Question

Care este ultima cifră a numărului:

$N = 6^{4\cdot 10^{10}}+1^{3\cdot 10^{11}}-45^{2\cdot 10^{10}}$ ?

Answer 1

Răspuns:

8

Explicație pas cu pas:

$N = 6^{4\cdot 10^{10}} + 1^{3\cdot 10^{11}} - 45^{2\cdot 10^{10}} \\ \\ N\:mod\:10 = \Bigg(6^{4\cdot 10^{10}} + 1 - 45^{2\cdot 10^{10}}\Bigg) \:mod\:10 \\ \\ = \Bigg(6^{4\cdot 10^{10}}\:mod\:10 + 1\:mod\:10-45^{2\cdot 10^{10}}\:mod \:10\Bigg)\:mod \:10 \\ \\= \Bigg(6^{4\cdot 10^{10}}\:mod\:10 + 1 - (45\:mod\:10)^{2\cdot 10^{10}}\:mod\:10\Bigg)\:mod\: 10 \\ \\ = \Bigg(6^{4\cdot 10^{10}}\:mod\:10 + 1 - 5^{2\cdot 10^{10}}\:mod\:10\Bigg)\:mod\:10 \\ \\ = \Bigg(6^{4\cdot 10^{10}}\:mod\:10 + 1 - 5^{2\cdot 10^{10}}\:mod\:10\Bigg)\:mod\:10$

$6^{4\cdot 10^{10}} = 36^{2\cdot 10^{10}} < 45^{2\cdot 10^{10}} \implies N < 0$

$6^1 \: mod \: 10 = 6\\ \\ 6^2 \:mod \:10 = ((6^1\:mod\: 10) \cdot (6^1\:mod\:10))\:mod \: 10 = 36\:mod \: 10 = 6\\ \\ 6^3 \: mod\:10 = ((6^2\:mod \:10)\cdot (6^1\:mod\:10))\:mod \:10 = (6\cdot 6)\:mod\:10 = 36\:mod 10 = 6 \\ \\ \vdots \\ \\ 6^{4\cdot 10^{10}} \: mod \: 10 = 6\\ \\ 5^1\:mod \:10 = 5\\ \\ 5^2\:mod 10 = ((5^1\:mod \:10) \cdot (5^1\:mod\:10))\:mod\: 10 = (5\cdot 5)\:mod\: 10 = 25\:mod \:10 = 5 \\ \\ \vdots \\ \\ 5^{2\cdot 10^{10}} \: mod \: 10 = 5$

$\implies N \: mod \: 10 = (6 + 1 - 5)\:mod \:10 = 2 \: mod\:10 = 2\\ \\ \text{Deoarece N este negativ, avem ultima cifra} = 10 - 2= 8$