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Magoosh Brain Twister: Zeroing In on the Answer

brain_twister

This week’s Brain Twister is a numeric entry question. Ready to give it a go?
 

30^30 x 29^29 x 28^28…2^2 x 1^1 = n.

How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

Numeric Entry [________________]

 

Wondering how you fared? Check back on Thursday for the answer and explanation! 🙂

 

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27 Responses to Magoosh Brain Twister: Zeroing In on the Answer

  1. Mohammad July 15, 2017 at 11:48 pm #

    30^30 yields 30 zeros
    20^20 yields 20 zeros
    10^20 yields 10 zeros

    Total=60 zeros

    The other numbers which when multiplied gives a 10 are 2 and 5. There are plenty of ‘2s’ there; wise would be to count ‘5s’.

    5^5 yields five 5s
    15^15 yields fifteen 5s
    25^25 yields twenty five 5s

    All together there are seventy 5s. There are definitely seventy 2s there in the product.

    Total=70 zeros from this product

    TOTAL = 70+30=130

    • Mohammad July 15, 2017 at 11:49 pm #

      Sorry a small mistake 10^10 yields 10 zeros

      • Magoosh Test Prep Expert
        Magoosh Test Prep Expert July 17, 2017 at 4:07 pm #

        Hi Mohammad,

        You are exactly correct here! Great job! Your work almost exactly aligns with our explanation. Keep it up, and have a great day! 🙂

  2. pavan November 27, 2014 at 8:58 am #

    hi chris,
    105 is answer but it’s much time consuming question.can you give us short cut for this type of question?

    A garment order consists of jackets costing $36 each and shirts costing $26 each. If the total
    cost of the order is $1,200, and if the average (arithmetic mean) cost per garment is $30, how
    many more shirts than jackets are in the order?

    this kind of question also take much time

    Regards,
    pavan.

    • pavan November 27, 2014 at 10:58 am #

      130 but to deal with problem take lot of time in exam.

      • Chris Lele
        Chris Lele December 4, 2014 at 11:27 am #

        Hi Pavan,

        Once you see that you are just counting zeroes (30 zeros in 30, 20 zeroes in 20, etc. and realizing that you don’t need to count any ‘2’s just the fives, 50 fives in 25^25, 15 fives in 15, etc.), this problem can be done in less than two minutes.

        Of course, that’s no easy feat!

  3. pavan November 27, 2014 at 7:58 am #

    I thought 90 would be

    • Chris Lele
      Chris Lele December 4, 2014 at 11:30 am #

      The answer is 130. The explanation is up on the blog 🙂

  4. Olha November 26, 2014 at 8:59 am #

    63 zeros:

    1) 0 in power of 30 yields 30 zeros
    2) 20 ^20 – 20 zeros
    3) 10 ^10 – 10 zeros
    now you have 60 zeros

    4) 5^5 (a number that ends in 5), multiplied by an even number (2^2 or 4^4) – 1 zero
    5) same with 15^15 multiplied by an even number (e.g.14^14) – 1 zero
    6) and 25 ^ 25 multiplied by an even number – 1 zero

    Total 63 zeros

    • Chris Lele
      Chris Lele December 4, 2014 at 11:33 am #

      Olha,

      The part with the zeroes is correct. With the ‘5’s, though, each five (meaning the five 5’s in 5^5) can be multiplied by some even number–and there are plenty of those spread throughout this sequence of numbers. So you want to count all the fives. There are 80 of those, so the answer is 130.

      Hope that helps!

  5. Shanthi November 25, 2014 at 11:35 pm #

    3^30 2^20 10^10 are seemingly he only candidates that could give rise to trailing zeros.

    (3×10)^30
    (2×10)^20
    (1×10)^10

    powers of 3 are 3, 9, 12, 15, 18, 27
    powers of 2 are 2, 4, 8, 16, 32, 64,128
    any power of 1 is 1

    Therefore we can ignore 3^30, 2^20 and 1^10.

    So what we have left is 10^30, 10^20 and 10^10.

    So I am guessing the answer is 30 + 20 + 10 = 60.

    • Chris Lele
      Chris Lele December 4, 2014 at 11:34 am #

      What about the ‘5’s 🙂

      Any 5 multiplied by an even number will give you a zero.

