Question 24 Not yet swered If m, n, and p are positive...

Question 24 Not yet swered If m, n, and p are positive integers such that m+n is even and the value of $(m+n)^2 + n + p$ is odd, which of the following must be true? A. m is odd Points out of 1 B. p is odd C. n is even D. If n is even, p is odd E. If p is odd, n is odd

Transcript

00:01 Say that we are given three distinct odd integers p, q, and r.

00:07 We can represent these as follows p equals 2x plus 1, q equals 2y plus 1, r equals 2z plus 1 for some integer x, y, and z.

00:19 Right? this might help later on when we're starting to look at these different statements.

00:26 So which of the following is not always true? first we have the statement that p times q squared times r cubed is odd.

00:38 It's important to note here that an odd integer times an odd integer will always give us an odd integer.

00:48 Since these are all odd integers then we can say that this product is indeed odd.

00:55 Q squared is odd, p times q squared is odd, r cubed will be odd, and so on.

01:02 Next we have the statement that p plus q squared times r cubed is even.

01:10 So here we'll use our representation 2x plus 1, 2y plus 1 to evaluate what p plus q is.

01:18 2x plus 1 plus 2y plus 1.

01:22 So that's the same as 2x plus 2y plus 2, which that is even because they all have the common factor of 2.

01:31 You can also say that an odd integer plus an odd integer is even, but it is nice to see it algebraically.

01:43 Therefore this inner sum is even and we are taking the square of that.

01:48 So we have an even times an even which is even.

01:55 Therefore we are taking this even integer and multiplying it by the cube of an odd integer which we found was odd in the last problem and even times an odd is always even.

02:11 That's because an even has the factor of 2, so when you're multiplying that by any number it's still going to withhold its factor of 2.

02:20 So this statement is also true.

02:26 Let's move on.

02:27 We now have p minus q plus r squared times q plus r is even.

02:34 So if we want to look at this algebraically like we did the last time, we have 2x plus 1 minus 2y plus 1 plus 2z plus 1.

02:45 That is going to give us 2x minus 2y plus 2z, 1 minus 1 plus 1 plus 1.

02:54 That plus 1 shows us that this is odd.

03:01 So this sum squared is an odd times an odd which is odd.

03:12 Next we have q plus r and as we found previously an odd plus an odd is even.

03:24 Therefore we have an odd times an even which is going to give us an even...

Question

Question 24 Not yet swered If m, n, and p are positive integers such that m+n is even and the value of $(m+n)^2 + n + p$ is odd, which of the following must be true? A. m is odd Points out of 1 B. p is odd C. n is even D. If n is even, p is odd E. If p is odd, n is odd

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