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10 Jan 2017, 03:50
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bazu
\(2*2 = (2+0)*(2 - 0) = 2^2 - 0^2\)
Of course, here we need both a and b to be positive integers so it won't work. Anyway, we need to pick 2 of the given 4 numbers. Their product will be more than 4.
ShashankDave
Joined: 03 Apr 2013
Last visit: 26 Jan 2020
Posts: 215
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Location: India
Concentration: Marketing, Finance
Schools: McDonough '21 (A) Kelley '21 ISB '21 (I)
GMAT 1: 740 Q50 V41
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13 Jul 2017, 00:34
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IanStewart
IanStewart As an extension to your method, this is how I did it. Please see if my reasoning is correct.
First of all
\(a^2-b^2 = (a+b)(a-b)\)
Here, both a and b are positive integers and so will be their sum or product.
For any product that can be rewritten in this manner, there has to be two integers (a+b) and (a-b) whose product the number will be.
Now, what happens if we add these two integers?
\((a+b) + (a-b) = 2a\)
Thus, For it to be possible for a number to rewritten in this form, it must be possible for the number to be written as a product of any two numbers whose sum is an even integer. In other words, the two multiplied numbers must be either both even or both odd. Taking examples.
2,5
2*5
There is no other method to write this product, and since the sum is 7, this doesn't count.
2,7
2*7
sum 9, doesn't count
2,8
2*8
sum 10, yes this counts.
5,7
5*7
sum 12, yes.
5,8
5*8
another way
10*4
sum 14, yes.
see where I'm getting at?
7,8
7*8
another way
14*4
sum 18, yes.
This out of 6 possible selections, 4 count as favorable.
Final answer
=\(\frac{4}{6}\)
==\(\frac{2}{3}\)
(A)
13 Jul 2017, 08:00
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ShashankDave
You're definitely on the right track if you want to prove exactly when it's possible to do this, but when you take examples, unless I've misunderstood what you're saying, I think you're assuming the numbers in the product are equal to a and b. But if you look, say, at this product:
10*4 = (7 + 3)(7 - 3)
then the values of a and b are 7 and 3, respectively, and not 10 and 4. That's why the conclusion you've written, "the two multiplied numbers must be either both even or both odd" is technically not right, as your example of 5*8 demonstrates (there we have one odd and one even, but we can still write the product as a difference of squares). Your conclusion is very close to being right, though - it's right if you rephrase it: "it must be possible to somehow write the product as a product of two even or two odd numbers".
You could modify your proof just slightly, and get a cleaner statement of when it is possible to do this. As you correctly point out, the sum of a+b and a-b must be even. So the two numbers a+b and a-b must either both be odd, or both be even. Thus the product (a+b)(a-b) is either odd, or it is the product of two even numbers, and is thus divisible by 4. And that's it: those are the two circumstances where it's possible to write a product of two integers as a difference of squares. Either our product must be odd, or it must be a multiple of 4. So we can do this in any of these situations:
product of two odds: e.g. 5*17 = (11 - 6)(11 + 6)
product of two evens: e.g. 14*24 = (19 - 5)(19 + 5)
product of one odd and one multiple of 4 (by moving one 2 from the even number to the odd number, to get two even numbers) : e.g. 11*16 = 22*8 = (15 + 7)(15 - 7)
and we can always do it in those situations, because when we have a product of two odds or of two evens, the median of our two numbers will always be an integer, and then we can always use the 'trick' above.
But we cannot do it in this situation:
product of one even number that is *not* divisible by 4 and one odd number: 14*5
because a^2 - b^2 must always be either odd or divisible by 4. There are a few other ways to prove that - for example, when a and b are both even, a^2 and b^2 are both divisible by 4, so a^2 - b^2 must be divisible by 4, and when a and b are both odd, you can just plug a = 2s + 1 and b = 2t + 1 into a^2 - b^2, and you'll see in that case a^2 - b^2 is also divisible by 4. So whenever a^2 - b^2 is even, it is always divisible by 4, which is what we established in a different way above.
