enigma123 wrote:
Hi Fluke and everyone,
If x,y and z are integers and xyz is divisible by 8, is x even?
1. yz is divisible by 4.
2. x,y and z are all NOT divisible by 4.
I am struggling to understand this question. The approach that I am trying to use is :
Statemenen 1 alone: INSUFFICIENT. Why? Because it doesn't say anything about x. x could be 1 and yz could be 4. x could be 8 and yz could e 4. So x can be ODD and EVEN both. Therefore, not sufficient. Is my understanding and concept correct for declining Statement 1?
Statement 2: If x,y and z are not divisible by 4, but they are divisible by 8 i.e they are divisible by at least 2^3. I am stuck after that. Can someone please help to tell me how to approach this mathematically?
The correct answer is B i.e statement 2 alone is sufficient.
Just think in terms of prime factors;
xyz is divisible by 8.
means; x*y*z must have at least 3 2's, doesn't matter from where those 2 come; the 3 2's are there.
Then think what possible ways can I get those three 2's.
x=8, y=1,z=1
We got 3 2's; this time from x.
x=1,y=8,z=1
I got 3 2's; this time from y.
x=1,y=1,z=8
I got 3 2's; this time from z.
There are many such combinations.
x=2,y=2,z=2
x=4,y=2,z=1
x=1,y=2,z=4
Q: Is x=even?
1. yz is divisible by 4.
This means that
yz must have at least two 2's because 4 has two 2's.
Now,
if y=2, z=2; yz=4;
But from the stem we know xyz contains three 2's
Thus, the one additional 2 must come from x AND x becomes even. If "x" contains one 2, it becomes even.
But, we don't know whether y=2, z=2;
y may be 4 or 8 or 16
Then the odd/even for x or z doesn't make a difference. Because, y alone took care of both statements.
If y=8; y has 3 2's
Now even if x=1; z=1; xyz will be divisible by 8 AND yz will be divisible by 4.
Thus, we saw two different scenario where x may be an even or an odd.
Not sufficient.
2. x,y and z are all NOT divisible by 4.
This statement tells us that neither of x, y or z is divisible by 4. What does it mean?
It means;
x does not contain two 2's. It may contain 0 2's or 1 2's.
y does not contain two 2's. It may contain 0 2's or 1 2's.
z does not contain two 2's. It may contain 0 2's or 1 2's.
However, we already know that xyz contain 3 2's.
Combining both the stem condition and statement 2, we can conclude that each variable has one AND only one 2 in its factors.
x=2
y=2
z=2
OR
x=2*3*7*11*13*13*13
y=2*5*43
z=2
The point is x will contain exactly one 2 in its factor. And if x contains one 2 in its factors, it must be even.
Sufficient.
Ans: "B"