Made Easy test series. Doubt: My output for this question is => 1,2 and this isn't present in any of the options. Can someone please help me understand this, as per my logic, I am terminating the loop at i= 4 because of short circuit due to OR operation. Please let me know if this approach is correct or I am missing out on something.

Question

Answer 1

i value can't be 4.

In question, we can see i is incremented as i=i*(i+3)/2.

The condition is

!(  ( (i+1) mod 5==0 ) || ( (i+7) mod 6==1 ) || ( (i-27)mod 7 ==2 ) )

The iteration of the above loop is explained using the table below.

Initially, count value is 0 and each time when condition is true, then c is incremented as c++.

When condition will be true, print value = i mod 17.

When condition will be true, then i is incremented as i=i*(i+3)/2.

 

i value	condition	print value	incremented i	count
1	True	1mod17=1	1(1+3)/2=2	1
2	True	2mod17=2	2(2+3)/2=5	2
5	True	5mod17=5	5(5+3)/2=20	3
20	True	20 mod 17=3	20(20+3)/2=230	4
230	True	230 mod 17=9	230(230+3)/2= 26795	5
26795	True	26795 mod 17=3	26795(26795+3)/2=359026205	6
359026205	True	359026205 mod 17=9	359026205(359026208+3)/2=64449908476890320	7
64449908476890320

 

So till now it is okay, but it will take huge calculations further. So we cant say which option is correct.

So see the logic of this code but don't worry that how to do this in exam,because in GATE they are not going to give this type of “bakwasss” questions.

 

Comments

This indeed helped. Thanks a lot, Prajjwal!

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How condition is true at i = 1?

First two sub-conditions are FALSE but (1-27) mod 7 == 2 is TRUE so entire inner condition becomes TRUE as it is joined by OR and outside of that we have NOT so !TRUE = FALSE. So loop should not execute.

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I don't know why they are framing such type of question.