A number of 50 digits has all its digits equal to 1 except the 26th digit. If the number is divisible by 13, then find the digit in the 26th place.
(In reply to
Puzzle Answer by K Sengupta)
We know that:
10^3 = -1(mod13)
=> 10^6=1 (mod 13)
=>10^(6t) =1 (mod 13)
Substituting t=4, we have:
10^24 =1(mod 13) .........(i)
Again, 10^(6t)=1(mod 13)
=> 10^(6t)-1 = 0(mod 13)
=>{10(6t)-1}/9 =0(mod 13)
=> R(6t)=1, where R(x) is the xth repunit.
Substituting t=4, we have:
R(24)= 0(mod 13)........(ii)
Let the 50-digit number be denoted by N, and the 26th digit of N be denoted by d.
Then, we must have:
N= R(24)*10^26 + (10+d)*10^24 + R(24)
Now, from (i), 10^24 =1(mod 13)
=> 10^26=9(mod 13)
Then, N (mod 13)= {0*9+ (10+d)*1+0}
=10+d
Since N is divisible by 13 and d is a base ten digit, it follows that:
10+d=13, giving: d=3
Consequently, the required digit in the 26th place is 3.
Explanation for Puzzle Answer
