A basic 2-equation GMAT/GRE Problem

If `x+y=a` and `x-y=b`, then `2*x*y=`
(A) `(a^2-b^2)/2`
(B) `(b^2-a^2)/2`
(C) `(a-b)/2`
(D) `(a*b)/2`
(E) `(a^2+b^2)/2`

A classic problem that I see all the time on both the GMAT and GRE, take a look at one way to solve it:
[spoiler below]




we start out with two equations:
1: `x+y=a`
2: `x-y=b`
after squaring both sides of (1):
3: `(x+y)^2=x^2 + 2*x*y + y^2 = a^2`
after squaring both sides of (2):
4: `(x-y)^2=x^2 -2*x*y+y^2=b^2`
now lets subtract (4) from (3) to get:
5: `(x^2 + 2*x*y + y^2) - (x^2 -2*x*y+y^2) = 4*x*y = a^2 - b^2`
now if we divide both sides of (5) by 2 we get:
6: `2*x*y=(a^2-b^2)/2`

which is answer choice (A)

second smallers integer?

What's the second smallest integer with a units digit of 5, which is the square of an integer, and the cube of an integer?

[Spoiler Below]


Video:

Text:
Let's call the number we are looking for `n`.

we know it's the square of an integer, so:
`n = "int"_a^2`
and we know it's the cube of an integer, so:
`n = "int"_b^3`

since any integer can be prime-factorized, let's consider the prime factorization of `"int"_1` and `"int"_2`:
`"int"_a = "prime"_1^("int"_(a1)) * "prime"_2 ^("int"_(a2)) ...`

`"int"_b = "prime"_1^("int"_(b1)) * "prime"_2 ^("int"_(b2)) ...`

`n = ("int"_a)^2 = ("prime"_1^("int"_(a1)) * "prime"_2^("int"_(a2))...)^2 = "prime"_1^(2*"int"_(a1)) * "prime"_2^(2*"int"_(a2))...`

and

`n = "int"_b^3 = ("prime"_1^("int"_(b1)) * "prime"_2^("int"_(b2))...)^3 = "prime"_1^(3*"int"_(b1)) * "prime"_2^(3*"int"_(b2))...`


so `"prime"_1^(2*"int"_(a1))*"prime"_2^(2*"int"_(a2))... = "prime"_1^(3*"int"_(b1))*"prime"_2^(3*"int"_(b2))...`
and so we can say:
`"prime"_x^(2*"int"_(ax)) = "prime"_x^(3*"int"_(bx))` for all x.
and therefore: `2*"int"_ax = 3*"int"_bx` this is the number of `"prime"_x` in the prime factorization of `n`.
since it has a factor of `2` and `3`, then the number of `"prime"_x` in `n` must be a multiple of 6.

to summarize, for any prime factor of `n`,there are a multiple of 6 of them in `n`.


we also know that the units digit of `n` is `5`, which means it has a factor of `5`, so:
`n = 5 * "int"_3`

so `n` has at least one `5` in it's prime factorization, and consequently it must have at least 6.

so `5^6` is the least integer that fits the description, but we are looking for the 2nd smallest.

well `5^12` would fit the description, and it's not the smallest, but maybe it's not the 2nd smallest,

`5^6 * 2^6` is worth considering, but `5^6 * 2^6 = 10^6`, which is one million, which doesn't have a `5` as the units digit.  This does not fit the description in the problem.

`5^6 * 3^6` fits the description and is the second smallest integer that does.
`n = 5^6 * 3^6 = 11390625`