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The answer is 135.

15 times 9 equals 135

27 times 5 equals 135

Q: What is a 3 digit number divisble by 5 and 9?

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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.

The only way to get an even number is to put the "4" to the right. To get the largest 3-digit number, you must put the largest digit (5) to the left, so that leaves us with 534.

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.

Related questions

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.