Question

prove that (10²⁵-7) is divisible by 3​

Answer 1

Answer:

${10}^{25} = (1 + 9) {}^{25} = \sum_{r = 0} ^{25} \binom{25}{r} {9}^{r} \\ \\ = \binom{25}{0} {9}^{0} + \binom{25}{1} {9}^{1} + \binom{25}{2} {9}^{2} + ... \\ \\ = 1 + 9 \bigg \{ \binom{25}{1} + \binom{25}{2} {9}^{1} + \binom{25}{3} {9}^{2} + ... \binom{25}{25} {9}^{24} \bigg \} \\ \\ = 1 + 9k \\ \\ where \: \: k \: is \: an \: integer \\ \\ now \\ \\ {10}^{25} - 7 \\ = 9k + 1 - 7 \\ = 9k - 6 \\ = 3(k - 2) \\ = 3m \\where \\ m = k - 2 \: is \: integer \: as \: well$

So it's divisible by 3