BTGmoderatorDC wrote:What is the remainder if 7^10 is divided by 100?
A] 1
B] 43
C] 19
D] 70
E] 49
Source: Manhattan Prep
First note that:
> 1 is the remainder of 101 (=1*100+1) divided by 100
> 32 is the remainder of 532 (=5*100+32) divided by 100
> 47 is the remainder of 7847 (=78*100+47) divided by 100
$${7^{10}} = K \cdot 100 + R{\mkern 1mu} {\mkern 1mu} \,\,{\mkern 1mu} {\mkern 1mu} \left( {K\,\,{\mathop{\rm int}} \,\,,\,\,\,0 \le R \le 99\,\,{\mathop{\rm int}} } \right){\mkern 1mu} $$
$$? = R$$
$${7^{10}} = {\left( {{7^2}} \right)^5} = {49^5}$$
$${49^2} = {\left( {50 - 1} \right)^2} = {5^2} \cdot {10^2} - 100 + 1 = M \cdot 100 + 1\,\,\,,\,\,\,M\,\,{\mathop{\rm int}} \ge 1\,\,\,\,\,\,\,\,\,\,\left( {M = {5^2} - 1} \right)$$
$${49^4} = {\left( {M \cdot 100 + 1} \right)^2} = {M^2} \cdot {10^4} + M \cdot 200 + 1 = N \cdot 100 + 1\,\,\,,\,\,\,\,N\,\,{\mathop{\rm int}} \,\, \ge 1\,\,\,\,\,\,\,\,\left( {N = {M^2} \cdot {{10}^2} + 2M} \right)$$
$${49^5} = \left( {N \cdot 100 + 1} \right) \cdot 49 = K \cdot 100 + 49\,\,\,,\,\,\,\,K\,\,{\mathop{\rm int}} \,\, \ge 1\,\,\,\left( {K = 49N} \right)$$
$$? = 49$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
