Difference between revisions of "2006 Seniors Pancyprian/2nd grade/Problem 2"
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== Problem == | == Problem == | ||
| − | Find all three digit numbers <math>\overline{xyz}</math>(=100x+10y+z) for which <math>\frac {7}{4}(\overline{xyz})=\overline{zyx}</math>. | + | Find all three digit numbers <math>\overline{xyz}</math>(<math>=100x+10y+z</math>) for which <math>\frac {7}{4}(\overline{xyz})=\overline{zyx}</math>. |
| + | == Solution == | ||
| + | <math>175x+\frac{35}{2}y+\frac{7}{4}z=100z+10y+x</math> | ||
| + | <math>174x+\frac{15}{2}y-\frac{393}{4}z=0</math> | ||
| − | = | + | <math>232x+10y-131z=0</math> |
| + | 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 == | |
| − | |||
* [[2006 Seniors Pancyprian/2nd grade/Problem 3|Next Problem]] | * [[2006 Seniors Pancyprian/2nd grade/Problem 3|Next Problem]] | ||
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* [[2006 Seniors Pancyprian/2nd grade|Back to Exam]] | * [[2006 Seniors Pancyprian/2nd grade|Back to Exam]] | ||
| − | |||
* [[Seniors Pancyprian Competition]] | * [[Seniors Pancyprian Competition]] | ||
* [[Seniors Pancyprian Problems and Solutions]] | * [[Seniors Pancyprian Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
Latest revision as of 06:23, 29 October 2009
Problem
Find all three digit numbers 
(
) for which 
.
Solution
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 
, then 
and 
. If 
, then 
and 
. If 
, then 
and 
. If 
, then 
, but there are no solutions for 
. Thus only 
, 
, and 
work.
