The formal answer is that 18 goes into the number without remainder.
A simple test for divisibility by 18 is as follows:
(a) the number must be even.
(b) the sum of all the digits of the number must be divisible by 9.
(a) ensures the number is divisible by 2 and (b) that it is divisible by 9. So, together they ensure divisibility by 2*9 = 18.
Note that (b) only works with 3 and 9.
The smallest number that is divisible by 18 and 48 is 144.
If the number is even and the sum of its digits is divisible by nine then the number is divisible by 18.
Any even number divisible by 9, if the sum of the digits is divisible by 9. If it's an even number divisible by 9, it's divisible by 18.
18
Fifty-two is divisible by 18, but it does not end up a who,e number.
What we want to find is the number in that list that is not divisible by 4. This number would be 18, your answer.
A number is divisible by 9 if a whole number of 9s can go into it and leave no remainder.
18
No; for example, 6 is not divisible by 12, nor is 18.
First we have to find a prime number by which 18 is divisible. 18 is divisible by both 2 and 3. However, we can start with any number. 18/3 = 6, now we have to look for a prime number by which 6 is divisible. 6/3 = 2, 2 is divisible by itself. So, prime factorization of 18 = 2x3x3.
The LCM of both numbers is 18
Any number that is a multiple of 18 (that is, any number that is a whole number multiplied by 18)So:1x18=182x18=363x18=544x18=725x18=90etc...Because there is no limit to the size of the number we are multiplying by 18 (ie. 1 000 000 x 18 = 18 000 000) there are infinite numbers divisible by 18. However, not all numbers are divisible by 18, only those you can get by multiplying a whole number by 18.