Table of Contents
How many 5 digit numbers can be formed using 0 5 which are divisible by 3?
1, 2, 3, 4, 5 –> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!= 120.
How many three digit numbers are divisible by 5 can be formed using?
Hence any of 8 digits (1,2,3,4,6,7,8,9) can be placed at hundreds place….Permutation and Combination #57.
| 57. How many three digit numbers divisible by 5 can be formed using any of the digits from 0 to 9 such that none of the digits can be repeated? | |
|---|---|
| A. 136 | B. 124 |
| C. 112 | D. 108 |
How many 5-digit odd numbers can be formed from the digits 12345?
Answer: The number of 5 digits odd number can be made from 12345 is 1875.
How many 5-digit numbers can be formed which are divisible by 3 using digits 0 6?
So, we can have total 2 * 3C2 * 2C2 = 2 * 3 * 1 = 6 possible sets of digits. So, can make 6 * 5! = 6! = 720 possible permutations.
What can be remainder when a three digit number ABC is divided by 5?
Step-by-step explanation: Smallest 3 digit number that leaves remainder 2 when divided by 5 is 102. Largest 3 digit number that leaves remainder 2 when divided by 5 is 997.
How many three digit numbers are there multiple of 5 without repetition?
The total is 136. How many 3 digit numbers can be formed using the digit 2, 3, 4, 5 and 6 without repetitions? How many of these are even numbers?
What are the conditions for a number to be divisible by 5?
Recommended: Please try your approach on {IDE} first, before moving on to the solution. For a number to be divisible by 5, the only condition is that the digit at the unit place in the number must be either 0 or 5. Check if the given digits contain both 0 and 5.
Why do we not use all the possible combinations of numbers?
Not using all digits allows more possibilities. (So if 2 is the last digit, and there is no 1 used, there are 3!=6, 3! for the swap = 12 more. 12 more for no 3, 12 more for no 5. If there is no 4 then just 6 ways, same 6 more if there is no 2. +48 more ways). Not using 2 digits is left to the reader.
How many possibilities does each digit in the number 6 have?
Two answers are possible. One is with the repetition of digits. The other is without repetition of digits. 1st digit can be filled by any of the 6 numbers. Similarly 2nd, 3rd and 4th digits. Hence each digit will have 6 possibilities. Now the first digit can be filled by any of the six numbers. Therefore there are 6 possibilities
How many choices can be made for each digit at units place?
Since the number is supposed to be even, the digits at units place must either be 2 or 4 leaving us with 2 choices for the digit at units place. 1 choice for the digit at thousands place. Rest of the 2 digits can be any one from 1, 2, 4, 5 and 9 i.e. 5 choices each