While browsing my news feed, I ran across a math problem at https://www.popularmechanics.com/science/math/a30985891/solve-hard-math-problem-nonlinear-equations/ and it goes something like this:
a+e = 11
a(a)-f = 15
f+c = 6
c(a) - e = 13
Then the article author raved about how evil and wonderful it was at the same time. What am I missing?
The first equation tells us that NOT (a&e are even) and NOT (a&e are odd).
a^2 - f = 15 so NOT (a^2 & f are even) and NOT (a^2 & f are odd).
f+c = 6 EITHER (f&c are even) OR (f&c are odd).
a^2 + c = 21 (this equation is the result of the previous two equations added together).
Assuming that the 9-year-old problem creator is using whole numbers, we know that a^2 must be between 21 and 15. What number fits that other than a^2 = 16 aka a=4?
Therefore, c=5 and f=1
Looking at the final equation, we know that (5)(4)-e = 13 so e must be 7.
If we are right, then the first equation will work too.
4+7 = 11
Right.
Why is this supposedly so difficult? I gleaned the answer within 30 seconds.
a+e = 11
a(a)-f = 15
f+c = 6
c(a) - e = 13
Then the article author raved about how evil and wonderful it was at the same time. What am I missing?
The first equation tells us that NOT (a&e are even) and NOT (a&e are odd).
a^2 - f = 15 so NOT (a^2 & f are even) and NOT (a^2 & f are odd).
f+c = 6 EITHER (f&c are even) OR (f&c are odd).
a^2 + c = 21 (this equation is the result of the previous two equations added together).
Assuming that the 9-year-old problem creator is using whole numbers, we know that a^2 must be between 21 and 15. What number fits that other than a^2 = 16 aka a=4?
Therefore, c=5 and f=1
Looking at the final equation, we know that (5)(4)-e = 13 so e must be 7.
If we are right, then the first equation will work too.
4+7 = 11
Right.
Why is this supposedly so difficult? I gleaned the answer within 30 seconds.
