Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 8"
(→Solution) |
|||
(3 intermediate revisions by 3 users not shown) | |||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
We let <math>a=(x+y)</math> and <math>b=(20x+21y)</math> to get the new system of equations <cmath>a^2b=12 \qquad (1)</cmath> <cmath>ab^2=18 \qquad(2).</cmath> Multiplying these two, we have <math>(ab)^3=12 \cdot 18</math> or <cmath>ab=6 \qquad (3).</cmath> We divide <math>(3)</math> by <math>(1)</math> to get <math>a=2</math> and divide <math>(2)</math> by <math>(1)</math> to get <math>b=3</math>. Recall that <math>a=x+y=2</math> and <math>b=20x+21y=3</math>. Solving the system of equations <cmath>x+y=2</cmath> <cmath>20x+21y=3,</cmath> we get <math>y=-37</math> and <math>x=39</math>. This means that <cmath>21x+20y=20x+21y+x-y=3+39-(-37)=\boxed{79}.</cmath> ~samrocksnature | We let <math>a=(x+y)</math> and <math>b=(20x+21y)</math> to get the new system of equations <cmath>a^2b=12 \qquad (1)</cmath> <cmath>ab^2=18 \qquad(2).</cmath> Multiplying these two, we have <math>(ab)^3=12 \cdot 18</math> or <cmath>ab=6 \qquad (3).</cmath> We divide <math>(3)</math> by <math>(1)</math> to get <math>a=2</math> and divide <math>(2)</math> by <math>(1)</math> to get <math>b=3</math>. Recall that <math>a=x+y=2</math> and <math>b=20x+21y=3</math>. Solving the system of equations <cmath>x+y=2</cmath> <cmath>20x+21y=3,</cmath> we get <math>y=-37</math> and <math>x=39</math>. This means that <cmath>21x+20y=20x+21y+x-y=3+39-(-37)=\boxed{79}.</cmath> ~samrocksnature | ||
+ | |||
+ | ==Solution 2== | ||
+ | Each number shares are factor of <math>6</math>, which means <math>(x+y)(20x+21y)=6</math>, or <math>x+y=2</math> and <math>20x+21y=3</math>. We see <math>y=-37</math> and <math>x=39</math>, so <math>39(21)-20(37)=\boxed{79}</math> | ||
+ | |||
+ | ~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== | ||
+ | #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]] | ||
+ | #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 10:07, 12 July 2021
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.