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What is the 3 digit numbers between 101 and 170 whose digits total 9?
Wiki User
∙ 14y ago
108,117,126,135,144,153,162.
171, 180
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∙ 14y ago
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The total number of five digit numbers that can be be formed digits 12345 is?
It is 120 if the digits cannot be repeated.
What are the three digit numbers between 101 and 170 whose digits total 9?
108 117 126 135 144 153 and 162
What is the total number of digits in the numbers from 1 to 1000000?
In the numbers 1-9 each number has 1 digit and there are 9 of them, so that's 9.In 10-99 each number has 2 digits, and there are 90 of them: 2x90 = 180There are 900 three digit numbers [100 through 999]: 2700 digits.There are 9000 four digit numbers: 36000 digits.90,000 numbers with five digits: 450,000 digits.900,000 numbers with six digits: 5,400,000 digits.Then 1 number with seven digits: 7 digits.Add them up and you have 5,888,896 digits.
How many four digit numbers can be formed using all digits if same digit on four place is not allowed example 5555 is not allowed?
Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991
What is the number of 5 digit telephone numbers having atleast one of their digits repeated?
5 digit telephone numbers having at least one of their digits repeated is = total possible 5 digit telephone numbers - 5 digit telephone numbers without any digit being repeated. =(10*10*10*10*10)-(10*9*8*7*6) =100000-30240 =69760
The total number of five digit numbers that can be be formed digits 12345 is?
It is 120 if the digits cannot be repeated.
What are the three digit numbers between 101 and 170 whose digits total 9?
108 117 126 135 144 153 and 162
What is the total number of digits in the numbers from 1 to 1000000?
In the numbers 1-9 each number has 1 digit and there are 9 of them, so that's 9.In 10-99 each number has 2 digits, and there are 90 of them: 2x90 = 180There are 900 three digit numbers [100 through 999]: 2700 digits.There are 9000 four digit numbers: 36000 digits.90,000 numbers with five digits: 450,000 digits.900,000 numbers with six digits: 5,400,000 digits.Then 1 number with seven digits: 7 digits.Add them up and you have 5,888,896 digits.
How many 3 digit number can be made using the digits 2 4 7 9?
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
Find all the 2 digit numbers whose digits total 11?
All two digit numbers which the digits add up to 11 are :- 29, 38, 47, 56, 65, 74, 83 & 92.
How many four digit numbers can be formed using all digits if same digit on four place is not allowed example 5555 is not allowed?
Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991
What is the number of 5 digit telephone numbers having atleast one of their digits repeated?
5 digit telephone numbers having at least one of their digits repeated is = total possible 5 digit telephone numbers - 5 digit telephone numbers without any digit being repeated. =(10*10*10*10*10)-(10*9*8*7*6) =100000-30240 =69760
How many different three digit numbers can you form using the digits 1 2 3 5 6 7 and 9 without repetition?
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
How many numbers are there between 100 and 1000 such that one of their digits is 6?
I suggest you calculate (1) how many numbers are in total, and (2) how many numbers in that range do NOT have 6 as a digit. Then you can subtract the result of (1) minus the result of (2).
How many five digit numbers can be formed using the digits 02345 when repetition is allowed such that the number formed is divisible by 2 or 5 or both?
There are 2000 possible five digit numbers that can be formed from the digits 02345 that are divisible by 2 or 5 or both. To be divisible by 2, the last digit must be even, namely 0, 2 or 4 (in the digits allowed). To be divisible by 5, the last digit must be 0 or 5. Thus to be divisible by 2 or 5 or both, the last digit must be 0, 2, 4 or 5 (a choice of 4). Presuming that a 5 digit number must be at least 10000, then: For the first digit there is a choice of 4 digits (2345); for each of these there is a choice of 5 digits (02345) for the second, making a total so far of 4 x 5 numbers; for each of these choices for the first and second digits there is a choice of 5 digits (02345) for the third digit making the total so far (4 x 5) x 5 numbers; for each of these choices for the first three digits there is a choice of 5 digits (02345) for the fourth digit making the total so far (4 x 5 x 5) x 5 numbers; for each of these choices for the first four digits there is a choice of 4 digits (0245 - as discussed above) for the last digit, giving a total of (4 x 5 x 5 x 5) x 4 numbers. So the total number of five digit numbers so formed is: number = 4 x 5 x 5 x 5 x 4 = 2000.
How many ten digit palindromes are there?
90000. With 10 digit palindromes, the last 5 digits are the same as the first 5 digits in reverse, eg 12345 54321. So it comes down to how many 5 digit numbers are there? They are the numbers "10000" to "99999", a total of 99999 - 10000 + 1 = 90000.
How many 2-digit or 3-digit numbers can be formed using the digits 134568 and 9 which are diisible by 4?
48. Assuming no digit can be used more than once, the two digit numbers divisible by 4 are: 16, 36, 48, 56, 64, 68, 84, 96 8 of them. For any number to be divisible by 4, only the last two digits need be divisible by 4; so for three digit numbers, each of the two digit numbers above can be preceded by any of the remaining 5 digits and still be divisible by 4. → 5 x 8 = 40 three digit numbers are divisible by 4 → 40 + 8 = 48 two or three digit numbers made up of the digits {1, 3, 4, 5, 6, 8, 9} are divisible by 4. If repeats are allowed, there are an extra 2 two digit numbers (44 and 88) and each of the two digit numbers can be preceded by any of the 7 digits, making a total of 7 x 10 + 10 = 80 two and three digits numbers divisible by 4 make up of digits from the given set.
