Question

This question was posted in Mathematics earlier today. An unusual question:

in the matrix equation
[a b][a]=[20]
[c d][e] [43]
each variable is positive, an integer, and less than 10. What is a+b+c+d+e?

Answer 1

a^2 + be = 20
ac +de = 43

a must be less than 5, If a= 4, Then be = 4 and e must be 1,2,or4
4c + 4d = 43, not possible 43 not divisible by 4
4c + 2d = 43, not possible 43 not divisible by 2
Thus, e = 1 and b = 4
4c + d = 43 ..... c= 9, d = 7

a+b+c+d+e = 4+4+9+7+1 = 25

Answer 2

yup - that is pretty much how I solved it as well.

1st equation implies 'a' < 5.
2nd equation implies 'a' and 'e' cannot BOTH be even.
so:
a = 1 => be = 19 (reject as it implies b or e > 9)
a = 2 => be = 16 (reject as 8*2 and 4*4 both lead to 'a' and 'e' being even)
a = 3 => be = 11 (reject as it implies b or e > 9)
a = 4 => be = 4 (2*2 rejected as 'a' and 'e' would be even, 4*1 accepted)

so we get a=4, b=4, e=1
then use 2nd equation with these values to get:
4c + d = 43
max 'd' = 9 => min 'c' = (43-9)/4 = 8.5
min 'd' = 1 => max 'c' = (43-1)/4 = 10.5
therefore c=9 which leads to d=7

so solution is:
a=4, b=4, c=9, d=7, e=1

a+b+c+d+e=25

Answer 3

Yes, asnaseer nailed it when it was originally posted as well.

Unusual problem.

Answer 4

Nice job Asnaseer!

Answer 5

As James pointed out it's also a great example of the fallibility of Wolfram
http://www.wolframalpha.com/input/?i=solve%7B%7B%7Ba%2Cb%7D%2C%7Bc%2Cd%7D%7D.%7B%7Ba%7D%2C%7Be%7D%7D%3D%7B%7B20%7D%2C%7B43%7D%7D%2Ca%2Cb%2Cc%2Cd%2Ce%7D+
Oh how the mighty have fallen :)

Answer 6

but do keep in mind that even Wolfram was written by mere mortals :-)
we are ALL fallible - you must admit though Wolfram is a very impressive piece of software?

Answer 7

It is. But I think it's a double-edged sword in the hands of a mathematical beginner.

Answer 8

true - especially if you treat it like the "gospel truth". there is no substitute for the human mind (at least not in my opinion)