11
An odd number is written 2n+1 with n any number (2n1+1)+(2n2+1)+...+(2n9+1) = 2(n1+n2+...+n9)+9 = 50 is not possible as 2(n1+n2+...+n9) = 41 is not possible with integers
Even numbers : they can be written 2n 2n1+2n2+....+2nm = 2(n1+n2+....+nm) so it's always an even number 2n1x 2n2x....x2nm = 2(n1+n2+....+nm) which is always an even number Odd numbers: they canbe written 2n+1 (2n1+1) + (2n2+1) + ....+ (2nm +1) = 2(n1+n2+....+nm) + m which is even if m is even and odd if m is odd, so it depend of the number of terms (2n1+1) x (2n2+1) x ....x (2nm +1) is always odd cause the last term of the expansion is always +1 and other terms have at least 2 as factor
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
the value of the exponent n1
11
No, as 2 is a prime number, the product of 2 prime numbers can be even But all other prime numbers (except 2) are odd then their product is odd (2 x n1+1) x (2 x n2+1) = 2 x (2 x n1 x n2 + n1+ n2) +1
The formula for calculating the frequency of a Lissajous figure is f = (n2-n1) * (f1-f2) / (2 * (n1+n2)), where f is the frequency of the Lissajous figure, n1 and n2 are the integer ratios of the frequencies f1 and f2 on the x and y axes respectively.
An odd number is written 2n+1 with n any number (2n1+1)+(2n2+1)+...+(2n9+1) = 2(n1+n2+...+n9)+9 = 50 is not possible as 2(n1+n2+...+n9) = 41 is not possible with integers
Even numbers : they can be written 2n 2n1+2n2+....+2nm = 2(n1+n2+....+nm) so it's always an even number 2n1x 2n2x....x2nm = 2(n1+n2+....+nm) which is always an even number Odd numbers: they canbe written 2n+1 (2n1+1) + (2n2+1) + ....+ (2nm +1) = 2(n1+n2+....+nm) + m which is even if m is even and odd if m is odd, so it depend of the number of terms (2n1+1) x (2n2+1) x ....x (2nm +1) is always odd cause the last term of the expansion is always +1 and other terms have at least 2 as factor
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2-1) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
the value of the exponent n1
the value of the exponent n1
void main() { int i; float n1,n2; abc: printf("Enter two nos "); scanf("%f%f",&n1,&n2); printf("\n %f + %f = %f " ,n1,n2,n1+n2); printf("\n %f - %f = %f " ,n1,n2,n1-n2); printf("\n %f x %f = %f " ,n1,n2,n1*n2); printf("\n %f / %f = %f " ,n1,n2,n1/n2); printf("\npress 5 to make another calculation"); scanf("%d",&i); if (i==5) goto abc; }
The sum of the first 10 positive integers, using the formula N1 + (N1 + 1) + ... + N2 = N2 * (N2 + 1) / 2 - (N1 - 1) * N1 / 2 is: 55