# MULTHREE Question code giving wrong answer

My logic - since every digit from 3rd digit onwards is equal to the sum of the previous digits mod 10.
and a number is divisible by 3 if its sum of the digit is divisible by 3.

The sum of all k digits of a number is given by

s = (2^(k-2)-1)X((d0+d1)%10) + (d0+d1)

Therefore,
s%3 = ((((2^(k-2)%3-1)%3)X(((d0+d1)%10)%3))%3 + (d0+d1)%3)%3

my code - https://ideone.com/yuVmkF

Thanks.

what if (d0+d1)%5==0 2 Likes

you must have noted something that next number in sequence starting from 4th one is
(2*unitdigit of previous number)%10
so if we remove corner cases like (d0+d1)%5==0 which will return false except for 15 we can say that
a repeated sequence of 2,4,8,6 will occur so we can find sum till the pattern’s first occurance then we simply add to the sum ((remaining length)/4)*20 and add remaining few numbers of repeated sequence on the basis of len=(remaining length)%4
if len =1 we add 2
if len =2 we add 6
if len =3 we add 14
i hope you get the idea i want to tell here

1 Like

Yea I got this idea thanks bro. I also realized my formula which i wrote for s is not correct.