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Thread: Digit sum

  1. #1
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    Post Digit sum

    What is the sum of the digits of the decimal form of the product below?

    1999 2001
    2 x 5


  2. #2
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    Sep 2006
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    Re: Digit sum

    Hi,

    Your problem is not clear. what you want actually? what is the meaning of the following...

    1999 2001
    2 x 5

    ------------------
    suresh


  3. #3
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    Re: Digit sum

    I think you need the solution for the following....

    (2^1999 * 5^2001) Sum of the results of this product...

    Am i correct?

    ----------------
    suresh


  4. #4
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    Re: Digit sum

    Yes. You R Correct


  5. #5
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    Re: Digit sum

    The answer for your problem is 7.

    Solution:

    Your problem is (2^n * 5^(n+2))

    Let's take n=1
    result is 2*125 = 250 = 7

    Let's take n=2
    result is 4*625 = 2500 = 7

    Let's take n=3
    result is 8*3125 = 25000 = 7

    Let's take n=4
    result is 16*15625 = 250000 = 7
    .
    .
    .
    So finally we got the result like 250...... and the sum is 7

    ---------------------
    suresh


  6. #6
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    Re: Digit sum

    That's a Great Logic!
    Thnks.
    2^n*5^(n+2)
    = 2^n*5^n*5^2
    =(2*5)^n*25
    =10^n*25
    Sum is always 7.


  7. #7
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    Re: Digit sum

    great thinking suresh..


  8. #8
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    Aug 2009
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    Re: Digit sum

    2^1999=(2^2000)/2

    5^2001=(5^2000)*5

    lhs={(2*5)^2000}*5/2=(10^2)^1000*2.5=(100^1000)*2.5=(1 followed by 20 zeroes)*2.5=25 followed by 19 zeroes=>this gives sum of digits=2+5=7


  9. #9
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    Smile Re: Digit sum

    2^1999=(2^2000)/2

    5^2001=(5^2000)*5

    lhs={(2*5)^2000}*5/2

    (10^2000)*2.5
    {(10^2)^1000}*2.5

    (100^1000)*2.5
    (1 followed by 20 zeroes)*2.5

    25 followed by 19 zeroes

    hence, sum of digits =2+5=7

    just getting the advantage of particular set of 2 & 5


  10. #10

    Re: Digit sum

    Quote Originally Posted by rahul14904 View Post
    2^1999=(2^2000)/2

    5^2001=(5^2000)*5

    lhs={(2*5)^2000}*5/2

    (10^2000)*2.5
    {(10^2)^1000}*2.5

    (100^1000)*2.5
    (1 followed by 20 zeroes)*2.5

    25 followed by 19 zeroes

    hence, sum of digits =2+5=7

    just getting the advantage of particular set of 2 & 5
    Morning
    I'm trying to do same problem guys
    How did you get that:
    (100^1000) = 1 followed by 20 zeros
    Because what I got from 10^n
    Letting n=1, 10
    n=2, 100
    .
    .
    .
    n=10, 1 followed by 10 zeros


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