Question

If x is an odd integer, y and z are whole numbers, and x(y+z)+x is odd, which of the following must be true?

Answer 1

z is even
z is odd
y and z are either both odd or both even
if y is odd, then z is even
if y is even, then z is odd

Answer 2

Well we know the product of an odd and an even gives an even, and the product of an odd with an odd gives us an odd. We know that the sum of two odds is even, and the sum of an odd and an even gives us an odd.

Answer 3

Since \(x\) is odd and \(x+x(y+z)\) is also odd we can infer \(x(y+z)\) must then be even.

Answer 4

y and z are either both odd or both even.

Answer 5

that was a joke

Answer 6

If \(x(y+z)\) is even, since we know \(x\) is odd, we can infer \(y+z\) must be even. When is the sum of two numbers even? Well when either both are odd or both are even! Therefore we know \(y,z\) are either both odd or both even.

Answer 7

i mean that wasn't a joke

Answer 8

@oldrin.bataku I have question: x(y+z) +x = x((y+z+1)
x odd * whatever = odd, right? so the answer is not one of the options

Answer 9

no; an odd multiplied with an even is even

Answer 10

yes, my bad.

Answer 11

so, y +z +1 must be odd y +z must be even iff y and z both odd or both even, right?

Answer 12

Consider an odd number \(2m+1\) for integer \(m\) and an even number \(2n\) for integer \(n\):$$2n(2m+1)=2n\cdot2m+2n=4nm+2n=2(2nm+n)$$ which is divisible by \(2\) and is therefore even.

Answer 13

another question: we discuss and give out the answer only, does it a real help?

Answer 14

@Loser66 if \(y+z+1\) is odd then \(y+z\) must be even. If \(y+z\) is even then \(y,z\) must either be even or odd.

Answer 15

Consider arithmetic \(\mod 2\); an even number is congruent to \(0\) and an odd number congruent to \(1\). We have \(0+0=0\pmod2\) and \(1+1=0\pmod2\)

Answer 16

like it. it's more professional,

Answer 17

:-p and well I mean our discussion is public! hopefully OP will read our replies