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 |
Option: D Explanation : Click on Discuss to view users comments. |
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 |
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. |