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rattan123 Just gettin' started!
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remainder Post Sat Apr 14, 2012 8:26 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7

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    killer1387 GMAT Destroyer! Default Avatar
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    Post Sat Apr 14, 2012 9:07 am
    rattan123 wrote:
    What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
    ITS 0.

    seal4913 Really wants to Beat The GMAT! Default Avatar
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    Post Sat Apr 14, 2012 10:21 am
    killer1387 wrote:
    rattan123 wrote:
    What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
    ITS 0.
    That's helpful to him I'm sure... if he's asking i'm sure he wants an approach to how to figure it out.

    Shalabh's Quants Really wants to Beat The GMAT!
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    Post Sat Apr 14, 2012 11:57 am
    rattan123 wrote:
    What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
    => (2^100+3^100+4^100+5^100)/7

    => (2.2^99 + 3.3^99 + 2^200 + 5.5^99)/7

    => [2*(2^3)^33 + 3*(3^3)^33 + 2^2*(2^3)^66 + 5*(5^3)^33]/7

    => [2*8^33 + 3*27^33 + 2^2*8^66 + 5*125^33]/7

    => [2*(7+1)^33 + 3*(28-1)^33 + 2^2*(7+1)^66 + 5*(126-1)^33]/7

    => Taking remainders, [2*1^33 + 3*(-1)^33 + 2^2*1^66 + 5*(-1)^33]/7

    => [2+ 3*(7-1)+ 4 + 5*(7-1)]/7; For [(-1)^n]/m; if n is odd, remainder is m - n.

    => [2 + 3*6 + 4 + 5*6]/7

    => 54/7 => Remainder is 5.

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    seal4913 Really wants to Beat The GMAT! Default Avatar
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    Post Sat Apr 14, 2012 12:17 pm
    Shalabh's Quants wrote:
    rattan123 wrote:
    What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
    => (2^100+3^100+4^100+5^100)/7

    => (2.2^99 + 3.3^99 + 2^200 + 5.5^99)/7

    => [2*(2^3)^33 + 3*(3^3)^33 + 2^2*(2^3)^66 + 5*(5^3)^33]/7

    => [2*8^33 + 3*27^33 + 2^2*8^66 + 5*125^33]/7

    => [2*(7+1)^33 + 3*(28-1)^33 + 2^2*(7+1)^66 + 5*(126-1)^33]/7

    => Taking remainders, [2*1^33 + 3*(-1)^33 + 2^2*1^66 + 5*(-1)^33]/7

    => [2+ 3*(7-1)+ 4 + 5*(7-1)]/7; For [(-1)^n]/m; if n is odd, remainder is m - n.

    => [2 + 3*6 + 4 + 5*6]/7

    => 54/7 => Remainder is 5.
    That is incorrect the answer is zero...

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    Post Sat Apr 14, 2012 12:19 pm
    Quote:
    What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
    consider breaking down the question this way:-

    Remainder of 2^100 / 7 = 2*(2^99) / 7
    or 2*(2^3)^33 /7
    or 2*8^33/7 = 2 (as remainder of 8*8*8..../ 7 = 1*1*1.... =1 only)

    similarly R. of 3^100 / 7 or (3^2)^50 / 7 or 9^50 / 7 --- (1)
    as remainder of 9 / 7 is 2 we can say R of (1) above = R of 2^50 / 7 or (2^2)*(2^3)^16 / 7
    or 4*8^16 / 7 thus by same logic as above R. here is 4*1 = 4

    Similarly, R. of 4^100 / 7 = 2^200 / 7 = 4*(2^3)^66 / 7 = 4*8^66 / 7 = 4

    similarly, R. of 5^100 / 7 = R. of 25^50 / 7 = R. of 4^50 / 7 = R. of 2*(2^3)^33 = 2
    adding all four together 2+4+4+2 = 12 thus finally remainder is 12 / 7 = 5

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    Post Sat Apr 14, 2012 12:25 pm
    Neo Anderson wrote:
    Quote:
    What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
    consider breaking down the question this way:-

