Sage-Tips

How many 3 digit numbers are there exactly?

In the tens place and the one’s place, there can be any digit from 0-9. There is a possibility of 10 digits in the hundredth place and 10 digits in the unit’s place. Thus, there are 9 × 10 × 10 = 900 three-digit numbers in all. Therefore there are 900 three-digit numbers in all.

How many three digit numbers are there such that the digits are repeated at least once ie all the digits do not occur only once?

Ans :- 648 numbers . Originally Answered: How many three-digit numbers can be formed without repeating digits? It depends whether you allow leading zeroes. If you don’t, there are 9 choices for the first digit (1–9), 9 choices for the second digit (0–9 less the one you already used), and 8 for the third digit.

How many 3 digit numbers and its reverse are divisible by 7 with no repetition of the numbers?

Originally Answered: All three digit numbers are different and are divisible by 7. If they are reversed, they are also divisible by 7. How many such numbers are there? There are 16 such numbers: 161, 168, 252, 259, 343, 434, 525, 595, 616, 686, 700, 707, 770, 777, 868 & 959.

Which numbers will have exactly 3 factors?

We know that the numbers between 1 and 100 which have exactly three factors are 4, 9, 25 and 49.

How many 3 digit odd No are there?

The three digit numbers are 100 to 999 inclusive so there are 999-100+1 = 999-99 = 900 So, 900 three digit numbers If half of all numbers is odd then half of 900 is 450, there are 450 odd positve 3 digit numbers.

How many times can 7 occur exactly in a 3 digit number?

So, we have 3 ways in which 7 can occur exactly once in the 3 digit number: i. 7 in the hundreds place: With 7 in the hundreds place, we can choose the other two digits as follows: one of {0,1,2,3,4,5,6,8,9} for the tens place and one of the same set for the ones place.

How many ways can 1st digit of a 3 digit number be chosen?

Let’s count how many numbers have their 1st digit (Hundredth place digit) as 7. so, 1st digit can be chosen in only 1 way. for both the remaining digits, we have 9 choices excluding 7. Thus, we have 1*9*9 = 81 ways wherein 1st digit of a 3 digit number is 7.

How many digited numbers can be formed in 9 x 8?

Hence, with 7 in the tenth place, the 3 digited numbers can be formed in 9 x 8 = 72 ways. Lastly, putting 7 in the unit place, the hundredth place can be formed in 8 ways (as 0 and 7 cannot be put here) and for each of these 8 ways, the tenth place can be formed in 9 ways.

How many possible values can a three digit number accept?

Here the number of three number is (999-100)+1=900. Now, if no digit is 3, then the highest digits can be one of 0 to 9 excluding 0,3 as 0 will reduce the number to 2-digit. So, the highest can accept 8 possible values. The rest two digits each can accept one of 0 to 9 excluding 3.