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