samirchaudhary wrote:
Dear Samir,
Your analysis still leaves a gap in my understanding of the problem as in explanation of answer choice D , you explain the difference between squares of consecutive integers part but then what about (p + q + 1). Same for 5 . Kindly explain and thanks.
Dear royQV,
(p + q + 1) always gives us an odd number and the point I wanted to make from the analysis is that any odd number (read odd number we get by adding p,q and 1) can be expressed as the difference of the squares of consc integers.
Also for the 5th part, though we get even numbers as well as the difference of squares, all such even number may not be express-able as the difference of the squares of integers. <Not very clear though.>
Dear Samir & RoyQV,
Yes, there are a few tricky number property shortcuts hidden in this problem. Appreciating these shortcuts could save time on a particularly challenging number property question on the test.
First of all, as Mr. Samir pointed out,
any odd number can be expressed as the difference of squares of consecutive integers. You see, if we know (n^2), then all we have to do is add (n), then (n + 1), and that will result in (n + 1)^2.
For example,
7^2 = 49
49 + 7 + 8 = 64 = 8^2
20^2 = 400
400 + 20 + 21 = 441 = 21^2
This can be a very handy trick for finding squares close to square you already know (e.g. a multiple of ten). It also means that, given any odd number, we can easily express the odd number as a difference of squares. Any odd number can be written as the sum of two consecutive integers. For example, 71 = 35 + 36. Well, 71 must be the difference between those two squares, because
(35^2) + 35 + 36 = (36^2)
(35^2) + 71 = (36^2)
Verify with a calculator that this pattern works for these numbers and for other numbers. That's the first pattern to know.
Now, the second pattern has to do with the difference of squares of two consecutive even numbers or two consecutive odd numbers. Those differences will always be multiples of 4. If you think about it algebraically, the difference between (n + 2)^2 = n^2 + 4n + 4 and n^2 is 4n + 4, which of course is divisible by four. Of course, all it takes is trying two odd prime numbers such as 3 and 5 to see that 3^2 + 5^2 = 9 + 25 = 34 is not something always divisible by 4, and therefore cannot be the difference between two odd or even consecutive numbers.
In fact, any odd number squared is a multiple of 4 plus 1, and any even number squares is a multiple of 4. Thus, the difference of any two odd squares or any two even squares would have to be a number divisible by 4. Again, it's good to check this with a calculator in hand until you verify for yourself that these patterns work.
Does all this make sense?
Mike