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Magoosh Brain Twister: Variable Cloudiness

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It’s Brain Twister week again! Get ready to train your brain.
 

If a – j are distinct positive integers, what is the units digit of the lowest possible value of a^b x c^d x e^f x g^h x i^j?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 8

 

Good luck, and don’t forget to check back on Thursday for the answer and explanation. 🙂

 

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13 Responses to Magoosh Brain Twister: Variable Cloudiness

  1. Katrina kaif March 6, 2015 at 11:39 pm #

    For lowest possible value the answer is to allocate lowest power value to get in turn least porduct
    ie 9^2*8^3*7^4*6^5*1^10. (10^1 not used so as to get the lowest value -1<100)

    the answer comes out to be 4.

    • Chris Lele
      Chris Lele March 9, 2015 at 11:44 am #

      Hi Katrina,

      Your logic seems correct, but the final answer should be (C) 2. What final step did you use to get to 4?

  2. Christine Fahey March 4, 2015 at 4:48 pm #

    0 is the units digit of the lowest possible value

    Lowest possible value = 10^1 * 9^2 * 8^3 * 7^4 * 6^5 * 5^6
    Multiply the units digit of each of these pairs, then get units digit of that product
    However, 10^1 = 10 –> units digit = 0
    0*anything = 0
    Therefore, the answer is 0 (without having to multiply to get the actual lowest possible value)

    • Christine Fahey March 4, 2015 at 4:53 pm #

      Oops – I added an extra pair!
      First sentence should read:
      Lowest possible value = 10^1 * 9^2 * 8^3 * 7^4 * 6^5

    • Chris Lele
      Chris Lele March 5, 2015 at 2:42 pm #

      Hi Christine,

      You are very close, except for one little twist :). It should be 1^10 not 10^1, since 1 is less than 10. So, zero will not be amongst any of the units digits. So there’ll actually be a little more grunt work required to get the answer.

      Hope that helps!

  3. Jenil March 3, 2015 at 6:33 am #

    a – j -> distinct positive integers

    #) Lowest possible value of –

    a^b x c^d x e^f x g^h x i^j

    a=1
    b= any positive integer

    a^b = 1

    c=2
    d=3
    c^d = 2^3 = 8

    e=5
    f=4
    e^f = 5^4 = 625

    g=7
    f=6
    g^f = 7^6 =117,649

    i= 9
    j= 8
    i^j = 9^8

    #) Units Digit

    1^any value x 2^3 x 5^4 x 7^6 x 9^8
    Units digit – 1 x 8 x 5 x 9 x 1 = 360 = 0

    Ans : (A) = 0

    • Chris Lele
      Chris Lele March 4, 2015 at 11:22 am #

      Hi Jenil,

      Your works with the units digit is correct, but you want to revisit how you arranged the variables. There’s a little logic twist here (remember the question is asking for the “least integer”).

      Hope that helps 🙂

  4. aditya March 2, 2015 at 6:14 pm #

    6

    • Chris Lele
      Chris Lele March 4, 2015 at 11:26 am #

      Hmm…how did you get ‘6’? Can you show the steps you took to get there? 🙂

  5. keveri March 2, 2015 at 12:00 pm #

    I think the lowest possible value is 1^10*9^2*8^3*7^4*6^5.

    I don’t know where I would go from here to get units digit besides doing the calculation.
    Is there an easier method?

    • Chris Lele
      Chris Lele March 4, 2015 at 11:25 am #

      Hi Keveri,

      That is correct in terms of the way the variables should be arranged (points for logic :))

      As for the unit digits, you only have to consider the last number. For instance, 9^2 = 1 and 6^5 = 5. Multiply the results of those five pairs of numbers, focusing on the units and that’ll get you the right answer 🙂

      • Keveri March 4, 2015 at 6:51 pm #

        That makes sense! However, I think you meant to say that the units digit of 6^5 = 6

        And since 6^2=36, this gives a unit digit of 6. So for any power that 6 is raised to,the units digit will always be 6.

        For example,

        6^3= 36 * 6
        6^4= _ _ 6 * 6

        Thanks for the informative answer Chris.

        -Keveri

        • Chris Lele
          Chris Lele March 5, 2015 at 2:35 pm #

          Yes, that’s correct with 6^x always ending in a ‘6’. Sorry for any confusion 🙂


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