The sum will be a positive integer.It will be at least as large as double the smaller integer and at most as large as double the larger integer.
some are but most are not. There are actually an infinite number that are, 13, 26, 39, 52, etc. are all divisible by 13.
There is no limit to the number of obtuse angles a figure can have. A regular n-gon has n obtuse angles where n is any positive integer greater than 4.
It most certainly can.
No because natural numbers are positive whole numbers greater than 0
840 with 32 factors
The smallest positive integer is 1. 1 is the multiplicative identity; ie anything times 1 is itself. The greatest negative integer is the most positive negative integer which is -1. Therefore the product of the greatest negative integer and the smallest positive integer is the greatest negative integer which is -1.
The sum will be a positive integer.It will be at least as large as double the smaller integer and at most as large as double the larger integer.
That is a list of the proper divisors of 588. Most definitions of proper factors do not include the number 1.
It might seems like it, but actually no. Proof: sqrt(0) = 0 (0 is an integer, not a irrational number) sqrt(1) = 1 (1 is an integer, not irrational) sqrt(2) = irrational sqrt(3) = irrational sqrt(4) = 2 (integer) As you can see, there are more than 1 square root of a positive integer that yields an integer, not a irrational. While most of the sqrts give irrational numbers as answers, perfect squares will always give you an integer result. Note: 0 is not a positive integer. 0 is neither positive nor negative.
It is not - so the question is seriously flawed. If you stated that -1 was the greatest (most positive) negative integer you would be correct. However, numbers are infinite so you cannot state a greatest integer. Integers are both positive and negative numbers, but have to be whole numbers
It doesn't. 48 has the most factors of positive numbers under 50.
Sometimes the number 1 is included in the proper factors, and sometimes it is not. In the list of factors I have included the number 1. The following numbers have 10 or 11 proper factors, depending on whether 1 is allowed as a proper factor. The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30. The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36. The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42. The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45. The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48.
Hi... Every integer can be expressed as the product of prime numbers (and these primes are it's factors). Since we can multiply any integer by 2 to create a larger integer which can also be expressed as the product of primes, and this number has more prime factors than the last, we can always get a bigger number with more prime factors. Therefore, there is no definable number with the most primes (much like there is no largest number)!
I suppose you could call them improper factors, but the term most often used for factors that aren't proper is "trivial."
some are but most are not. There are actually an infinite number that are, 13, 26, 39, 52, etc. are all divisible by 13.
It is most commonly called an unsigned integer, but some programming languages have other terms for it.