with 6. $= 1$

Unit place filled with either 0 or 5 because its divided by 5. $= 2$

Hundredth place filled with remaining 8 digits. $= 8$

Similarly, tenth place filled with 7 digits. $= 7$

So, total positive integer $= 1 \times 2 \times 8 \times 7 = 112$

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How many numbers can be formed using digits 0,1,2..........9 which is more than or equal to 6000 and less than 7000 and is divisible by 5 whereas any number can be repeated as many times?

Thousand Place must be filled with 6. =1 Unit place filled with either 0 or 5 because its divided by 5. =2Hundredth place filled with remaining 8 digits. =8 Similarly, tenth place filled with 7 digits. =7 So, total positive integer =1×2×8×7=112

How many numbers can be formed using digits $0, 1, 2..........9$ which is more than or equal to $6000$ and less than $7000$ and is divisible by $5$ whereas any number can be repeated as many times?

Thousand Place must be filled with 6. $= 1$

Unit place filled with either 0 or 5 because its divided by 5. $= 2$

Hundredth place filled with remaining 8 digits. $= 8$

Similarly, tenth place filled with 7 digits. $= 7$

So, total positive integer $= 1 \times 2 \times 8 \times 7 = 112$