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Mathematics 8 Online
OpenStudy (anonymous):

i have ,7,7,7,7,7,7,7,7,7 5,5,5,5,5,5,5,5 3,3,3,3,3,3,3,3 1,1,1,1,1,1,1,1 Using 10 numbers exactly. How do i total exactly 37

OpenStudy (anonymous):

7a+5b+3c+d=37 a+b+c+d=10

OpenStudy (anonymous):

And you have some extra restrictions, like the amount a < some amount and they are all natural numbers. You can solve this mathematically but probably you just want to try just plugging in some numbers.

OpenStudy (anonymous):

Forgive an old man but in layman's terms can you assist my great grandson in plain English. I left full time education a very long time ago.

OpenStudy (anonymous):

My point is that it can be solved mathematically, there are ways to do it, but it's not well known. Just try out various combinations, like I am doing right now :)

OpenStudy (anonymous):

I have been trying all evening

OpenStudy (anonymous):

I can get to 37 with 9 figure combinations and 11 figure combinations easily. But not 10 figures and my old brain tends to shut down at this time of night

OpenStudy (anonymous):

Yep. It might not be possible

OpenStudy (anonymous):

I'll try to work it out mathematically now, tired of trying xD. Not sure if I'll be able to solve it though

OpenStudy (anonymous):

Let's see what I have so far: \[A,B,C,D \in N\]\[A \le 9 \]\[B,C,D \le 8\]\[7A + 5B + 3C + D = 37\]\[A+B+C+D=10\] \[6A+4B+2C = 27\]

OpenStudy (anonymous):

\[6A = 27 -4B-2C\]\[A=\frac{27-4B-2C}{6}\]\[A=4.5 -\frac{2}{3}B - \frac{1}{3}C\]

OpenStudy (anonymous):

I'm not sure where to go from here, I can create a whole set of possible equations, but it would take me ages to check

OpenStudy (anonymous):

There is probably a better way to solve this :D

OpenStudy (anonymous):

You can't solve this any combination of ten odd numbers brings out an even number as a result.

OpenStudy (anonymous):

\[10(2n+1) = 20n + 10\] Sorry, I had a mistake there

OpenStudy (anonymous):

in fact we can generalize it to any combination of an even number of odd numbers has an even number

OpenStudy (anonymous):

Exactly xcrypt but thanks for working on it any way. I guess the answer he will give is " The answer is it cannot be done" Thank you

OpenStudy (anonymous):

that's an evil question :D

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