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A, B and C aren't numbers, they're letters. Probably variables. Without knowing what numbers they represent, we can't calculate their LCM.
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∙ 8y ago
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∙ 8y agoSince variables can be any number, the LCM possibilities are infinite.
Wiki User
∙ 8y agoA variable can be any number. The LCM possibilities are infinite.
Wiki User
∙ 8y agoVariables can stand for any number. The LCM possibilities are infinite.
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∙ 8y agoWe can't answer that definitively without knowing the values of A and B.
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∙ 5y agoA common denominator is A*B*C. There is not enough information to determine whether or not it is the least such number: that will depend on whether or not the three numbers are co-prime.
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∙ 8y agoThe GCF is 1.
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∙ 9y agoABC
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What is the highest common multiple of B and C?
The greatest common multiple of any set of integers is infinite.
What is the greatest common factor C and B?
The greatest factors of A, B, and C, respectively, are the absolute values of A, B, and C. The greatest common factor of A, B, and C is 1.
What is the least common multiple of 8 C A?
LCM(8, C, A) = 8*C*A.
Least common multiple in C programming?
The following code for example is a solution (you could do it with less variables, but this is more readable):int GCD(int a, int b){int n, k, c;n = (a>b)?a:b;k = (a>b)?b:a;while (k){c = n%k;n=k;k=c;}return n;}
What number is a multiple of 36 a 200 b 108 c 76 d 48?
b 108
What is the highest common multiple of B and C?
The greatest common multiple of any set of integers is infinite.
What does LCM satand for?
L= lowest C= common M= multiple Example: The lowest common multiple of 5 and 2 is:..... (list the factors) 5= 5, 10, 15, 20, 25 2= 2, 4, 6, 8, 10 The lowest common multiple would be 10, since it appears on both of the sequences, AND its is the LOWEST number.
How can you find the Lowest common denominator of four integers?
lcm(a,b,c,d) = lcm(lcm(a,b,c),d) = lcm(lcm(a,b),lcm(c,d))
If A B and C are counting numbers and both A and B are multiples of C what can you say about A plus B?
A + B is also a multiple of C. ------------------------------------------- let k, m and n be integers. Then: A = nC as A is a multiple of C B = mC as B is a multiple of C → A + B = nC + mC = (n + m)C = kC where k = n + m kC is a multiple of C. Thus A + B is a multiple of C.
What is the least common multiple of 40 and 80 a 120 b 80 c 40?
It is b: 80
When A B and C are counting numbers and both A and B are multiples of C what can you say about A plus B?
If A and B are multiples of C, then A + B is also a multiple of C: If A is a multiple of C then A = mC for some integer m If B is a multiple of C, then B = nC for some integer n → A + B = mC + nC = (m + n)C = kC where k = m + n and is an integer → A + B is a multiple of C
Is 18 a least common multiple A 18 B 36 C 9 D 4?
C is this your homework??
Which of the follwing pairs have a least common multiplyer a is 4 and 6 b is 3 and 8 c is 2 and 12 d is 3 and 6?
a). The least common multiple of 4 and 6 is 12 . b). The least common multiple of 3 and 8 is 24 . c). The least common multiple of 2 and 12 is 12 . d). The least common multiple of 3 and 6 is 6 . Gosh, I guess they all have.
What are two real-world problems that can be solved using the GCM of two numbers?
There are none because there is no such thing as a Greatest Common Multiple (GCM). If {a, b, c, ... x} is any set of integers, then a*b*c*...*x is a common multiple. Then twice that number is also a common multiple and is greater. And then, twice THAT number is a common multiple and greater still. It is easy to show that this process can go on for ever and so there is no such thing as a GCM.
What is the least common multiple of C B and A?
To answer that, you'll need to have a numerical value for the letters.
What is the least common multiple of decimals and fractions in C?
To calculate the least common multiple (lcm) of decimals (integers) and fractions you first need to calculate the greatest common divisor (gcd) of two integers: int gcd (int a, int b) { int c; while (a != 0) { c = a; a = b % a; b = c; } return b; } With this function in place, we can calculate the lcm of two integers: int lcm (int a, int b) { return a / gcd (a, b) * b; } And with this function in place we can calculate the lcm of two fractions (a/b and c/d): int lcm_fraction (int a, int b, int c, int d) { return lcm (a, c) / gcd (b, d); }
What is pseudo code for LCD of two numbers?
The LCD (least common denominator) is better known more generally as the LCM (least common multiple). The LCM of any two integers, typically denoted LCM (a, b), is the lowest positive integer that is evenly divisible by both integers. That is, LCM (a, b) = c such that c/a = c/b. Division by zero is undefined thus it stands to reason that neither a nor b can be zero. However, many people regard LCM (a, 0) = a to be valid even though it is technically undefined, as is LCM (0, 0). Note that either a or b may be negative but LCM (a, b) must always be positive.The product of a and b is obviously a common multiple of a and b. However, it is not necessarily the lowest common multiple. For instance, 20 is the product of 2 and 10 and is therefore a common multiple of 2 and 10. However, the lowest common multiple of 2 and 10 is 10 because 10/2 = 5 and 10/10 = 1. From this we can surmise that LCM (a, b) can be no greater than the product of a and b and it cannot be any less than the largest of a and b.There are several ways to calculate the LCM of two integers, however one of the simplest is by reduction by the greatest common divisor (GCD). That is, GCD (a, b) = c such that a and b are evenly divisible by c. The GCD and LCM are similar types of problem, however the GCD is much easier to calculate and can be implemented efficiently using Euclid's algorithm.In pseudo-code, we can write the GCD (Euclid) algorithm as follows:algorithm GCD (a, b) is:while (a b) doif a > b then a := a - b else b := b - aend whilereturn aFrom this we can now write the LCM algorithm in terms of the GCD algorithm:algorithm LCM (a, b) is: return a / GCD (a, b) * bNote that a * b / GCD (a, b) produces the same result, however it is more efficient to perform the division before the multiplication. This is because a / GCD (a, b) is guaranteed to be an integer that is less than or equal to a and is therefore guaranteed not to overflow. Multiplying that integer by b may still overflow, however the chances of overflow are reduced compared to that of multiplying a and b prior to the division.
