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Magoosh Brain Twister: Get Your Factorials Straight!


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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18 Responses to Magoosh Brain Twister: Get Your Factorials Straight!

  1. Shuddha August 6, 2014 at 12:12 am #

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

    • Chris Lele
      Chris Lele August 6, 2014 at 11:26 am #

      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 🙂

  2. Siddharth Jain July 29, 2014 at 10:10 am #

    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).

    • Chris Lele
      Chris Lele July 29, 2014 at 1:04 pm #

      That definitely works–good job 🙂

  3. anshg July 29, 2014 at 8:10 am #

    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.

    • Chris Lele
      Chris Lele July 29, 2014 at 1:08 pm #

      Hi Anshg,

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

  4. Jiyo Hebe July 28, 2014 at 12:17 pm #

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

  5. Sriram July 28, 2014 at 12:11 pm #

    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.

    • Chris Lele
      Chris Lele July 29, 2014 at 1:13 pm #

      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 🙂

  6. Shahadat Hasan REZA July 28, 2014 at 11:55 am #

    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.

    • Chris Lele
      Chris Lele July 29, 2014 at 1:14 pm #

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

  7. Muhammad Zubair July 28, 2014 at 11:39 am #

    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)

    • Chris Lele
      Chris Lele July 29, 2014 at 1:16 pm #

      Good quick way to get the answer 🙂

      • Sriram July 29, 2014 at 1:45 pm #

        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.

    • anshg July 29, 2014 at 2:39 pm #

      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. 🙂

      • Chris Lele
        Chris Lele July 30, 2014 at 10:52 am #

        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 🙂

  8. Guri Kejriwal July 28, 2014 at 9:58 am #

    (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

    • Chris Lele
      Chris Lele July 29, 2014 at 1:19 pm #

      That’s an interesting solution! Thanks for sharing 🙂

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