Question

consider a sequence of 500 zeroes it is decided to modify the sequence by the following steps
in step 1,to every position in the sequence we add 1
in step 2 , to every even position in the sequence we add 2
in step 3 , to every position which is a multiple of 3 we add 3
this continous upto 500 th step after the 500 th step what will be the value in 120 th position

Answer 1

So let me explain you in a detailed way

Your first aim in such problems must be to find the multiples of 120

Multiple of 120 - 1,2,3,4,5,6,8,10,12,15,20,30,40,60,120

At stating -         0,0,0,0,0,0,0,0,0 ,0  ,0,   0,   0,  0,  0

Now performing the step1 - add 1

1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

Now performing step 2 - Add 2 on even

1, 1+2=3 ,1 ,1+2=3, 1, 1+2=3, 1+2=3,  1+2=3, 1+2=3 , 1 ,  1+2=3 ,  1+2=3,  1+2=3,  1+2=3 ,  1+2=3

Now performing step 3 - Add 3 on multiple of 3

1 , 3 , 1+3=4 , 3 , 1+3=4 , 3 , 1 , 3+3=6 , 3 , 3 , 3+3=6 , 1+3=4 , 3 , 3+3=6 , 3 , 3+3=6 , 3+3= 6

So we get 6 at 120th position.