[DS]If a, b, and c are integers and abc ≠0

[anilbhat85]

If a, b, and c are integers and abc ≠ 0, is a2 - b2 a multiple of 4?

(1) a = (c - 1)2

(2) b = c2 - 1


I have read the explanation. By i approached this problem by putting values for c

For c = 0 and 1 i got a^2-b^2 as 0. Should we consider o as a multiple of 4 ?

0*4 = 0 ?

[mithunsam]

The fact is that you cannot pick 0 or 1 for c.

Question states abc≠ 0. We cannot ignore this statement.

If c is 0, then abc will become 0.

If c is 1, then
stmt 1 => a = 0. Which means abc = 0
stmt 2 => b = 0. Which again means abc = 0

That means, c cannot be either 0 or 1.

[tim]

good point about c not being 0. nevertheless, 0 is indeed a multiple of 4. it is a multiple of every integer, in fact..

[yousuf_azim]

What is the ANS?

[jnelson0612]

What is the ANS?

The answer is C.

[acchi369]

I have a question on this problem. If you were to multiply out statement 1, you would get:

a = (c-1)^2
a = (c-1)(c-1)
a = c^2 -2c + 1

So, if you were to plug that in for a in the original equation, you would get:

a^2 - b^2
(a+b)(a-b)
((c^2 -2c + 1) + b)((c^2 -2c + 1) -b)

So, without solving that equation, wouldn't you know that the term has to be divisible by 4 given that the middle term -2c is in there twice so 2c * 2c is 4c^2?

I am probably making the wrong assumption but I feel like that was enough information for me.

[jlucero]

You're missing the fact that you need to add/subtract all those variables first, before you multiply them together. If you FOILed that entire equation, you would get an absolute mess of an equation (4 terms multiplied by 4 terms = 16 total terms), but it would be much more complicated than 4c(some big equation).

((c^2 -2c + 1) + b)((c^2 -2c + 1) -b)

[explorer31]

Can we not simply substitute random integers for C and deduce a, b value and solve instead of algebra. I tried this during the CAT and did the algebraic way and spent too much time on it.What would be the best approach and how do we decide to go by 'Testing cases' strategy and not any other?

[RonPurewal]

there's no "best" approach.
at any point, if you know EXACTLY what you're doing AND EXACTLY why you're doing it... you should keep going.
if you don't... you should stop.

that's basically all there is to time management on this exam.

I tried this during the CAT and did the algebraic way and spent too much time on it.
What would be the best approach and how do we decide to go by 'Testing cases' strategy and not any other?

the prescription here is very simple -- although you need a solid dose of self-discipline (and an awareness that there is no "partial credit"!) to do what you need to do consistently.

basically, the deal is this:

if you're doing algebra, and you are 100.0000% sure that you understand EXACTLY WHAT you're doing, and EXACTLY WHY you're doing it... keep going.

if you are LESS than 100% sure... DO NOT KEEP GOING. because at that point you're basically just doing random steps -- which is what you'd do in school in the hopes of getting "partial credit", but, "partial credit" is not a thing here.

thought experiment:
imagine that you're building some furniture out of VERY expensive, VERY rare wood.
think about how conservative you would be about drilling holes in this wood: you would ONLY drill holes if you were 100.000000% sure you were drilling them in EXACTLY the right places (and also understood WHY the holes go in those places, in terms of the finished product).
...otherwise, you wouldn't drill the holes.

you should approach algebra on this exam just as conservatively.

...BUT... if you DO know exactly what you're doing, and exactly why... then DON'T HESITATE! just DO it!
even if you end up writing out (c^2 – 2c + 1)^2 and expanding it out, and then writing out (c^2 – 1)^2, expanding it, and subtracting those terms out... this won't take THAT long. the former expansion has six products; the latter has four. that's definitely not a crazy number of steps -- unless YOU talk YOURSELF into thinking that it is!

[RonPurewal]

Can we not simply substitute random integers for C and deduce a, b value and solve instead of algebra.

^^ yes, although you need to try enough cases to be confident that you've found a pattern.

since this problem is about divisibility by 4, you can construct a proof by trying any 4 consecutive integers for "c" (as long as they're allowed -- i.e., not 0 or 1).
with regard to divisibility by 4, all the multiples of 4 (4, 8, 12, 16, etc) act the same way; all the multiples of 4 plus one (1, 5, 9, 13, etc) act the same way; all the multiples of 4 plus two (2, 6, 10, 14, etc) act the same way; and all the multiples of 4 plus three (3, 7, 11, 15, etc) act the same way.
the other variables are both functions of "c". so, if you try 4 consecutive integers for "c", you're actually covering ALL possible cases with regard to divisibility by 4.

[explorer31]

Perfect! I knew there had to be an easier strategy than algebra and that is trying 4 consecutive integers for C for testing the divisibility of 4! This is going on my Flash cards for sure! Thank you very much.

[RonPurewal]

Perfect! I knew there had to be an easier strategy than algebra and that is trying 4 consecutive integers for C for testing the divisibility of 4! This is going on my Flash cards for sure! Thank you very much.

you're welcome.
BUT just remember that the most important thing here is NOT to just try to memorize such incredibly specific things.
keep in mind -- the likelihood that you'll actually get a problem specifically about divisibility by 4 AND consecutive integers is ... very, very small. so, there isn't much utility in trying to commit these sorts of specifics to memory.
if you have an AMAZING memory -- in other words, if it's literally zero effort for you to remember something like this -- then, sure. but otherwise you'll be much better served by focusing on the actual point of these kinds of exercises, which is to INVESTIGATE situations by AGGRESSIVELY TESTING NUMBERS.

in other words -- unless you have an absolutely rigorous, foolproof approach using theory, you should just start throwing numbers at the situation... early and often.
wherever there are PATTERNS, they will emerge very quickly -- because this test is designed to reward investigation and pattern recognition pretty much above all else. on top of that, this test is ALSO designed so that you don't need to KNOW much stuff to get the highest scores.

so, just throwing a whole bunch of numbers at the problem, and carefully analyzing the results, is worth more than trying to memorize hundreds or even thousands of these super-specific case-by-case rules.

[explorer31]

Agreed ! Thank you for pointing that out Ron!

[RonPurewal]

you're welcome.