quadratic equations with perfect squares

[AnkurK544]

Hey Guys!

I created a flash card for myself sometime ago and now looking back, it does not make sense!!!

The question is : If n=(3^8) - (2^8) what are the factors of n?

My hint to myself was : Use perfect squares to create a quadratic equation then solve.

Please help.

Thanks

[RonPurewal]

well, you can't make any equations, since nothing here is an equation in the first place.
the point is to manipulate this expression, to get it into a different form.
same idea as "simplifying"; i wouldn't use that term here, though, since the final form won't be any simpler than the initial form.

the pattern that's going to come in handy here is...
this^2 – that^2 = (this + that)(this – that)
we can actually apply this three times in a row.

[RonPurewal]

3^8 – 2^8
= (3^4 + 2^4)(3^4 – 2^4)

(3^4 + 2^4)(3^4 – 2^4)
= (3^4 + 2^4)(3^2 + 2^2)(3^2 – 2^2)

(3^4 + 2^4)(3^2 + 2^2)(3^2 – 2^2)
= (3^4 + 2^4)(3^2 + 2^2)(3 + 2)(3 – 2)

= (81 + 16)(9 + 4)(3 + 2)(3 – 2)

= 97 • 13 • 5
... and those are your factors.
they're all prime. (it's easy to see that 13 and 5 are prime; you might need to try dividing 97 by a few numbers, such as 7, before deciding that it's also prime.)

[RonPurewal]

by the way—
you can use the pattern up to three times (as shown above), but, really, there's no need to use it more than once.

by the time you get here ...

= (3^4 + 2^4)(3^4 – 2^4)

... it's more efficient just to evaluate these:
(81 + 16)(81 – 16)
97 • 65
97 • 13 • 5
unless you are BOTH incredibly lightning-fast at algebra AND hideously slow at arithmetic—an unlikely combination—this is going to be less laborious than continuing with the factoring.

[RonPurewal]

and, finally (and most importantly)—
if you get stumped by something like this, IMMEDIATELY start using old-fashioned arithmetic.

it's really not that hard just to multiply out the powers, subtract them, and then factor.
annoying? sure.
incredibly time-consuming? nah.

3^8 is 3^4 • 3^4, or 81 • 81 = 6561.
2^8 is 2^4 • 2^4, or 16 • 16 = 256.
so, the difference here is 6561 – 256 = 6305.

this number is clearly divisible by 5, so start by doing that. if you divide 6305 ÷ 5, you get 1261.

now, just try dividing 1261 by odd numbers (or use divisibility tests on it, if you know them) until you strike gold.
• dividing by 3 gives a decimal
• the last digit is not 0 or 5, so dividing by 5 will clearly be a decimal (xxxx.2); no reason to try it
• dividing by 7 gives a decimal
• we already know it's not divisible by 3, so no point in trying to divide by 9
• dividing by 11 gives a decimal
• if we divide by 13, we'll find out that 6305/13 = 97, and then we're done.

yes, it's probably annoying to do this, but you can DEFINITELY do it within a couple of minutes—unless YOU tell yourself that you can't.

[RonPurewal]

also, apropos of the above, it's worth nothing that GMAC specifically engineers problems that work out "nicely" with long-hand arithmetic.

my all-time favorite example is #199 in the OG (13th or 2015 edition) multiple-choice section.
sure, there's a cute shortcut with powers of 10. but, if you don't IMMEDIATELY see how to do that, there's always your old friend, long division.

if you tell yourself "this will take too long", then you might chicken out of doing the long division.
if you just pick up the proverbial shovel and start digging, though, you'll be surprised at how little time it takes.

the point is, as usual, that there are 2 keys to success on this exam:
1/ do stuff.
2/ don't NOT do stuff.