# C Programming MCQ - Decision Making & Looping

46:

Consider the program fragment

j = 2;
while ((i % j) ! = 0)
j = j + 1;
if (j < i) printf ("%d", j);

If i >= 2, then the value of j, will be printed only if

 A. i is prime B. j does not divide i C. j is odd D. i is nor prime Answer Report Discuss Option: D Explanation : Click on Discuss to view users comments. Write your comments here:
47:

Consider the following program fragment

d = 0;
for(i = 1; i < 31; ++i)
for(j = 1; j < 31; ++j)
for(k = 1; k < 31; ++k)
if(((i + j + k) % 3)  == 0)
d = d + 1;
printf("%d", d );

The output will be

 A. 9030 B. 27000 C. 3000 D. none of the above Answer Report Discuss Option: A Explanation : a + b +c  3 will be 0 if a + b+c is a multiple of 3. This will happen in one of the following ways. All three - a, b, and c are multiples of 3. This can only happen if a, b, and c take one of the 10 values. - 3 , 6, 9 , ..... , 30, independent of one another. So, there are 10 x 10 x 10 = 1000 ways this can happen. Another possibility is that a, b, and c all leave a remainder 1 so that a + b + c is evenly divisible by 3. Considering all the different possibilities and adding. we get 9000. That will be the integer that gets printed. Click on Discuss to view users comments. Write your comments here: