BTGmoderatorDC wrote:If x and y are positive integers, what is the remainder when x^y is divided by 10?

(1) x = 26
(2) y^x = 1
Source: Manhattan Prep

$$x,y\,\, \ge 1\,\,\,{\rm{ints}}$$
$${x^y} = 10M + R$$
$$M,R\,\,{\mathop{\rm int}} \,\,\,,\,\,\,0 \le R\,\, \le 9$$
$$? = R\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,?\,\,\,:\,\,\,{\text{units}}\,\,{\text{digit}}\,\,{\text{of}}\,\,{x^y}\,\,\,}\,$$
$$\left( 1 \right)\,\,{x^y} = {26^y}\,\,\,\,\left( {y \ge 1\,\,{\mathop{\rm int}} } \right)\,\,\,\, \Rightarrow \,\,\,\,? = 6\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\rm{SUFF}}.$$
$$\left( 2 \right)\,\,{y^x} = 1\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{1}}\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{2}}\, \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,{\rm{INSUFF}}{\rm{.}}$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

When a number is divided by 10, unit digit of that number will be remainder.

Statement 1 : x = 26
Any number which has 6 as its unit digit, when raised to any positive integer power, will give a number with 6 as a unit digit.
So x^y will always have 6 as unit digit.
So remainder = 6
---> sufficient

Statement 2 :
For y = 1, multiple values of x are possible.
Take x = 2 => Remainder = 2
Take x = 3 => Remainder = 3
---> insufficient

Ans. A