Question: Consecutive Integers p, q, r and s

Comment on Consecutive Integers p, q, r and s

Hi Brent,
Your videos are highly insightful. Please, why did you not represent p,q, r and s with consecutive integers, eg, 1,2,3,4 respectively,& simplify. I did that and I got 1 also. It seems faster.
greenlight-admin's picture

If we choose only one set of values (e.g., 1, 2, 3, and 4 to represent p, q, r and s) then we risk the possibility not handling every possible case.

For example, IF the expression for Quantity A were something like (p+1)(q+2)(r+1)(s+2), then just plugging in 1, 2, 3, and 4 to represent p, q, r and s would suggest that Quantity B is greater. However, if we ALSO let 2, 3, 4 and t represent p, q, r and s, then we'd see that the correct answer is actually D.

Now, for the original question, it happens to be the case that plugging in ANY 4 consecutive numbers will always yield the correct answer of C. So, your approach works. However, if you take a more general approach, then you can add more certainty to your response.

Will odd number divided by 2 always yield remainder 1?
greenlight-admin's picture

Yes, that's correct.

In one of the videos, you explained that 2K + 1 is odd. Does that apply here, i.e., taking the '1' as the remainder?. Overall, thank you for the amazing product.
greenlight-admin's picture

That's correct!

If k is an integer, we know that 2k must be divisible by 2.
So, when we divide 2k by 2, the remainder is 0.

Since 2k+1 is ONE GREATER THAN 2k, we know that we'll get a remainder of 1 when we divide 2k+1 (aka an odd number) by 2.

PS: Thanks for the kind words about the course!

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