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pavelpimen
pavelpimen
Student
High School - 10th Grade

The expression E(n)= 2^(7n+3) + 3^(2n + 1) X 5^(4n+1), n belongs N* is divided by : a)7 ; b)11 ; c)23 ; d) 17 ; e) 5?

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Posted by pavelpimen on Friday July 17, 2009 at 6:35 AM and tagged with math.

Answers:

  1. giorgiana1976
    giorgiana1976 Teacher
    Doctorate

    eNotes Editor

    First of all, we'll verify the expression for n=1, so that

    E(1)=2^(7*1+3)+3^(2*1+1)*5^(4*1+1)=2^10+3^3*5^5=85399

    E(1)=23*3713 so E(1) is divisible by 23.

    Let's suppose that for k>1, E(k)=2^(7*k+3)+3^(2*k+1)*5^(4*k+1) is divisible by 23

    We'll verify if

    E(k+1)=2^(7*(k+1)+3)+3^(2*(k+1)+1)*5^(4*(k+1)+1)

    is divisible by 23, too.

    E(k+1)=2^7*2^(7*k+3)+3^2*3^(2k+1)*5^4*5^(4k+1)=

    =2^7*2^(7*k+3)+3^2*5^4*3^(2k+1)*5^(4k+1)=

    =2^7*2^(7*k+3)+5625*3^(2k+1)*5^(4k+1)=

    =128*(2^(7*k+3)+3^(2k+1)*5^(4k+1)+5497*3^(2k+1)*5^(4k+1))=

    But, from induction hypothesis, we know that the paranthesis frpom the first term is divisible by 23 and 5497 is divisible by 23, also, so E(k+1) is divisible by 23, too.

    From this, E(n) is divisible by 23, so the right answer is c).

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    Posted by giorgiana1976 on Friday July 17, 2009 at 7:13 AM