uniphonous
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Frog style logical problem
- Thread starter uniphonous
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lev888
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What is "Frog style logical problem"?
I suspect it is one of those stupid problems where you are supposed to find THE correct answer when there is no such thing. Here it is to determine a unique "frog" operation, a binary operation on two integers.
I've never seen one of these, but is it something like:
To get the first two digits of the answer, take the first two-digit no. of the question + (third digit of qn -1) [MATH]\times[/MATH] (fourth digit of question +1)
e.g. 59 + (1*2) =61?
If so, [MATH][/MATH]I'll leave the last 2 digits as an exercise!
To get the first two digits of the answer, take the first two-digit no. of the question + (third digit of qn -1) [MATH]\times[/MATH] (fourth digit of question +1)
e.g. 59 + (1*2) =61?
If so, [MATH][/MATH]I'll leave the last 2 digits as an exercise!
HallsofIvy
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You are given some operation that gives that data and are asked to find how the operation would be applied to the final data. There are infinitely many perfectly valid answers!
One method is to treat this as values given by a function of two variables, z= f(x,y).
Then we are given that f(59, 21)= 6146, f(83, 61)= 9351, and f(77, 42)= 8633.
Given any n data points, there always exist a polynomial of degree n-1 that gives those n data points: two points determine a line, three points determine a parabola, etc. Since we have three data points, I would look for a function of the form \(\displaystyle f(x,y)= ax^2+ bxy+ cy^2\). We have three equations to solve for a, b, and c.
\(\displaystyle 59^2a+ (59)(21)b+ 21^2c= 6246\)
\(\displaystyle 83^2a+ (83)(61)b+ 61^2c= 9351\)
\(\displaystyle 77^2a+ (77)(42)b+ 42^2c= 8633\)
To answer the question, solve those three linear equations for a, b, and c then set x= 92, b= 62 in \(\displaystyle ax^2+ bxy+ cy^2\).
I don't say that is THE answer you are looking for but it certainly is one valid answer.
One method is to treat this as values given by a function of two variables, z= f(x,y).
Then we are given that f(59, 21)= 6146, f(83, 61)= 9351, and f(77, 42)= 8633.
Given any n data points, there always exist a polynomial of degree n-1 that gives those n data points: two points determine a line, three points determine a parabola, etc. Since we have three data points, I would look for a function of the form \(\displaystyle f(x,y)= ax^2+ bxy+ cy^2\). We have three equations to solve for a, b, and c.
\(\displaystyle 59^2a+ (59)(21)b+ 21^2c= 6246\)
\(\displaystyle 83^2a+ (83)(61)b+ 61^2c= 9351\)
\(\displaystyle 77^2a+ (77)(42)b+ 42^2c= 8633\)
To answer the question, solve those three linear equations for a, b, and c then set x= 92, b= 62 in \(\displaystyle ax^2+ bxy+ cy^2\).
I don't say that is THE answer you are looking for but it certainly is one valid answer.
@HallsofIvy
I see! Thanks.
I see! Thanks.