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Difference between revisions of "2006 Seniors Pancyprian/2nd grade/Problem 2"
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<math>232x+10y-131z=0</math>
 
<math>232x+10y-131z=0</math>
  
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We can now make some conclusions. z is even, so z can be only 2, 4, 6, or 8, since 0 is an extraneous solution. Now if <math>z=2</math>, then <math>x=1</math> and <math>y=3</math>. If <math>z=4</math>, then <math>x=2</math> and <math>y=6</math>. If <math>z=6</math>, then <math>x=3</math> and <math>y=9</math>. If <math>z=8</math>, then <math>x=4</math>, but there are no solutions for <math>y</math>. Thus only <math>123</math>, <math>246</math>, and <math>369</math> work.
+
We can now make some conclusions. z is even, so z can be only 2, 4, 6, or 8, since 0 is an extraneous solution. Now if <math>z=2</math>, then <math>x=1</math> and <math>y=3</math>. If <math>z=4</math>, then <math>x=2</math> and <math>y=6</math>. If <math>z=6</math>, then <math>x=3</math> and <math>y=9</math>. If <math>z=8</math>, then <math>x=4</math>, but there are no solutions for <math>y</math>. Thus only <math>132</math>, <math>264</math>, and <math>396</math> work.
  
 
== See also ==
 
== See also ==

Latest revision as of 07:23, 29 October 2009

Problem

Find all three digit numbers $\overline{xyz}$($=100x+10y+z$) for which $\frac {7}{4}(\overline{xyz})=\overline{zyx}$.

Solution

$175x+\frac{35}{2}y+\frac{7}{4}z=100z+10y+x$

$174x+\frac{15}{2}y-\frac{393}{4}z=0$

$232x+10y-131z=0$

We can now make some conclusions. z is even, so z can be only 2, 4, 6, or 8, since 0 is an extraneous solution. Now if $z=2$, then $x=1$ and $y=3$. If $z=4$, then $x=2$ and $y=6$. If $z=6$, then $x=3$ and $y=9$. If $z=8$, then $x=4$, but there are no solutions for $y$. Thus only $132$, $264$, and $396$ work.

See also

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