Question

Mă ajutați vă rog la problemele astea?​

Answer 1

Răspuns:

Explicație pas cu pas:

a modulo b = restul impartirii lui a la b

1.

$5^{1+2+3+\cdots + 2014} = 5^{n+1007}\implies 1+2+\cdots + 2014 = n + 1007\implies n = \frac{2014\cdot 2015}{2} - 1007 = 1007\cdot 2015 - 1007 = 1007\cdot 2014 = 2028098$

$7^{2+4+6+\cdots +100} = 7^{n(n+1)}\implies 2(1+2+3+\cdots +50) = n(n+1)\implies 50\cdot (50 + 1) = n(n+1) \implies n = 50$

2.

$7000^{671} = {(7\cdot 10^3)}^{671} = 7^{671}\cdot 10^{2013}\\\\\implies \text{Ultimele 2013 cifre sunt egale cu 0}.\\\\\textrm{Sirul dat de formula } (7^n)\:modulo\: 10:\\\\1, 7, 9, 3, 1, 7, 9, 3, 1, 7, 9, \cdots\\\\ \text{Cifra imediat inaintea 0-urilor este :}\\\\ 671 \:modulo \:4 = 3 \implies \text{Cifra imediat inaintea 0-urilor este } 3$

3.

$a = 2^{2013}\cdot 5^{2014} + 7^5 \\\\ a = (2\cdot 5)^{2013}\cdot 5 + 7^5 = 5\cdot 10^2013 + 7^5\\\\ \text{a modulo 10 se poate gasi in sirul de la 2.}\\\\ 5\:modulo \: 4 = 1\implies a\:modulo\:10 = 7$

4.

$25^{2013} = {(5^2)}^{2013} = 5^{2\cdot 2013} = 5^{4026}. \\ \\ \text{Exponent par} \implies \text{Numarul este patrat perfect}$

5.

$n^2 = 49 \iff n = 7\\\\(n+3)^2 = 81 \iff n+3 = 9 \iff n = 6\\\\n^2 = 2^4\cdot 5^6\iff n^2 = {(2^2\cdot 5^3)}^2\iff n = 4\cdot 125 \iff n = 500\\\\(4n+1)^2 = 3^4\cdot 5^2 \iff 4n+1 = 3^2 \cdot 5 \iff 4n+1 = 45 \iff 4n = 44\iff n = 11\\\\n^3 = 27 \iff n = \sqrt[3]{27} \iff n = 3\\\\(n+1)^3 = 125\iff n+1 = \sqrt[3]{125} \iff n+1 = 5\iff n = 4\\\\n^3 = 7^3 \cdot 5^6 \iff n = 7\cdot 5^2 \iff n = 175\\\\(n-1)^3 = 2^3\cdot 3^6\iff n-1 = 2\cdot 3^2 \iff n-1 = 18 \iff n = 19$

6.

$27^{27} = {(3^3)}^{27} = 3^{81}\\\\2^{50}\: si \: 3^{30}\\\\\text{Scoatem radical de ordinul 10 din ambele parti si obtinem:}\\\\ 2^5\: si \: 3^3\\\\2^5 = 32\\\\3^3 = 27\\\\ \implies 2^5 > 3^3 \implies 2^{50} > 3^{30}$