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a-b+c=-b+c when a=0 and not equal for other values.. So both are insufficient.. Am I missing something here?

If # represents one of the operations +, - and *, is \(a#(b-c)=(a#b)-(a#c)\) for all numbers \(a\), \(b\) and \(c\).

(1) \(a#1\) is not equal to \(1#a\) for some numbers \(a\).

\(#\) is neither addition (as \(a+1=1+a\)) not multiplication (as \(a*1=1*a\)), so \(#\) is a subtraction. Then \(LHS=a#(b-c)=a-b+c\) and \(RHS=(a#b)-(a#c)=(a-b)-(a-c)=c-b\), so the question becomes "is \(a-b+c=c-b\) for all numbers \(a\), \(b\) and \(c\)?" --> "is \(a=0\)". So when \(a=0\) (and \(#\) is a subtraction) then \(a#(b-c)=(a#b)-(a#c)\) holds true but not for other values of \(a\), so not for all numbers \(a\), \(b\) and \(c\). Answer to the question is NO. Sufficient.

(2) \(#\) represents subtraction --> the same as above. Sufficient.

Re: If # represents one of the operations +,- and *, is a # [#permalink]

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21 Sep 2010, 10:55

zisis wrote:

If # represents one of the operations +,- and *, is a # (b-c) = (a#b) – (a#c) for all numbers a, b and c.

(1) a#1 is not equal to 1#a for some numbers a

(2) # represents subtraction

Mods, please to DS section...posted by mistake in PS - apologies

You're told that "#" is either addition, subtraction, or multiplication, and then asked if "#" satisfies the distributive property. Of these three, distribution only holds for multiplication, so if "#" is "*", it holds, and if "#" isn't "*", then it does not hold.

All we really need to know is what operation "#" really is.

(1) This is only true of subtraction, so # is subtraction and the distributive property does not hold. Sufficient. (2) Same as above. Sufficient.

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