what os the set of all integers divisible by 5
Oh, dude, you want all the numbers divisible by 10? Alright, so it's like 10, 20, 30, 40, and so on. Basically, any number that ends with a zero when you divide it by 10. But hey, who's really keeping track of all those numbers anyway, am I right?
You test by := All EVEN Numbers are divisible by '2' All numbers ending in '5' or ''0' are divisible by '5' All numbers whose digits add to '9' are divisible by '9'. e.g. 234 2+3+4 = 9 All numbers whose form is ' m(m+n)n ' e.g. 132 are divisible by '11'. This leaves only '3' and '7' to test by actual division.
No. For example, 21 has factors 1, 3, 7 and 21 making it composite, but is odd (nondivisible by 2).
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.
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).
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.
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.
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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
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.
I get 2,520. I could be wrong.
All numbers of the form 855*k where k is any integer.
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.
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.