    Remainder of 2^100 / 7 = 2*(2^99) / 7
    or 2*(2^3)^33 /7
    or 2*8^33/7 = 2 (as remainder of 8*8*8..../ 7 = 1*1*1.... =1 only)

    similarly R. of 3^100 / 7 or (3^2)^50 / 7 or 9^50 / 7 --- (1)
    as remainder of 9 / 7 is 2 we can say R of (1) above = R of 2^50 / 7 or (2^2)*(2^3)^16 / 7
    or 4*8^16 / 7 thus by same logic as above R. here is 4*1 = 4

    Similarly, R. of 4^100 / 7 = 2^200 / 7 = 4*(2^3)^66 / 7 = 4*8^66 / 7 = 4

    similarly, R. of 5^100 / 7 = R. of 25^50 / 7 = R. of 4^50 / 7 = R. of 2*(2^3)^33 = 2
    adding all four together 2+4+4+2 = 12 thus finally remainder is 12 / 7 = 5
    Mr. Anderson (Very Happy),

    When I run this thru excel or a calculator there is no remainder... so what I think (which is prob 99% wrong) is you can factor out the power and get 2 + 3 + 4 + 5 which is 14; so 14/2 = 7 no remainder

    rattan123 Just gettin' started!
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    Post Sat Apr 14, 2012 6:38 pm
    thnx....the answer is 5 only...

    tomada GMAT Destroyer!
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    Post Sun Apr 15, 2012 8:36 pm
    (2^3)/7 -> Remainder = 1
    (2^4)/7 -> Remainder = 2
    (2^5)/7 -> Remainder = 4
    (2^6)/7 -> Remainder = 1
    (2^7)/7 -> Remainder = 2
    (2^8)/7 -> Remainder = 4
    …and this pattern continues…
    (2^9)/7 -> Remainder = 1
    (2^10)/7 -> Remainder = 2

    When the exponent is 3,6,9,12, etc., the Remainder = 1
    When the exponent is 4,7,10,13, etc., the Remainder = 2
    When the exponent is 5,8,11,14, etc., the Remainder = 4

    An exponent of 100 is part of the set that begins with 4,7,10,13 so, for (2^100)/7, the Remainder = 2

    That's as far as I got.

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    Post Sun Apr 15, 2012 9:58 pm
    seal4913 wrote:
    Shalabh's Quants wrote:
    rattan123 wrote:
    What is the remainder when 2^100+3^100+4^100+5^100 is divided by 7
    => (2^100+3^100+4^100+5^100)/7

    => (2.2^99 + 3.3^99 + 2^200 + 5.5^99)/7

    => [2*(2^3)^33 + 3*(3^3)^33 + 2^2*(2^3)^66 + 5*(5^3)^33]/7

    => [2*8^33 + 3*27^33 + 2^2*8^66 + 5*125^33]/7

    => [2*(7+1)^33 + 3*(28-1)^33 + 2^2*(7+1)^66 + 5*(126-1)^33]/7

    => Taking remainders, [2*1^33 + 3*(-1)^33 + 2^2*1^66 + 5*(-1)^33]/7

    => [2+ 3*(7-1)+ 4 + 5*(7-1)]/7; For [(-1)^n]/m; if n is odd, remainder is m - n.

    => [2 + 3*6 + 4 + 5*6]/7

    => 54/7 => Remainder is 5.
    That is incorrect the answer is zero...
    I am sure answer is 5 only. Hope you both too agree now. Smile

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    killer1387 GMAT Destroyer! Default Avatar
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    Post Tue Apr 17, 2012 2:13 am
    The answer is 5 i made mistake while solving initially;

    2^100+3^100+4^100+5^100
    = 2^100+3^100+(7-3)^100+(7-2)^100

    now taking remainders:

    R{2*(2^100+3^100)}%7
    = R{2(2+4)}%7
    = 5

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    Post Tue Apr 17, 2012 11:29 am
    2^100 = 2^99*2 = 8^33*2, R(2^100/7) = R(2*8^33/7) = 2
    3^100 = 9^50, R(9^50/7) = R(2^50/7) = R(4*8^16/7) = 4
    4^100 = 2^200, R(2^200/7) = R(4*8^66/7) = 4
    R(5^100/7) = R((-2)^100/7) = R(2^100/7) = R(2*8^33/7) = 2

    R(Expression/7) = R((2+4+4+2)/7) = 5

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