Am in need of some help from the collective with my son's 1st year maths homework... yes, I know, I know... 1st year.

Here's his homework:

Diophantus' Arithmetica

How can you find four numbers when you know all of the possible sums of three of them?
For example, let's say that the various possible sums are 20, 22, 24 and 27.
What are the four numbers we are looking for?



Now, I've worked out (using a process of elimination) the 4 numbers are : 9, 11, 7, 4

My question is : How can you possibly do this without just plugging number in willy nilly - there MUST be an equation which supports this question.

I've googled and found this : http://math.ucr.edu/~res/math153/history05a.pdf
but I'm not convinced i'm going down the right road.

Help please, before I go put on a big red hat with a D on it and sit in the corner for the rest of the evening.

add them all together
20+22+24+27 = 93

now this must be 3 x all four numbers

so all four add up to 93/3 = 31

so the four numbers are

31 - 20 = 11
31 - 22 = 9
31 - 24 = 7
31 - 27 = 4

My son said "Thank you"..

Cheers pal, you're a genius !

ERRRRRRRRRRRRRR

Isnt this cheating ?????

quote:

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Isnt this cheating ?????

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Oh,

Thats ok then..................

Well, I at least got the answer, but I knew there must have been an easier way than just plugging in numbers.

My son was happy that I got the answer, but he's only 12 so I had to explain to him that the workings are the important bit.

I did A level maths mechanics some 20 years ago, so I'm a little rusty.

I just love this forum - the knowledge here is unbelievable !

"I just love this forum - the knowledge here is unbelievable !"

I agree, but I got lost on line 2, i understand what but not why"now this must be 3 x all four numbers "?

a+b+c=20
a+b+d=22
a+c+d=24
b+c+d=27


add them all together you have 3a+3b+c3+3d = 93
divide by 3 and you have a+b+c+d = 31

so
d = (a+b+c+d) - (a+b+c)
d = 31 - 20 = 11
and so on.


[Edited on 5/3/13 by Grimsdale]