If x and y are positive integers, what is the remainder when 10^x+y is divided by y?
1) x=50
2) y=2
The OA is B
Source: Official Guide
If x and y are positive integers, what is the remainder when
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$$x,y\,\, \ge 1\,\,{\rm{ints}}\,$$swerve wrote:If x and y are positive integers, what is the remainder when 10^x+y is divided by y?
1) x=50
2) y=2
Source: Official Guide
$$?\,\,\,:\,\,\,\left( {{{10}^x} + y} \right)\,\,{\rm{remainder}}\,\,{\rm{when}}\,\,{\rm{divided}}\,\,{\rm{by}}\,\,y$$
$$\left( 1 \right)\,\,x = 50\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,y = 2\,\,\,\, \Rightarrow \,\,\,\,\,{10^{50}} + 2\,\,{\rm{even}}\,\,\,\, \Rightarrow \,\,\,\,\,? = 0\, \hfill \cr
\,{\rm{Take}}\,\,y = 3\,\,\,\, \Rightarrow \,\,\,\,\,{10^{50}} + 3\,\,\,{\rm{not}}\,\,{\rm{divisible}}\,\,{\rm{by }}\,{\rm{3}}\,\,\,\,\left( {\sum {\,{\rm{digits}}\,\,\,{\rm{ = }}\,\,{\rm{4}}} } \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? \ne 0\,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,\,y = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,{10^{\,x\, \ge \,1}} + 2\,\,\,{\rm{even}}\,\,\,\, \Rightarrow \,\,\,\,\,? = 0\,$$
This solution follows the notations and rationale taught in the GMATH method.
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Fabio.
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Hi Swerve,swerve wrote:If x and y are positive integers, what is the remainder when 10^x+y is divided by y?
1) x=50
2) y=2
The OA is B
Source: Official Guide
To avoid ambiguity, you might want to add some spaces or brackets to the original expression.
As it stands, 10^x+y could be interpreted as either 10^(x+y) or (10^x) + y
Cheers,
Brent
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Statement 1
$$when\ x=0;\ 10^0=1;\ sum\ of\ 1=1$$
$$when\ x=1;\ 10^1=10;\ sum\ of\ 1+0=1$$
$$when\ x=2;\ 10^2=100;\ sum\ of\ 1+0+0=1$$
$$when\ x=3;\ 10^3=1000;\ sum\ of\ 1+0+0+0=1$$
$$\since\ the\ sum\ of\ 10^xis\ always=1,$$
$$Then,\ the\ remainder\ of\ \frac{10^x+y}{y}\ is\ only\ dependent\ on\ the\ value\ of\ y.$$
$$if\ y\ =\left(multiple\ of\ \ 3\ +1\right)\ then\ 10^x+y\ will\ yield\ the\ remainder\ 2\ $$
$$if\ y\ is\ =\left(miltiple\ of\ 3+2\right);\ then\ \frac{10^x+y}{y}\ will\ remain\ 0$$
$$hence;\ when\ x=50\left(from\ statement\ 1\right)$$
$$\frac{10^{50}+y}{y}\ \ \ remainder\ is\ unknown\ without\ the\ value\ of\ y,$$
Hence, Statement 1 is NOT SUFFICIENT
statement 2
$$y=2$$
$$with\ the\ sum\ of\ \ 10^x\ as\ 1$$
$$\frac{1+y}{y}=\frac{1+2}{2}=\frac{3}{2}$$
$$when\ y=2;\ \ 10^x+y\ \ will\ remain\ 1$$
$$Statement\ 2\ is\ \ SUFFICIENT$$
$$when\ x=0;\ 10^0=1;\ sum\ of\ 1=1$$
$$when\ x=1;\ 10^1=10;\ sum\ of\ 1+0=1$$
$$when\ x=2;\ 10^2=100;\ sum\ of\ 1+0+0=1$$
$$when\ x=3;\ 10^3=1000;\ sum\ of\ 1+0+0+0=1$$
$$\since\ the\ sum\ of\ 10^xis\ always=1,$$
$$Then,\ the\ remainder\ of\ \frac{10^x+y}{y}\ is\ only\ dependent\ on\ the\ value\ of\ y.$$
$$if\ y\ =\left(multiple\ of\ \ 3\ +1\right)\ then\ 10^x+y\ will\ yield\ the\ remainder\ 2\ $$
$$if\ y\ is\ =\left(miltiple\ of\ 3+2\right);\ then\ \frac{10^x+y}{y}\ will\ remain\ 0$$
$$hence;\ when\ x=50\left(from\ statement\ 1\right)$$
$$\frac{10^{50}+y}{y}\ \ \ remainder\ is\ unknown\ without\ the\ value\ of\ y,$$
Hence, Statement 1 is NOT SUFFICIENT
statement 2
$$y=2$$
$$with\ the\ sum\ of\ \ 10^x\ as\ 1$$
$$\frac{1+y}{y}=\frac{1+2}{2}=\frac{3}{2}$$
$$when\ y=2;\ \ 10^x+y\ \ will\ remain\ 1$$
$$Statement\ 2\ is\ \ SUFFICIENT$$