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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.
Program to generate a pattern a aba abcba?
#include<iostream.h> void main() { cout<<' '<<' '<<"a"<<'\n'; cout<<' '<<"a"<<"b"<<"a"<<'\n'; cout<<'a'<<'b'<<'c'<<'b'<<'a'<<"\n"; }
C program to print all combinations of a 4-digit number?
#include<stdio.h> int main() { int a,b,c,d; for(a=1; a<5; a++) { for(b=1; b<5; b++) { for(c=1; c<5; c++) { for(d=1; d<5; d++) { if(!(a==b a==c a==d b==c b==d c==d)) printf("dd\n",a,b,c,d); } } } } return 0; }
How do you factor n2 plus 7n - 44?
To factor a trinomial (three-term expression) of the form n2+bn+c, find the two numbers h and k that are factors of c and add up to b. Then, write those numbers in this template: (n+h)(n+k) For the trinomial n2+7n-44, c = -44 and b = 7. The two numbers that are factors of -44 and add up to 7 are 11 and -4. So, the factored form would be (n+11)(n-4).
How do you make a pythagorean triple?
Use Euclid's formula. Take any two integers m and n that are coprime and let m > n (if not, just swap them around). Then a = m2 - n2 b = 2mn and c = m2 + n2 form a primitive Pythogorean triple. And since (a, b, c) is a Pythagorean triple then so is (k*a, k*b, k*c) for any integer k, except that the latter is not primitive.
