if x is a positive integer, is x^3 - 3x^2 + 2x...

[stevenstratmann]

if x is a positive integer, is x^3 - 3x^2 + 2x divisible by 4?
1. x = 4y + 4, where y is an integer

2. x = 2z + 2, where z is an integer

guide 1 p 98

1. makes sense and is sufficient because if 'y' was 0 then x would still be 4 which would make the integers 4,3,2 which is divisible by 4, guaranteed.

2. i am confused because if z is an integer of -0- then 'x' would be = 2...so the integers are 2, 1, 0 which is not divisible by 4. So the answer is no.

yes answers would be any other number. If 'z' was 1 then 'x' would be 4, 3, 2 which is divisible by 4.

Please explain. Thanks!



guide 1 p 98

[dmitryknowsbest]

z=0 still works because 0 is divisible by 4. To quote our definition of divisibility (page 13): "An integer is said to be divisible by another number if the integer can be divided by that number with an integer result (meaning that there is no remainder."

0/4= 0 with no remainder. Therefore 0 is divisible by 4. This general rule comes up frequently in odd/even problems. Zero is even because it is divisible by 2.

[stevenstratmann]

thanks for your response. Before I went into the response I redid the question to familiarize myself. As I was working the problem for question 2.) 0I noted that when 'z' is -0- x ='s 2, 1, 0 and in the stem it states 'x' is a positive number which -0- does not abide by. I am assuming this is another way to conclude that 'd' is the proper answer.

[dmitryknowsbest]

No, that doesn't work because z=0 yields x=2. It's the initial xpression (rephrased as (x(x-1)(x-2)) that would equal 0, and there's no rule against that.

0 is a valid value for z, and it doesn't change the answer one way or another.