An n digit number is a positive number with exactly n digits. nine hundred distinct n digit numbers are to be formed using only the three digit 2, 5 & 7. the smallest value of n for which this possible is? (2024)

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An n digit number is a positive number with exactly n digits. nine hundred distinct n digit numbers are to be formed using only the three digit 2, 5 & 7. the smallest value of n for which this possible is? ›

Nine hundred distinct n digit numbers are to be formed using only the three digits 2,5 and 7. The smallest value of n for which this is possible is. 6.

We have 9 choices for the first digit (which cannot be 0). Then we have 9 choices (including 0) for the second digit. Then 8 choices for the third digit. Therefore, we can form 9*9*8 = 648 3-digit numbers of distinct digits.

Hence, 9 digit numbers of different digits can be formed in 3265920 ways.

Hence, the required number of numbers = (6×3)=18. Q.

A five digit number divisible by 3 is to be formed using the digits 0,1,2,3,4 and 5, without repetition. The total number of ways this can be done, is. 216.

While doing this, we keep in mind that order doesn't matter, so we don't use repeats such as {1,2,3} and {3,2,1}. These are all of the possible combinations of three numbers, and we see that there are seven combinations in this list. Therefore, there are 7 combinations possible with 3 numbers.

You can also think it this way: You have 9 ways to choose the first digit (it cannot be zero), 9 ways to choose the second (it can be zero but it cannot be the same as the first one) and 8 ways to choose the third one (it can be zero but it cannot be the same as the first or second one). Total: 9×9×8=648.

9 digit numbers are those numbers that have 9 digits in them and they start from the number 100000000 and end on 999999999. There are a total of 900 million, 9-digit numbers.

How many combinations are there in a 9-digit password? - Quora. I'm going to assume you mean traditional digits (from 0 to 9). In this case, there are 10 to the power of 9 possible passwords, since there are 9 spots, and we can select 10 possibilities in each of them. This is equal to 1,000,000,000 (1 billion).

The number 916238457 is an example of nine digit number which contains each of the digit 1 to 9 exactly once.

How many numbers can be formed with the digits 2, 3, 4, 5, 4, 3, 2 so that the odd digits occupy the odd places? ›

Required number of numbers = ( 3 × 6 ) = 18 . Was this answer helpful?

Thus, 3-digit numbers can be formed in 60 ways without repetition. Q.

Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digitsi) = 5*4*3*2*! = 5! = 120.

Hence, 180 three digit natural numbers are divisible by 5. Q.

Number of such numbers =120.

By using the concept of fundamental principle of multiplication we get the total such numbers as 4 × 3 = 12. So a total of 12 two digit number can be generated by any of the 4 digits. Hope it helps!

The units place can be filled with 0 to 9 ways. i.e., 10 ways. So a total of 9 * 10* 10 = 900 three digit numbers can be formed.

If what you want are all possible three digit numbers with no repetition of the digits then you have 10 choices for the first digit, you have 9 choices for the 2nd digit, and you have 8 choices for the 3rd digit giving you 10x9x8 = 720 in all.

If you are meaning simply a 3 digit number using digits 0–9 then the answer is 10•10•10 = 1,000 with repetition. 10•9•8 = 720 without repetitions.

for each digit you have 9 possibilities to use a number between 1 and 9, so the total number of 3-digits numbers is 9 * 9 * 9 = 729. when you take all numbers between 1 and 999 (which is the largest 3-digit number), you get 999 in total, obviously.