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;}
#include<stdio.h> #include<conio.h> void main() { int x,y,z,i; clrscr(); printf("Enter Two Numbers = "); scanf("d%d",&x,&y); for(i=1;i<=x*y*z;i++) { if(i%x==0&&i%y==0) { printf("LCM is %d\n",i); break; } } getch(); }
If the rational expressions have large exponent, then you need to factor out this way: (a + b)ⁿ = (a + b)(a + b)...(a + b) [So there are n "(a + b)" factors.] Here are the examples... (a + b)³ = (a + b)(a + b)(a + b) (a + b)4 = (a + b)(a + b)(a + b)(a + b)
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It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n where nCr = n!/[r!*(n-r)!]
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The power of a quotient is the quotient of the power! (a/b)^n = (a^n) / (b^n) where a/b is the quotient and n is the power.
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In algebra this is a method of solving equations for a variable. It stands to reason that if A=B then A+1=B+1. In fact it is true that if N is a number then A+N=B+N A-N=B-N A x N = B x N A/N = B/N We can do anything we want as long as we do the same thing to both sides.
If a divides b then a is a factor of b and b is a multiple of a.Either of them could be the denominator. In a/b, b is the denominator while in b/a, a is the denominator.------------------------------------------------------------------------------------------------------------------If a divides b then b=n*a for some number n. Thus n=b/a
b is equal to 5
The number of links are: L=b-(n-1)=b-n+1 Where b=Number of branches n=number of nodes
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