P Cubed And Eighty (Advanced Quant Problems Bank)

[AlexandraM39]

Hello,

I have re-read the provided solution for this problem several times but still don't get it. If anyone could explain it differently, that would be appreciated.

Thanks in advance!

If p 3 is divisible by 80, then the positive integer p must have at least how many distinct factors?

Answers are: 2,3,6,8,10
Correct Answer is 6.

Provided solution:
The prime factorization of 80 is (2)(2)(2)(2)(5) = 2 45 1. Thus, p 3 = 2 45 1 x, where x is some integer.

Assigning the factors of p 3 to the prime boxes of p will help us see what the factors of p could be.

(There is a chart here showing p 3 times and it's possible factors)

The prime factors in ( ) above are factors not explicitly given for p 3, but which must exist. We know that p 3 is the cube of an integer, and must have “triples” of the prime factors of p. Since p 3 has a factor of 2 3, p must have a factor of 2. The fact that p 3 has an “extra” 2 and a 5 among its factors indicates that p has additional factors of 2 and 5.

If p is a multiple of (2)(2)(5) = 20, then at the very least p has 1, 2, 4, 5, 10, and 20 as factors. So we can conclude that p has at least 6 distinct factors.

Alternatively, we can use this shortcut for computing the number of factors:
(2’s exponent + 1)(5’s exponent + 1) = (2 + 1)(1 + 1) = (3)(2) = 6.

The correct answer is C.

[RonPurewal]

Hi,
Please give us some specifics here.

What do you already understand?
What don't you understand?
Where in that explanation are you getting stuck?
Etc.

Thanks.

[AlexandraM39]

Thanks for your reply. Someone else has explained it to me and so I understand it now, however, I am still having trouble understanding the logic behind the short cut provided at the end. Does this hold in all cases (it does in all the cases I tested but just to be sure)? Is it something that I should just memorize or can it be explained further.

Cheers,
Ally

[RonPurewal]

Thanks for your reply. Someone else has explained it to me and so I understand it now, however, I am still having trouble understanding the logic behind the short cut provided at the end. Does this hold in all cases (it does in all the cases I tested but just to be sure)? Is it something that I should just memorize or can it be explained further.

Cheers,
Ally

If you use that with the exponents from a prime factorization, then it should work.

Here's an illustration. Say you're looking at factors of 120 (= 2^3 x 3^1 x 5^1).
2^3 actually gives you 4 choices for "How many 3's in my factor?": none, one, two, or all three of them.
3^1 gives you 2 choices for "How many 2's": none or one.
Same for 5^1.
So you'll have 4 x 2 x 2 = sixteen different factors.

[RonPurewal]

On the other hand, I personally don't see the point in memorizing "cute shortcuts" like this. To me, they're just mental clutter.
If I had to count the factors of 120, I'd just count them. Literally. I'd just start with 1 x 120, and break 120 up into as many products as possible. Once the smaller and bigger numbers "meet", I'll have all the factors.
1 x 120
2 x 60
3 x 40
4 x 30
5 x 24
6 x 20
8 x 15
10 x 12
Sixteen factors.
That was ... well, not painful. Not even a little bit.

[RonPurewal]

But still-- When you think about "cute shortcuts", consider the costs and benefits. Then make an economic decision, like basically any other economic decision.

You should measure "costs" in terms of effort. (Full disclosure: I have a very bad memory-- in fact, documented memory disorders, originating in childhood injuries-- so I am somewhat biased here.)

"Benefits", you should measure in terms of time saved.
This is the reason why I just don't see very many of these shortcuts as anything worth bothering with. How much time would you save by using this one? 5 seconds, maybe 10 seconds at the most?
Not much, for sure. Given that-- along with the absolutely microscopic chance that you'll even get to use it in the first place-- my decision is pretty clear: Not worth it.

On the other hand, there are psychological factors too.
I've had students who just feel more confident if they've accumulated all sorts of random mental clutter.
And ... feeling confident is, well, important. For a test like this one-- on which the actual amount of required knowledge is fairly minimal-- it's just as important as your actual preparation.
So, if this is you, then you might benefit from "the placebo effect", even if you never use the shortcuts at all.

Me, I'm the other way around. Mental clutter makes me less confident. I like to keep things simple. Like a fourth-grader.