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Question Number 877 by sagarwal last updated on 07/Apr/15

$$\mathrm{An}8\times 8\mathrm{checkerboard}\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{coloured}$$$$\mathrm{with}6\mathrm{colors}\mathrm{in}\mathrm{a}\mathrm{way}\mathrm{that}\mathrm{no}2\mathrm{adjacent}$$$$\mathrm{cells}\left(\mathrm{sharing}\mathrm{side}\right)\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{color}.\mathrm{In}\mathrm{how}$$$$\mathrm{many}\mathrm{ways}\mathrm{can}\mathrm{it}\mathrm{be}\mathrm{done}.$$

Commented by 123456 last updated on 12/Apr/15

$$\left[\begin{array}{cccccccc}6& 5& 5& 5& 5& 5& 5& 5\\ 5& 4& 4& 4& 4& 4& 4& 4\\ 5& 4& 4& 4& 4& 4& 4& 4\\ 5& 4& 4& 4& 4& 4& 4& 4\\ 5& 4& 4& 4& 4& 4& 4& 4\\ 5& 4& 4& 4& 4& 4& 4& 4\\ 5& 4& 4& 4& 4& 4& 4& 4\\ 5& 4& 4& 4& 4& 4& 4& 4\end{array}\right]$$$$k=6\times 5^{7+7}\times 4^{7\times 7}$$$$k=6\times 5^{14}\times 4^{49}$$$$k=\left(2\times 3\right)\times 5^{14}\times {\left(2^2\right)}^{49}$$$$k=2\times 3\times 5^{14}\times 2^{2\times 49}$$$$k=2\times 3\times 5^{14}\times 2^{98}$$$$k=2^{98+1}\times 3\times 5^{14}$$$$k=2^{99}\times 3\times 5^{14}$$$$\log k=\log \left(2^{99}\times 3\times 5^{14}\right)$$$$\log k=99\log 2+\log 3+14\log 5$$$$\log k=\frac{99\ln 2+\ln 3+14\ln 5}{\ln 10}$$$$\log k\approx 40,06$$$$40⪅\log k<41$$$${10}^{40}⪅k<{10}^{41}$$

Commented by prakash jain last updated on 13/Apr/15

$$\mathrm{If}\mathrm{we}\mathrm{have}\mathrm{only}2\mathrm{color}\mathrm{the}\mathrm{grid}\mathrm{approach}$$$$\mathrm{fills}\mathrm{with}0\mathrm{s}.\mathrm{I}\mathrm{think}\mathrm{what}\mathrm{is}\mathrm{missing}\mathrm{are}$$$$\mathrm{special}\mathrm{cases}\mathrm{when}\mathrm{all}\mathrm{adjacent}\mathrm{cells}\mathrm{are}$$$$\mathrm{filled}\mathrm{with}\mathrm{same}\mathrm{color}\mathrm{as}\mathrm{in}2-\mathrm{color}\mathrm{cases}.$$

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