All numbers divisible by 3 are NOT divisible by 9. As an example, 6, which is divisible by 3, is not divisible by 9. However, all numbers divisible by 9 are also divisible by 3 because 9 is divisible by 3.
Yes. All numbers rationally divisible by 4 are also rationally divisible by 2.All numbers that when divided by 4 result in answers that are integers will also give answers that are integers when divided by 2, and so forth.For example: Let x/4=p,x/2=2.(x/4)=>x/2=2p.from the above illustration,it is clear that all the multiples of 4 are the multiples of 2.hence every number which is divisible by 4 is divisible by 2.
Prime numbers are not divisible by any number other than 1 and themselves. They cannot be divisible by 6.
Yes 12 is divisible by 48. 12 is divisible by 48 because 12x4 is 48. Also if you divide 48 by 4 it would give you 12. It is that because 12 and 48 are composite numbers and even numbers.
In order for a number to be divisible by 4, it has to be even.As every multiple of 100 is divisible by 4, we only have to look at the rightmost two digits.If the number consisting of these digits is divisible by 4, then the original number is also divisible by 4.For example:353 is odd, therefore is not divisible by 4.1284 - We'll look at the rightmost digits giving us 84, which is divisible by 4, so 1284 is also divisible by 4.258 - The two rightmost digits give us 58, which is not divisible by 4. Hence 258 is also not divisible by 4.
Not to give a whole number. Only numbers which end in a 5 or a 0 are divisible by 5.
Any integer times 708 will give an answer that is divisible by 708. 708, 1416, 2124, 2832, ...
Square numbers can't be prime. They have too many factors.
24 1.2 x 105
The HCF of the given three numbers is 9 and because the digital sum of the given numbers equals 9 then the numbers are all divisible by 9
Any positive integer not including 0. Also known as "counting numbers." 1,2,3,4,5,6,7,8,9,10,11...
150 give or take a few
No, square numbers greater than 1 have more than two factors.
Ok look, it is divisible by 2, 5, and 10. :-)
1764 is divisible by 1 and 4. There are probably other divisors as well. 1764 = 2x2x3x3x7x7 so it is divisible by any product of these factors as well as 1. For example it is divisible by 2x2x3x7 to give 3x7 =21
a number or algebraic expression by which another is exactly divisible. 3 is a factor of 9
No! Not by a long shot. With larger numbers, you cannot tell whether an odd number is prime or composite just by the last digits. All prime numbers greater than 7 must end in one or these: 1, 3, 7 or 9. But not all numbers ending with 1, 3, 7 or 9 are prime.Extra stuff:Well for example 207 is divisible by 3 so the answer is no (207 is 69 times three). The simple test (ie to check for 3 as divisor or also called a factor) is to add all the individual numbers up (here it totals 9) and then if that number is divisible by three then you don't have a prime number. So 27 is also not a prime - it divides by three. (quick check again the total of the numbers is 9 ie divisible by three) and another example 57 can be divided by 3 to give 19 and the quick check is 5 and 7 totals 12 which is then divisible by three.ps don't forget you need to check for all the other numbers also! ie like 4, 5, 7, 9, 11, 13 ... etc
An odd number is a number that's not divisible by 2; it's every other number starting at 1 (for our purposes. It can also go into negatives but you don't need to know that yet.) Some odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, and so on, just keep adding 2.
It is divisible but not evenly divisible. It divides to give 163.2.
1, 2, 71, 142. Anything else does not give whole numbers. 142/4 is 35.5
The LCM of 7 and 9 is 63.
1,2,3,4 and 1 million.
As an example: 18000+18000 = 36000
Irrational numbers are decimal numbers that can't be expressed as fractions. An example is the square root of 2
There are many four digit numbers which are evenly divisible by 6. The smallest of these is 1,002. If we add 6 to this number then this will give us the next. 1,002 + 6 = 1,008. We can keep adding 6 to find all the remaining numbers if we wish.