Bunuel wrote:
amitgovin wrote:
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
(1) xz is even
(2) y is even.
Given:
x is a factor of y --> \(y=mx\), for some non-zero integer \(m\);
y is a factor of z --> \(z=ny\), for some non-zero integer \(n\);
So, \(z=mnx\).
Question: is z even? Note that \(z\) will be even if either \(x\) or \(y\) is even
(1) \(xz\) even --> either \(z\) even, so the answer is directly YES or \(x\) is even (or both). But if \(x\) is even and as \(z=mnx\) then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.
(2) \(y\) even --> as \(z=ny\) then as one of the multiples of z even --> z even. Sufficient.
Answer: D.
Perfect explanation. Remember the factor foundation rule.
Also, other properties of factors that might be helpful to have in mind.
Just to remind you, The factor foundation rule states that "if a is a factor of b, and b is a factor of c, then a is a factor of c"
Also, if 'a' is a factor of 'b', and 'a' is a factor of 'c', then 'a' is a factor of (b+c). In fact, 'a' is a factor of (mb + nc) for all integers 'm' and 'n'
If 'a' is a factor of 'b' and 'b' is a factor of 'a', then 'a=b'
If 'a' is a factor of 'bc' and gcd (a,b) = 1, then 'a' is a factor of 'c'
If 'p' is a prime number and 'p' is a factor of 'ab' then 'p' is a factor of 'a' or 'p' is a factor of 'b'
In other words,, any integer is divisible by all of its factors- and it is also divisible by all of the factors of its factors
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