0
What else can I help you with?
What is Euclid's division lemma?
Given Positive Integers a and b there exists unique integers q and r satisfying a=bq+r; 0 lesser than or equal to r<b
Is a plus b divisible by c?
Sometimes. It depends on the values given to the variables.
Is the cardinality of an infinitely countable set the same as the rational numbers?
Yes. There is an injective function from rational numbers to positive rational numbers*. Every positive rational number can be written in lowest terms as a/b, so there is an injective function from positive rationals to pairs of positive integers. The function f(a,b) = a^2 + 2ab + b^2 + a + 3b maps maps every pair of positive integers (a,b) to a unique integer. So there is an injective function from rationals to integers. Since every integer is rational, the identity function is an injective function from integers to rationals. Then By the Cantor-Schroder-Bernstein theorem, there is a bijective function from rationals to integers, so the rationals are countably infinite. *This is left as an exercise for the reader.
What which of these numbers are divisible by 13 is it a. 52 or b. 9 c. 30 d. 27?
a.
What the LCM of 14 and 48?
For any two numbers a and b if a is divisible by b then LCM(a,b) is a.48 is divisible by 6, so LCM(48,6) is 48.
Are all negative rational numbers integers?
No, they are not because fractions can be negative also. fractions aren't integers
The variables a and b represent positive integers neither of which are divisible by 10 yet the product of a and b is 100000 Find the sum of a and b?
641
Both a and b are positive integers neitherof them is divisible by 10 and a times b equals 10000 What are a and b?
10000 = 16 x 625
If the product of a b and c is 0 and a and b are integers greater than 10 what must c equal?
if abc is 0 then at least one of the factors must be zero. since a and b are both nonzero, c must be zero.
If a and b are positive integers neither of which is divisible be 10 and if ab equals 10000 what is the sum of a plus b?
625*16=10000625+16=641
What is least positive integer that is divisible by each of the integers 1 through 7 inclusive?
what is least positive integer that is divisible by each of the integers 1 through 7 inclusive ? a 420 b 840 c 1260 d 2520 e 5o40
If a and b integers and a b then a b?
If a and b are integers, then a times b is an integer.
Is ab a multiple of 2 or 5 how do you know?
The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.It gets a lot more complicated in the latter case if they can be non-integers.The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.It gets a lot more complicated in the latter case if they can be non-integers.The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.It gets a lot more complicated in the latter case if they can be non-integers.The answer depends on whether ab represents a 2-digit number or, as in algebra, represents the product of a and b.If ab represents a 2 digit number, so that, in fact, it is equivalent to 10*a+b, then it is divisible by 2 if b = 0, 2, 4, 6 or 8 and divisible by 5 if b is 0 or 5. The value of a makes no difference.If ab represents the product of a and b and they are both integers, then ab is a multiple of 2 or 5 if at least one of a or b is a multiple or 2 or 5.It gets a lot more complicated in the latter case if they can be non-integers.
What are the factors of 312 using the divisibility rule?
312 is divisible by 2 b/c the 2 at the end of 312 is an even number312 is divisible by 3 b/c 3 + 1 + 2 = 6 which is divisible by 3312 is divisible by 4 b/c 12 is divisible by 4312 is not divisible by 5 b/c the one's place isn't 0 or 5312 is divisible by 6 b/c 312 is divisible by 2 and 3312 is not divisible by 7 b/c 2*7=14, 31-14=17 which is not divisible by 7312 is divisible by 8 b/c 312/8=39 (Explanation: A way to find out how 8 is divisible its that its first three digits [the ones, tens, and hundreds place] have to be divisible by 8)312 is not divisible by 9 b/c 3 + 1 + 2 = 6 which is not divisible by 9312 is not divisible by 10 b/c the ones place is not 0
When multiplying a nonzero fraction by its reciprocal the product is call?
When you multiply a nonzero fraction by its reciprocal you get 1. Take this example.Lets say you have a fraction called (a/b). The reciprocal of that fraction would be (b/a). If you multiply the two you'd get (a/b)*(b/a)=(ab/ab). Anything divided by itself is 1.
The exponential expression b0 equals for any nonzero base b?
1
If A is the set of all positive integers less than 300 that are divisible by 3 and B is the set of all prime numbers how many elements has the intersection of set A and set B?
One. The only multiple of 3 that is prime is 3 itself.
