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While I do not agree with Skoper, that the result (ie -2) when s can be 2 is not odd or even -> since -2 is an even number; the reason is different.

s can also be 3 -> which means that choice B will give more than one answer , including Zero (k=2, s=3, k*(3-3)=0 ; but s<>3, k*(s-3) is always even. The condition k<3 gives only k=2.). Therefore B does not give us the decision on Yes or No for the question stem.

This exercise tests what ETS test writers love is that 2 is the only even prime number and the smallest , this exercise looks very easy but very tricky

first we know that S and k are prime number

for k(s-3) to be odd k have to be odd becoz whatever time an even is even

St 1 tell us that s IS ODD ins NO IDEA about K
st2 SYES becoz if k is A PRIME number smaller to 3 THIS is only be 2
sok(s-3) WILL be even no matter s Sufficient

This is the typical yes/no type of question. If you can give a 'yes' or a 'no' to the question: "Is K(S-3) odd ?", then the statement is sufficient.

We're told that S and K are prime numbers. For K(S-3) to be odd, K must not be even, or (S-3) must not be even, or both must not be even.

From statement 1), all we know is that S>10. All prime numbers greater than 10 are odd (2 is the only even prime number), and we know an odd - odd will always result in an even number. So (S-3) will be even and therefore the product of K and (S-3) will be even. Statement 1 is therefore sufficient.

2) From statement 2, we know K must be 2 since it has to be prime and has to be less than 3. The product of K and (S-3) will therefore be even.
Thus, statement 2 is sufficient as well.

This exercise tests what ETS test writers love is that 2 is the only even prime number and the smallest , this exercise looks very easy but very tricky

first we know that S and k are prime number

for k(s-3) to be odd k have to be odd becoz whatever time an even is even

St 1 tell us that s IS ODD ins NO IDEA about K st2 SYES becoz if k is A PRIME number smaller to 3 THIS is only be 2 sok(s-3) WILL be even no matter s Sufficient

to see that pick number S be 6 S be 7

This question is typical of many D.S questions. I call it the 'yes/no' type of question. It's basically asking you whether K(S-3) can be odd? If the statement allows you to say "Yes, K(S-3) can be odd" or "No, K(S-3) cannot be odd", then the statement is sufficient. I've posted my explanations in this thread. THe answer should be D.

This exercise tests what ETS test writers love is that 2 is the only even prime number and the smallest , this exercise looks very easy but very tricky

first we know that S and k are prime number

for k(s-3) to be odd k have to be odd becoz whatever time an even is even

St 1 tell us that s IS ODD ins NO IDEA about K st2 SYES becoz if k is A PRIME number smaller to 3 THIS is only be 2 sok(s-3) WILL be even no matter s Sufficient

to see that pick number S be 6 S be 7

Mandy,

If s and k are prime numbers, is k(s-3) odd?

1) s>10

We know that all prime numbers are odd except the first one : number 2
Which means that if S>10 and S is a prime number, it has to be odd
Then we know that any odd number - any other odd number = even number
From this point it doesn't matter if K is odd or even because in any case it will be multiplied by (S-3) who is always even...so the result will be even.
A suff

When S > 10, we know for sure that S is odd, and odd - odd = even. So S will always be even, and the product of an even number with an odd number will be even. S1 is sufficient

When K < 3, or K = 2, 2 * anything is even. Therefore S2 is sufficient. Even if S = 3, then 2 * 0 = 0 which is also even. S2 is Suff.