Yes, if x is an integer divisible by 3, then x^2 is also divisible by 3. This is because for any integer x, x^2 will also be divisible by 3 if x is divisible by 3. This can be proven using the property that the square of any integer divisible by 3 will also be divisible by 3.
If an integer's square is divisible by four, then it is even.
a positive integer evenly divisible only by itself and one.
There are no integer palindromes between 1,000 and 9,000 that are divisible by 20.
Do the division! If you get an integer (i.e., no decimals), then it is divisible.
An integer is odd if and only if it is not divisible by two.
Yes, because it can be divisible by two to get an integer.
In mathematical logic, An integer A if divisible by 100 iff the last two digits are 0. "iff" stands for "if and only if".
10230000648 is evenly divisible by 4. If the last two digits of an integer are divisible by 4, the entire number is divisible by four.
How can the following definition be written correctly as a biconditional statement? An odd integer is an integer that is not divisible by two. (A+ answer) An integer is odd if and only if it is not divisible by two
Yes, if x is an integer divisible by 3, then x^2 is also divisible by 3. This is because for any integer x, x^2 will also be divisible by 3 if x is divisible by 3. This can be proven using the property that the square of any integer divisible by 3 will also be divisible by 3.
Suppose x = sqrt(3*a) where a is an integer that is not divisible by 3. then x2 = 3*a which is divisible by 3. but x is not even rational and so is not an integer and is certainly not divisible by 3.
Yes, but you wont get an integer. 120 and 135 are divisible to an integer.
If an integer's square is divisible by four, then it is even.
a positive integer evenly divisible only by itself and one.
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.
When you divide a number by 1, it remains the same. A number is considered divisible is the resulting answer is an integer with no remainder left over. Thus, if the starting number is an integer, it will be divisible by 1. 15345 is an integer, and thus is divisible by 1.