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exponent problem

This topic has 1 expert reply and 6 member replies
ash_maverick Senior | Next Rank: 100 Posts Default Avatar
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exponent problem

Post Sun Jan 21, 2007 4:43 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the remainder when 43^43+ 33^33 is divided by 10?

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    chandra_adesh Newbie | Next Rank: 10 Posts Default Avatar
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    Post Sun Jan 21, 2007 8:21 am
    Last digit in the expansion of 43^43=7
    Last digit in the expansion of 33^33=3

    Last digit of 43^43+33^33=0

    Hence remainder=0

    g2000 Junior | Next Rank: 30 Posts Default Avatar
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    Post Sun Jan 21, 2007 3:35 pm
    i'm curious if it can be solved like below.
    The problem is to ask
    43^43 + 33^33 mod 10

    Modular Exponentiation
    X^a (mod n).

    43^43 mod 10
    First,
    43^1 mod 10 = 3
    43^2 mod 10 = 3^2 mod 10 = 9
    43^4 mod 10 = 9^2 mod 10 = 1
    43^8 mod 10 = 1^2 mod 10 = 1
    ....
    43^32 mod 10 = 1 mod 10 = 1

    43 = 32 + 8 + 2 + 1
    43^43 mod 10
    = 43^32 mod 10 * 43^8 mod 10 * 43^2 mod 10 + 43^1 mod 10
    = (1 * 1 * 9 * 3 ) mod 10
    = 27 mod 10
    = 7

    Same procedure is done on 33
    33^1 mod 10 = 3
    33^2 mod 10 = 3^2 mod 10 = 9
    33^4 mod 10 = 9^2 mod 10 = 1
    .....
    33^32 mod 10 = 1 mod 10 = 1

    33 = 32 + 1
    33^33 mod 10
    =33^32 mod 10 * 33^1 mod 10
    = (1 * 3 ) mod 10
    = 3 mod 10
    = 3

    Therefore, their sum must be 3 + 7 which gives the unit digit 0.
    The remainder is obviously 0.

    maxim730 Senior | Next Rank: 100 Posts Default Avatar
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    Post Sun Jan 21, 2007 5:20 pm
    chandra_adesh wrote:
    Last digit in the expansion of 43^43=7
    Last digit in the expansion of 33^33=3

    Last digit of 43^43+33^33=0

    Hence remainder=0
    how did you get the last digit? 8)

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    Stacey Koprince GMAT Instructor
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    Post Mon Jan 22, 2007 6:13 pm
    Whenever they ask you a "units digit" question with really high exponents, there will be a pattern and you will only have to follow the units digit through the problem (b/c it is all multiplication).

    43^43:

    3^1 = 3^2 = 3^3 = 23^4 = 83^5 = etc. so the units digit pattern is 3-9-7-1. The pattern repeats every 4th term. So 3^4, 3^8, 3^12, etc, will all have the units digit 1. 3^40 will be 1, 3^41 will be 3, 3^42 will be 9, 3^43 will be 7. Same pattern as above. 3^32 will be 1, 3^33 will be 3. units digit 7 + units digit 3 will equal units digit 0. Anything that ends in 0 will have a remainder of 0 when divided by 10.

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    telugupilla Newbie | Next Rank: 10 Posts Default Avatar
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    Post Fri Feb 02, 2007 2:00 pm
    Thanks for this cool tip!

    aim-wsc Legendary Member
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    Post Fri Feb 02, 2007 9:14 pm
    Sad thats my suicide note Laughing none but this problem is responsible for it. Twisted Evil

    haha i know such big numbers are justbubbles. i dealt with units 5 6 1 they are of course easy. but this with 3 as a unit must belong to ''tough'' category. Smile

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    limits660 Senior | Next Rank: 100 Posts
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    Post Sun Feb 04, 2007 6:36 am
    Stacey Koprince wrote:
    Whenever they ask you a "units digit" question with really high exponents, there will be a pattern and you will only have to follow the units digit through the problem (b/c it is all multiplication).

    43^43:

    3^1 = 3^2 = 3^3 = 23^4 = 83^5 = etc. so the units digit pattern is 3-9-7-1. The pattern repeats every 4th term. So 3^4, 3^8, 3^12, etc, will all have the units digit 1. 3^40 will be 1, 3^41 will be 3, 3^42 will be 9, 3^43 will be 7. Same pattern as above. 3^32 will be 1, 3^33 will be 3. units digit 7 + units digit 3 will equal units digit 0. Anything that ends in 0 will have a remainder of 0 when divided by 10.
    WOW, beautiful trick. I totally missed that one 8) Thanks Laughing

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