Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 8"
Geometry285 (talk | contribs) m |
|||
Line 12: | Line 12: | ||
~Geometry285 | ~Geometry285 | ||
+ | |||
+ | == Solution 3 == | ||
+ | Multiplying the equations together, we get | ||
+ | <cmath>(x+y)^3(20x+21y)^3=2^3 \cdot 3^3 \implies (x+y)(20x+21y)=6</cmath>Therefore, | ||
+ | <cmath>x+y=2 \implies 20x+20y=40</cmath><cmath>20x+21y=3</cmath>Subtracting the equations, we get <math>y=-37</math> and <math>x=39</math>, therefore, <math>21 (39) - 20 (37) =\boxed{79}</math> | ||
+ | |||
+ | - kante314 - | ||
==See also== | ==See also== |
Latest revision as of 09:07, 12 July 2021
Contents
[hide]Problem
Let and
be real numbers that satisfy
Find
.
Solution
We let and
to get the new system of equations
Multiplying these two, we have
or
We divide
by
to get
and divide
by
to get
. Recall that
and
. Solving the system of equations
we get
and
. This means that
~samrocksnature
Solution 2
Each number shares are factor of , which means
, or
and
. We see
and
, so
~Geometry285
Solution 3
Multiplying the equations together, we get
Therefore,
Subtracting the equations, we get
and
, therefore,
- kante314 -
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.