For this week’s challenge, to keep the fun with factorials theme alive, I’ve come up with this little diabolical question. To add a little variety, I’ve made it a quantitative comparison question, which many forget can also make for a challenging question.

Good luck!

__Column A__16!^100 – 9!^100

__Column B__

(4!^100 – 3!^100)(4!^100 + 3!^100)

Check back on Thursday for the explanation!

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Chris, Can you please kindly post a video explanation answer on that and explain more thoroughly, I still didn’t get it? Thanks

Hi Shudda,

How’s this sound? Read some of the explanations below to see if they help. If that doesn’t work, click on explanation above. If that still doesn’t work, let me know which part of my explanation needs more elaboration, and I’d be happy to do just that 🙂

Although I couldn’t really get it, but still I’ll give it a go.

If we do some approximations on each side, we’ll be left with 16!^100 in column A and 4!^200 in column B.

Column A= 9!^100 * ((16*15*14*….*10)^100 – 1). Neglecting the 1 inside the brackets, column A ~ 16!^100.

Column B= 4!^200 – 3!^200 = 3!^200 * (4^200 – 1). Neglecting the 1 inside the brackets, column B~ 4!^200.

Now, comparing both columns, option A should be the answer (column A is greater than column B).

That definitely works–good job 🙂

On the right we can use (x-y)(x+y). And then take 3!^100 out of the braces, to get 3!^200(4^200 – 1). Since 1 is very small as compared to the other quantity, it can be ignored. So we get 3!^200(4^200), or 36^100(16^100)

On the left side take the 9! Term common and we get 9!^100(16*15*14*13*12*11*10)^100.

Cancel the common terms on each side and you be left with 1 on the right side and a pretty huge factorial term on the left.

So, in my opinion, it should be A.

Hi Anshg,

There are many paths to the correct solution–you’ve provided one I’ve yet to see 🙂

First: Collumn B is equi to (4!^100)² -(3!^100)² which can be formed into

(9*4)^100(16^100-1)

Collumn B is equi to 9!^100*((16!/9!)^100 -1).

Since both factors of A are larger then any of the factors in B. A has to be the greater quantity. Unfortunately this took me 15 minutes:(

This may be a poor strategy, but I just mimicked the problem to a simpler form,that is,

4!^2 – 1!^2 and 2!^4 – 1!^4 and found the first column to be considerably larger.

Hi Sriram,

I like the simplicity of your solution. It brings out the fact that 2x! is massively bigger than x!, so you can just discount the other stuff. Good job 🙂

Column A simplified to (9!)^100*((16*15*14*….*10)^100 -1)) and column B simplified to 36^100*((16^100)-1). From this it is clear that Column A is greater, so A should be the answer.

Yep, that’s a good way to do it :). Another student also got it using that method.

As we know that,

16! > (4!)^2,

Similarly 9! > (3!)^2

Hence, (16!)^100 > ((4!)^2)^100

So Column A { 16!^100 – 9!^100 } > Column B { ((4!)^2)^100 – ((3!)^2)^100 }

Notice the use of identity in Column B to simplify the expression, which is ,

(a-b)*(a+b) = (a^2 – b^2)

Good quick way to get the answer 🙂

Hi Chris,

If we compare a-b and c-d , if a>c and b>d, how can we conclude a-b > c-d ? This would make sense if a > c and b < d which is apparently not the case.

– Sriram.

Not a correct way to solve. You are basis your assumption on the analogy,

10 > 5

9 > 1

hence, (10-9) > (5-1)

So just a heads up you might get it right here, but that is not the right way to approach. 🙂

Yeah, I picked up on that as well–he didn’t explicitly spell out that the difference between 16! – 9!, but I’m thinking he was aware of that. At least it seemed so :).

So thanks for pointing that out, since it might lead others astray 🙂

(16*15*……*6*5)^100*(4^100) – (9*8*…*5*4)^100*(3!^100) ? (24^100)*(4!^100) – (6^100)*(3!^100)

Clearly LHS > RHS

Hence A is answer

That’s an interesting solution! Thanks for sharing 🙂