For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.
13213
As there are 5 different digits, allowing for repetition of a digit, the first digit can be any of the 5, and for each of these, the second can be any of the 5, and so for the third, making: 5 x 5 x 5 = 125 different 3-digit numbers.
To find the largest three-digit even number using the digits 3, 4, and 5, you need to arrange these digits in descending order to maximize the number, ensuring that the last digit is even. The largest possible even number is 542
This is a question of permutations; the answer is equal to the factorial of 5 (number of digits) divided by the factorial of 3 (number used in each selection), written 5! / 3!. This equals 120 / 6, or 20 ways.
040
Any number is divisble by all of those. Did you mean what is a NATURAL number divisible by those? That I am too lazy to find out.
yes 552 is divisible by 6 and the result is 92... to find out if a number is divisible by 6 then that number must be divisible by 2 and 3.. 1.If a number is divisble by three then the last digit must be divisible by 2 (so that the last digit must be 0,2,4,6,8) in this number the last digit is 2 so that it is divisible by 2 2.If a number is divisble by 3 then the sum of its digit must be divisible by 3. In this case the sum of the digits of 552 is 5+5+2= 12 and 12 is divisible by 3.... If this two rules fit in then that number is divisible by 6
30
3 and 5 are divisable by 195
Any number that ends with 5 or 0 except 0 itself is divisble by 5
2520
yes for a number to be divisble by 3 you have to add the numbers together and see if you can divide it by 3 evenly 1+5+3= 9 9 can be divided by 3 evenly so that makes the number, 153 divisble by 3
All those that end with 0 or 5 are divisible by 5. That would be 1000, 1005, 1010, ... 9990, 9995. There are 900 such numbers.
There are 5 numbers which can make the 3 digit numbers in this example. Therefore each digit in the 3 digit number has 5 choices of which number can be placed there. Therefore number of 3 digit numbers = 5 x 5 x 5 = 125
yes it is because 90 has a 0 at the end if the end of the number is 0 or 5 it is divisble by 5
They are divisble by 5 because every number at the end if it ends in 0 or 5 then its divisble by 5.