How many numbers are there between 99 and 1000 such that atleast one of the digit is 5?

How many numbers are there between 99 and 1000 such that atleast one of the digit is 5?

= 900 – 648 = 252. Please mark my answer as brainliest if my answer was helpful to you . Marking my answer as brainliest will help you earn some extra points.

How many numbers between 999 and 10000 can be formed with the help of the digits 0 2 3 6 7 when the digits are not he repeated?

Ten thousand place can be filled with 2, 3, 6, 7, 8. Thus 5 digits can be used in constructing the unit place.

How many numbers are there between 101 and 999 such that at least one of the digits is 5?

Let this AP contains n terms. Then, Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.

How many numbers are there between 100 and 1000 such that at least one of digits is 6?

225 numbers
Thus, there are 225 numbers between 100 and 1000 that have exactly one of the digits as 6.

How many numbers are there between 99 and 1000 having atleast one of their digits is 7?

90 numbers
The middle digit can be any one of the 10 digits from 0 to 9. The digit in hundred’s place can be any one of the 9 digits from 1 to 9. Therefore by the fundamental principle of counting there are 10 × 9 = 90 numbers between 99 and 1000 having 7 in the unit’s place.

How many numbers are there between 99 and 1000 having at least one of their digits 77?

Divisibility rules for numbers 1–30

Divisor	Divisibility condition	Examples
12	Subtract the last digit from twice the rest. The result must be divisible by 12.	324: 32 × 2 − 4 = 60 = 5 × 12.
13	Form the alternating sum of blocks of three from right to left. The result must be divisible by 13.	2,911,272: 272 – 911 + 2 = -637

How many numbers lie between 999 and 10000 can be formed?

So the required numbers = 360-60 = 300.

How many numbers are there between 100 and 1000 such that at least one of their digits is 7?

Now all the possible ways to form a digit which is having at least one digit as $ 7 $ between $ 100 $ and $ 1000 $ is $ 72+72+81+8+9+9+1=252 $ . Note: We can simply solve this problem by calculating the number of possible ways to form a number which do not have $ 7 $ as a digit between $ 100 $ and $ 1000 $.

How many numbers are there between 100 and 999?

Hence there are 320 such numbers between 100 and 999 which are odd and have distinct digits. Note: Students should note that at hundreds place zero cannot be placed as no three digit number starts with zero and hence it can be filled by only 8 ways.

How many digits are there between 100 and 1000 such that every digit is either 2 or 9?

Now we will use the fact that numbers should have only 2 and 9 as their digits, so we will find all the possible arrangements that can be done using these conditions and that will be the final answer. Then the only possible number is 222.

How many numbers between 0 and 10000 have one and only one?

So, we can state that 2916 numbers between 0 and 10000 have one and only one digit as 5. The numbers looked for must have one and only one digit as 5. Thus, the three other digits may be any of 0, 1, 2, 3, 4, 6, 7, 8, 9. Having each digit nine possible values, a set of three of them can have 9 ∗ 9 ∗ 9 = 9 3 = 729 different values.

How many 1 digit numbers have only one 5 in them?

In the given range (0 to 10,000) we can have 1-digit numbers, 2-digit numbers, 3-digit numbers, and 4-digit numbers to consider numbers having only one 5 in 1 of 1, 2, 3, or 4 positions, respectively:1. 1-digit numbers: 5 in the units place: Number (s) formed = 1.

How many numbers between 99 and 1000 have 7 in place?

The digit in hundred’s place can be any one of the 9 digits from 1 to 9. Therefore, by the fundamental principle of counting, there are 10 × 9 = 90 numbers between 99 and 1000 having 7 in the unit’s place.

How many 3 digit 3 digit numbers are there between 99-1000?

Therefore, by the fundamental principle of counting, there are 10 × 9 = 90 numbers between 99 and 1000 having 7 in the unit’s place. (ii) Total number of 3 digit numbers having atleast one of their digits as 7 = (Total numbers of three digit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all).