Difference between revisions of "2009 AMC 10A Problems/Problem 16"
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Hence <math>|a-d|=|d|\in\{1,3,5,9\}</math>, and the sum of possible values is <math>1+3+5+9 = \boxed{18}</math>. | Hence <math>|a-d|=|d|\in\{1,3,5,9\}</math>, and the sum of possible values is <math>1+3+5+9 = \boxed{18}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Transforming the absolute values to <math>\pm</math> equations, we can now add: | ||
+ | \begin{tabular}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} | ||
+ | & <math>a</math> & <math>-</math> & <math>\cancel{b}</math> & <math>=</math> & <math>\pm2</math> &&\\ | ||
+ | & <math>b</math> & <math>-</math> & <math>\cancel{c}</math> & <math>=</math> & <math>\pm3</math>&&\\ | ||
+ | <math>+</math>& <math>c</math> & <math>-</math> & <math>\cancel{d}</math> & <math>=</math> & <math>\pm4</math>&&\\ | ||
+ | \hline | ||
+ | & <math>a</math> & <math>-</math> & <math>d</math> & <math>=</math> & <math>\pm2</math>&<math>\pm3</math>&<math>\pm4</math>\\ | ||
+ | \end{tabular} | ||
+ | The possible values can now be calculated by hand. | ||
== See Also == | == See Also == |
Revision as of 16:58, 2 August 2020
Problem
Let ,
,
, and
be real numbers with
,
, and
. What is the sum of all possible values of
?
Solution 1
From we get that
Similarly, and
.
Substitution gives . This gives
. There are
possibilities for the value of
:
,
,
,
,
,
,
,
Therefore, the only possible values of are 9, 5, 3, and 1. Their sum is
.
Solution 2
If we add the same constant to all of ,
,
, and
, we will not change any of the differences. Hence we can assume that
.
From we get that
, hence
.
If we multiply all four numbers by , we will not change any of the differences. (This is due to the fact that we are calculating |d| at the end ~Williamgolly) Hence we can WLOG assume that
.
From we get that
.
From we get that
.
Hence , and the sum of possible values is
.
Solution 3
Transforming the absolute values to equations, we can now add:
\begin{tabular}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c}
&
&
&
&
&
&&\\
&
&
&
&
&
&&\\
&
&
&
&
&
&&\\
\hline
&
&
&
&
&
&
&
\\
\end{tabular}
The possible values can now be calculated by hand.
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.