Question

What is the remainder when 77,777... up to 56 digits is divided by 19?

Answer 1

Answer:

Let, N= 77…(37times)

Note: If a natural number X = 19m + n where m,n belongs to natural numbers then

X mod 19 = n mod 19

WKT

N= 7(10^37–1)/9

=> 9N=7(10^37–1)

now,

N mod 19= (9N mod 171)/9 [ 9 divides 9N]

we have,

9N = 7[10{(1026–26)^12}-1]

therefore 9N mod 171= 7[10{26^12}-1] mod 171

[ removing terms which are divisible by 171 in the binomial expansion of (1026–26)^12]

=> 26^12=(684–8)^6

=>9N mod 171 = 7[10{8^6}-1] mod 171

=> 8^6=(513–1)^2

=> 9N mod 171 = 7[10{1}-1] mod 171 =63

therefore,

N mod 19 = 63/9 = 7

Hence the answer is 7