I did it the same way on the first go. Then, while watching, a slightly different way occurred to me, to give greater confidence. Here, due to the symmetry of the expression: five ways to assign 3, then four ways to assign 1, and two thirds of the remaining six possibilities are divisible by 3, simultaneously counting both the successes and failures. This yields 80 and 40 together, correctly summing to the required 120.
There are 5 ways to pick the number which has value 3 but it may be irrelevant in the end... Let's sat x1=3 then all products containing x1 will be divisible by 3 and for the whole sum to be dividible by we require X2*x3*x4+x3*x4*x5=x3*x4*(x2+x5) to be disicible by 3 which is only achevable when x2 and x5 add to a multiple of 3 which can be 1+5 ; 5+1; 2+4;4+2 total of 4 ways of choosing the pair and then 2! Ways to choose the last two numbers Answer 5*4*2=40 Random observation 40=5!/3
Your random observation leads one to ask if you can simply reason that you only expect one third of all arrangements to be divisible by 3. It would have to be made more rigorous, but perhaps there is an approach that works there.
I think for 4 you could argue having 1 term as 4 excludes 3 products as well leaving X3*X4*(X2+X5) but in this case we need X3 or X4 =2 and X2+X5 to be 1+3;3+1; 1+5; 5+1 which also leads to 40 ways . I think it is just happenstance and the properties of addition on such a small set of numbers
I did it the same way on the first go. Then, while watching, a slightly different way occurred to me, to give greater confidence.
Here, due to the symmetry of the expression:
five ways to assign 3, then
four ways to assign 1, and
two thirds of the remaining six possibilities are divisible by 3, simultaneously counting both the successes and failures.
This yields 80 and 40 together, correctly summing to the required 120.
There are 5 ways to pick the number which has value 3 but it may be irrelevant in the end...
Let's sat x1=3 then all products containing x1 will be divisible by 3 and for the whole sum to be dividible by we require
X2*x3*x4+x3*x4*x5=x3*x4*(x2+x5) to be disicible by 3 which is only achevable when x2 and x5 add to a multiple of 3 which can be
1+5 ; 5+1; 2+4;4+2 total of 4 ways of choosing the pair and then 2! Ways to choose the last two numbers
Answer 5*4*2=40
Random observation 40=5!/3
Your random observation leads one to ask if you can simply reason that you only expect one third of all arrangements to be divisible by 3. It would have to be made more rigorous, but perhaps there is an approach that works there.
I think for 4 you could argue having 1 term as 4 excludes 3 products as well leaving
X3*X4*(X2+X5) but in this case we need X3 or X4 =2 and X2+X5 to be 1+3;3+1; 1+5; 5+1 which also leads to 40 ways .
I think it is just happenstance and the properties of addition on such a small set of numbers
you also have 1+2; 2+1; 4+5; 5+4; as possibilities to achieve multiples of 3. so in total 8 ways. So you also get to the computation: 8*2*5 = 80