Is \(k>3\)?

(1) \((k-3)(k-2)(k-1)>0\)

The product of 3 numbers is positive if all three are positive (+++) OR two of them are negative and the third one is positive (+--).

Note that: out of 3 numbers \(k-3\) is the least one and \(k-1\) is the biggest one.

\((+)(+)(+)\) is when even the least one is positive so when \(k-3>0\) --> \(k>3\);
\((+)(-)(-)\) is when the biggest one is positive (\(k-1>0\) --> \(k>1\)) and the next one (hence the leas one too) negative (\(k-2<0\) --> \(k<2\)), so when \(1<k<2\);

So \((k-3)(k-2)(k-1)>0\) means that: \(k>3\) or \(1<k<2\) --> \(k\) may or may not be more than 3. Not sufficient.

(2) \(k>1\). Clearly insufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is the range we had in (1) \(k>3\) or \(1<k<2\), so \(k\) may or may not be more than 3. Not sufficient.

Answer: E.
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If (k-3)(k-2)(k-1)>0, doesnt that mean, K>3 or K>2 or K>1?
Why is it K>3 or 1>k>2?
Please explain. Thank you.

When you have inequalities like this one A*B*C>0
Either all three terms are greater than 0 (k>3)
Or exactly one term is greater than 0, and other two are less than 0 (1<k<2)
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Is there a reason for choosing K>3 ? Can it be K<3,K<2 and K>1?

check out the link inequalities-trick-91482.html
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St. 1 gives us that k>2 since the equation is +ve only when all 3 are +ve i.e. k>3 or when 2 -ve and 1+ve fig is there i.e. K>2 -- Not sufficient

St. 2 tells us k>1 -- Not sufficient

Merging St. 1 and St. 2 tells us that k>2 -- Not sufficient.

Therefore, answer is E

If the numbers are (k-3), (k-2), and (k-1), then these are consecutive numbers. Accordingly, (--+) won't exist as it needs to have 0 as well and we know the product is greater than zero.

Can any one of you explain why is my understanding incorrect?