B looks the most reasonable
I will be using the language C++. Given the problem specification, there are an large variety of solving the problem, ranging from simple addition, to more complicated bit testing and selection. But since the problem isn't exactly high performance or practical, I'll use simple addition. For a recursive function, you need to create a condition that will prevent further recursion, I'll use the condition of multiplying by 0. Also, you need to define what your recursion is.
To wit, consider the following math expression
f(m,k) = 0 if m = 0, otherwise f(m-1,k) + k
If you calculate f(0,k), you'll get 0 which is exactly what 0 * k is.
If you calculate f(1,k), you'll get 0 + k, which is exactly what 1 * k is.
So here's the function
int product(int m, int k)
{
if (m == 0) return 0;
return product(m-1,k) + k;
}
The factors of 55 are 1, 5, 11, and 55 .
The factors of 77 are 1, 7, 11, and 77 .
The factors that are common to both numbers are 1 and 11 .
The greatest one is <em>11</em> .
You asked for the "greatest", so you only get one of them. If there were
more than one greatest, then you'd want to know which greatest was
greater than the other greatest.
Answer:
add 99 to 6790
Step-by-step explanation:
6790 +99 = 6889 which is 83 squared
