The product of 80 whole numbers is 80. Their sum cannot be...?

Question

A. 80
B. 88
C. 90
D. 96

Please explain...

Answer 1

40 X 2 X 1 X 1 X (56 more 1's) X -1 X -1 X (18 more -1's) = 80
40 + 2 + 1 + 1 + (56 more 1's) -1 -1 (18 more -1's) = 80 the answer can't be A

2 X 2 X 2 X 2 X 5 X 1 X 1 X 1 X (72 more 1's) = 80
2 + 2 + 2 + 2 + 5 + 1 + 1 + 1 + (72 more 1's) = 88 the answer answer can't be B

40 X 2 X 1 X 1 X (56 more 1's) X -1 X -1 X (18 more -1's) = 80
40 + 2 + 1 + 1 + (64 more 1's) -1 -1 (10 more -1's) = 96 the answer can't be D

The answer is C. 90 (not divisable by 4) (I can make 88 or 92, but I can't make 90)

Answer 2

John S completely missed the mark here, because
he used negative numbers.
The problem asked for whole numbers only!
I will do this by elimination. Then I'll give a
proof.
To get 88: 75*1 + 2+2+2+2 + 5 = 88.
To get 90: 77*1 + 4 + 4 + 5 = 90.
To get 96: 78*1 + 8+10 = 96.
The answer is A. 80.
Let's show that 80 cannot be represented in this
fashion.
We have x_1 + ... + x_79 + x_80 = 80
and x_1 x_2... x_79 x_80 = 80.
First note that x_80 divides 80.
and that x_80 cannot equal 1
because that gives x_1 + ... + x_79 = 79,
But then all the remaining x_i are 1,
which is impossible, since the x_i are
whole numbers, none are 0 and their product is 80.
Next,
x_80 = 80/ (x_1 x_2... x_79) = 80/n say.
So x_1 + ... + x_79 = 80 -80/n.
But since n cannot equal 80, we must
have 1< n <80.
But then x_1 + ... + x_79 >= 79
and 80-80/n <= 78
which is impossible.
Hope that makes sense!