what os the set of all integers divisible by 5
All numbers divisible by 10 end in zero and all numbers that end in 0 are divisible by 10. ( I am assuming base 10 notation, of course)
No. For example, 21 has factors 1, 3, 7 and 21 making it composite, but is odd (nondivisible by 2).
All even numbers have 2 as a factor because they are all divisible by 2. Half of the numbers are odd and half of them are even. It makes sense that half of them would be divisible by 2.Because of the definition of an even number (an even number is an integer that can be divided by 2 evenly; with no remainer).
Not all even numbers are divisible by 6. These numbers are not evenly divisible by 6: Any number smaller than 6. Any number not divisible by 3. If a number is divisible by both 2 and 3, it is divisible by 6.
The number 2 is the only even prime number because any (actually, every) other even number can be divided by 2. And they can't be prime if they are divisible by an integer (a counting number) other than 1 and themselves. Said another way, all even numbers greater than 2 are divisible by 2. As they are divisible by 2, they cannot be prime.
All numbers of the form 6+12x are, for all integer x.
Not sure about the set builder notation, but Q = {0}, the set consisting only of the number 0.
All numbers divisible by 10 end in zero and all numbers that end in 0 are divisible by 10. ( I am assuming base 10 notation, of course)
Yes. 117/9=13. An easy test for divisibility by nine (for an integer of any length) is to add all of the digits. If the sum is nine or a number divisible by nine, then the integer is divisible by nine. In this case, 1+1+7=9, so 117 is divisible by nine. (Be careful, this test fo divisibility only works generally for divisibility by three -- i.e., an integer is divisible by three if and only if the sum of its digits equals three or a multiiple of three-- and for nine.) To find if an integer is divisible by 4, you can check if the last two digits are. If they are, it is. To check if an integer is divisible by 6, you must make sure that the integer is divisible by both 3 and 2. If you want to check if an integer is divisible by 2, just make sure it's even. Any integer divisible by 10 will end in zero. Any integer divisible by 5 will end in either 5 or 0. This is an important part of pre-algebra, as well as algebra.
There are infinitely many numbers that are divisible by 202. They are all numbers of the form 202*k where k is some integer.
It is divisibility by 3 and divisibility by 5.Divisibility by 3: the digital root of an integer is obtained by adding together all the digits in the integer, with the process repeated if required. If the final result is 3, 6 or 9, then the integer is divisible by 3.Divisibility by 5: the integer ends in 0 or 5.
2520
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Just carry out the division, and see if you get an integer! Also, all numbers that are divisible by 5 end with 0 or with 5.
All numbers of the form 855*k where k is any integer.
I get 2,520. I could be wrong.
In normalized scientific notation all numbers are written in the form a x 10^b (a times ten raised to the power of b) where a is a nonzero single-digit integer and b is an integer.