The LCM of 2, 3 and 5 is 30. The LCM of any three relatively prime numbers a, b and c is a*b*c, and here, 2 is a, 3 is b and 5 is c. Since 2, 3 and 5 are relatively prime, the LCM is 2*3*5=30.
question: find the percentage A is of B or, 7 (A) is what percente of 35 (B) (B-A)/B x 100 =C. then subtract C from 100 B-A 35-7=28 /B 28/35=0.8 x 100 =C 0.8 x 100=80 - C 100-80=20. so, 7 is 20 percent of 35
To find the least common multiple (LCM) of 2b^2 and 6c^3, we first need to factorize both terms. The prime factorization of 2b^2 is 2 * b * b, and the prime factorization of 6c^3 is 2 * 3 * c * c * c. To find the LCM, we take the highest power of each prime factor that appears in either term, which gives us 2 * 3 * b^2 * c^3 = 6b^2c^3. Therefore, the LCM of 2b^2 and 6c^3 is 6b^2c^3.
88 b/c 4x22=88 8x11=88 and88x1=88
The property states that for all real numbers a, b, and c, their product is always the same, regardless of their grouping: (a . b) . c = a . (b . c) Example: (6 . 7) . 8 = 6 . (7 . 8) The associative property also applies to complex numbers. Also, as a consequence of the associative property, (a . b) . c and a . (b . c) can both be written as a . b . c without ambiguity.
lcm(a,b,c,d) = lcm(lcm(a,b,c),d) = lcm(lcm(a,b),lcm(c,d))
That depends on the values of A, B and C.
If C is co-prime with 6 and with 7, then LCM(6, 7, C) = 42*C If not, the answer depends on the value of C.
The LCM of a^2b and b^2c is a^2b^2c
If a, b & c are prime numbers then their LCM is equal to their product i.e. a x b x c.Here, 5, 7 and 13 are prime numbers.Therefore, LCM(7,5,13) = 7 x 5 x 13 = 455.
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); }
The LCM of 2, 3 and 5 is 30. The LCM of any three relatively prime numbers a, b and c is a*b*c, and here, 2 is a, 3 is b and 5 is c. Since 2, 3 and 5 are relatively prime, the LCM is 2*3*5=30.
For this you will need a couple of helper algorithms. The first is the GCD (greatest common divisor) which is expressed as follows:procedure GCD (a, b) isinput: natural numbers a and bwhile ab doif a>blet a be a-belselet b be b-aend ifend whilereturn aThe second algorithm is the LCM (least common multiple) of two numbers:procedure LCM (a, b) isinput: natural numbers a and b return (a*b) / GCD (a, b)Now that you can calculate the GCD and LCM of any two natural numbers, you can calculate the LCM of any three natural numbers as follows:procedure LCM3 (a, b, c) isinput: natural numbers a, b and c return LCM (LCM (a, b), c)Note that the LCM of three numbers first calculates the LCM of two of those numbers (a and b) and then calculates the LCM of that result along with the third number (c). That is, if the three numbers were 8, 9 and 21, the LCM of 8 and 9 is 72 and the LCM of 72 and 21 is 504. Thus the LCM of 8, 9 and 21 is 504.
#include<stdio.h> main() { int a,b,i,lcm,gcf; printf("\n Enter two numbers"); scanf("%d%d",&a,&b); for(i=0;i<=a;i++) { if((b%i==0)&&(a%i==0)) { gcf=i; } } lcm=a*b/gcf; printf("\n GCF is %d and LCM is %d",gcf,lcm); }
1008
C. 12b3
Well, isn't that just a happy little math problem! If A is less than B and B plus C equals 10, then it must be true that A plus C is less than 10. Just remember, in the world of numbers, everything adds up beautifully in the end.