      See if you can get the answer now 🙂

  6. Pranav November 25, 2014 at 11:33 pm #

    60

    • Chris Lele
      Chris Lele December 4, 2014 at 11:34 am #

      You may have forgotten to count the ‘5’s 🙂

      Also, the explanation is now up!

  7. Napoleo November 25, 2014 at 11:20 am #

    In this case we have to find the pair (2*5) in the above sequence and we can break each number separately for example:
    30^30=(2*5*3)^30
    20^20=(2*5*2)^20
    10^10=(2*5)^10

    At this time we have obviously 60 zero’s however we can create more pairs of (2*5) especially the number of five’s it will determine the number of zero’s due to the number of two’s is bigger* :
    25^25=(5*5)25——->50 five’s
    15^15=(3*5)^15——>15 five’s
    5^5—->5 five’s
    Eventually if we add how many number we have, then we will find that we have 130 which can much with 130 2’s in order to fix the pair (2*5)

    *(Each even number is a multiple of two with something on this case you have 2^30,2^28,2^26,2^24,2^22,2^20,2^18,2^16,…….2^2)——>240 two’s however you will need only 130 from the total of 240.

    So, we have 130 zero’s!!!!

    • Chris Lele
      Chris Lele December 4, 2014 at 11:36 am #

      Perfect explanation…but when working this problem out in real time, you don’t want to count out every ‘2’. Instead, you should have the realization that there will be a superfluity of ‘2’ spread throughout the sequence with which to match with the ‘5’s.

      Again, good job 🙂

  8. hasan November 25, 2014 at 2:44 am #

    30

    • Chris Lele
      Chris Lele December 4, 2014 at 11:36 am #

      Don’t forget 20^20 and 10^10…the explanation is now up 🙂

  9. Vishal November 24, 2014 at 9:41 pm #

    105!
    5+10+15+20+25+30=105

    • Chris Lele
      Chris Lele December 4, 2014 at 11:37 am #

      Close, but I think you forgot that extra batch of 25’s in 25^25. That would give you a total of 130 zeroes. BTW, the full explanation is now up on our blog 🙂

  10. Jenil November 24, 2014 at 12:11 pm #

    130

    We have to look for the total number of factors of 2 and 5 (10 = 2*5).
    Since the factors of 5 is the most restrictive, once we know the number of factors of 5 we can easily determine the total number of zeros at the end

    (5)^5 (5 factors of 5)
    (10)^10 = (2 * 5)^10 = (2^10)*(5^10) (10 factors of 5)
    (15)^15 = (3*5)^15 = (3)^15 * (5)^15 (15 factors of 5)
    (20)^20 = (2*2*5)^20 = (2^2)^20 * (5)^20 = (2^40) * (5)^20 (20 factors of 5)
    (25)^25 = (5^2)^25 = (5)^50 (50 factors of 5)
    (30)^30 = (2*3*5)^30 = (2)^30 * (3)^30 * (5)^30 (30 factors of 5)

    Total no of factors of 5 = 5 + 10 + 15 + 20 + 50 + 30 = 130

    Similarly we can calculate the number of factors of 2 (which suffice 130) could be greater

    So total number of zero’s at the end = 130

    • Olha November 25, 2014 at 10:06 am #

      You start from 5, but what about factors of 2 and 4? Also, the product of even numbers and 5, 15 etc, will yield 0 as the last digit.

      • Chris Lele
        Chris Lele December 4, 2014 at 11:39 am #

        I think what Jenil is doing is matching the 5’s with 2’s, since 5 x 2 = 10. Thus each 5 x 2 pairing will result in an extra ‘2’.

        Hope that makes sense 🙂

  11. Samy November 24, 2014 at 11:08 am #

    I just want to confirm one thing — are those x’s in between the numbers or the multiplication sign? Assuming the latter.

    • Jenil November 25, 2014 at 9:47 am #

      Even I had that the same doubt..hover over the image the “alt” attribute shows multiplication

    • Chris Lele
      Chris Lele November 25, 2014 at 10:25 am #

      Hi Samy,

      Yes, those are multiplication signs 🙂

    • Chris Lele
      Chris Lele December 4, 2014 at 11:39 am #

      They are multiplication signs 🙂


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