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Question Number 154672 by talminator2856791 last updated on 20/Sep/21
how many positive x≤10 000 integers are such that 2^x −x^2 is divisible by 7?
$$\: \\ $$$$\:\mathrm{how}\:\mathrm{many}\:\mathrm{positive}\:{x}\leqslant\mathrm{10}\:\mathrm{000}\:\mathrm{integers}\:\mathrm{are}\:\: \\ $$$$\:\mathrm{such}\:\mathrm{that}\:\mathrm{2}^{{x}} −{x}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}? \\ $$$$\: \\ $$
Answered by MJS_new last updated on 20/Sep/21
2^(3n) =7k+1 2^(3n+1) =7k+2 2^(3n+2) =7k+4 (7n)^2 =7k+0 (7n+1)^2 =7k+1 (7n+2)^2 =7k+4 (7n+3)^2 =7k+2 (7n+4)^2 =7k+2 (7n+5)^2 =7k+4 (7n+6)^2 =7k+1 ⇒ n=0, 1, 2, 3,...: (2^n −n^2 )/7=7k+ 1 1 0 6 0 0 0 2 3 4 0 2 4 1 4 0 5 2 6 5 3 1 1 0 6... ⇒ remainders are 0 for n=21k+{2, 4, 5, 6, 10, 15} 21×476+5=10001 ⇒ ⇒ answer is 475×6+2=2852
$$\mathrm{2}^{\mathrm{3}{n}} =\mathrm{7}{k}+\mathrm{1} \\ $$$$\mathrm{2}^{\mathrm{3}{n}+\mathrm{1}} =\mathrm{7}{k}+\mathrm{2} \\ $$$$\mathrm{2}^{\mathrm{3}{n}+\mathrm{2}} =\mathrm{7}{k}+\mathrm{4} \\ $$$$\left(\mathrm{7}{n}\right)^{\mathrm{2}} =\mathrm{7}{k}+\mathrm{0} \\ $$$$\left(\mathrm{7}{n}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{7}{k}+\mathrm{1} \\ $$$$\left(\mathrm{7}{n}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{7}{k}+\mathrm{4} \\ $$$$\left(\mathrm{7}{n}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{7}{k}+\mathrm{2} \\ $$$$\left(\mathrm{7}{n}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{7}{k}+\mathrm{2} \\ $$$$\left(\mathrm{7}{n}+\mathrm{5}\right)^{\mathrm{2}} =\mathrm{7}{k}+\mathrm{4} \\ $$$$\left(\mathrm{7}{n}+\mathrm{6}\right)^{\mathrm{2}} =\mathrm{7}{k}+\mathrm{1} \\ $$$$\Rightarrow \\ $$$${n}=\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},…:\:\left(\mathrm{2}^{{n}} −{n}^{\mathrm{2}} \right)/\mathrm{7}=\mathrm{7}{k}+ \\ $$$$\mathrm{1}\:\mathrm{1}\:\mathrm{0}\:\mathrm{6}\:\mathrm{0}\:\mathrm{0}\:\mathrm{0}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{0}\:\mathrm{2}\:\mathrm{4}\:\mathrm{1}\:\mathrm{4}\:\mathrm{0}\:\mathrm{5}\:\mathrm{2}\:\mathrm{6}\:\mathrm{5}\:\mathrm{3}\:\mathrm{1}\:\mathrm{1}\:\mathrm{0}\:\mathrm{6}… \\ $$$$\Rightarrow \\ $$$$\mathrm{remainders}\:\mathrm{are}\:\mathrm{0}\:\mathrm{for} \\ $$$${n}=\mathrm{21}{k}+\left\{\mathrm{2},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{10},\:\mathrm{15}\right\} \\ $$$$\mathrm{21}×\mathrm{476}+\mathrm{5}=\mathrm{10001}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{475}×\mathrm{6}+\mathrm{2}=\mathrm{2852} \\ $$
Commented by mathdanisur last updated on 20/Sep/21
very nice
$$\mathrm{very}\:\mathrm{nice} \\ $$

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