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CP DML Q47

0. 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

  • 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.
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