Exercise # 89 ds, OG 11th edition:

given that
k º 1 is not equal to 1 º k for some numbers k

the operation º can be any of the operations : +, - or * (given)
lets take + and k be 2

2 + 1 = 1+ 2 (so the operation is not +. the condition is not equal)
2 * 1 = 1 * 2 (the operation is also not *)
but 2 -1 != 1 -2 (hence the operation must be subtraction)

Note that div is not given in the possible operations list
(If º represents one of the operations +,-, and x)

if we know the operation is subtraction, we can easily figure out the result of the given function.
kº ( l + m) = (kºl) + (kºm)
substitute - for º
so k - (l +m ) = (k - l ) + (k - m)
SUFFICIENT

B is given as subtraction (straight forward)
SUFFICIENT.

So the answer is D.

Thank you very much for your help.
Your explanation is much more clear

paojaram wrote:Can someone help me to understand this exercise? It is not very clear the explanation in the book

If º represents one of the operations +,-, and x, is kº ( l + m) = (kºl) + (kºm) for all numbers k,l, and m?

(1) kº is not equal to 1 º k for some numbers k
(2) º represents subtraction

for the expression to be valid º has to be either - or *


from one
º is not a multiplication thus suff and it is subtraction

from 2

suff

D