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Math Help - adding numbers !!!!

  1. #1
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    adding numbers !!!!

    i need to know fast (by 5 oclock tommorow eastern time) a faster way to do this problem. We need an alternate way other than adding 200 times. Help would be appreciated, thank you!!!!!! :Find the sum of the digits of the first 200 even natural numbers.
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  2. #2
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    Quote Originally Posted by jarny View Post
    i need to know fast (by 5 oclock tommorow eastern time) a faster way to do this problem. We need an alternate way other than adding 200 times. Help would be appreciated, thank you!!!!!! :Find the sum of the digits of the first 200 even natural numbers.
    (n^2 + n)/2 will come in handy; there are other ways to solve this too.

    S_n = n(2a_1+(n-1)d)/2

    a_1 is the first term; d is the difference between the two terms

    Sn=200(2*2+(199)2)/2

    = 40,200

    OR:

    (n^2 + 200)/2 = 20,100, but you want the even numbers;

    Thus, multiply by 2 = 40,200
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  3. #3
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    Hello, jarny!

    Here's a rather "primitive" approach . . .


    Find the sum of the digits of the first 200 even natural numbers.

    In a set of any ten consecutive natural numbers,
    . . the even ones end in: 2,4,6,8,0
    Their sum is: 20

    Among the first 200 natural numbers, there are 20 such sets-of-ten.
    . . Hence, the unit-digits add up to: 20\times20 = \boxed{400}


    Now consider the ten-digits.

    In the first hundred numbers, they are:
    . . 0,0,0,0,0,\;1,1,1,1,1,\;2,2,2,2,\;\hdots\;9,9,9,9,  9
    Their sum is: 5 \times (1 + 2 + 3 + \hdots + 9) \:=\:5(45) \:=\:225

    Since this happens again in the second hundred numbers,
    . . the ten-digits total:  2 \times 225 \:=\:\boxed{450}


    Now, the hundred-digits.
    . . They are all 0 in the first 99 numbers.
    . . They are all 1's in the next 100 numbers.
    . . Finally, the "200" has a 2.
    Hence, the hundred-digits add up to: 100 + 2 \:=\:\boxed{102}


    Therefore, the total of all the digits is: 400 + 450 + 102 \:=\:\boxed{952}

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  4. #4
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    whos right

    can somebody help? i dont know whos right, the first one i dont know but soroban, did you do the first 200 or 200 even ones and would it go up to 400 then?
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    trying to find solution but can't

    Soroban, you went up to 200 but it would be up to four hundred right? so it would be 0 as the third digit in the first 49 a 0, next 50 a 1, third 50 a 2, fourth 50 a 3, then last one 4 right? could you reexplain the work then?
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    When you said you wanted the sum of the digits, do you mean the actual digits, such as 2, 4, 6, 8 ,0, 2, 4, 6, 8, ... and so forth, or the actual numbers? To me it seems more realistic you'd want to know the numbers, in which case 40,200 is correct.

    That is, 2 + 4 + 6 + ... + 400 = 40,200
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  7. #7
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    Hello, jarny!

    You're absolutely right . . .

    The first 200 even numbers: 2,\,4,\,6,\,8,\,\hdots\,400

    I'll try again . . . *blush*


    In a set of any ten consecutive natural numbers,
    . . the even ones end in: 2,4,6,8,0
    Their sum is: 20

    Among the first 400 natural numbers, there are 40 such sets-of-ten.
    . . Hence, the unit-digits add up to: 40\times20 = \boxed{800}


    Now consider the ten-digits.

    In the first hundred numbers, they are:
    . . 0,0,0,0,0,\;1,1,1,1,1,\;2,2,2,2,\;\hdots\;9,9,9,9,  9
    Their sum is: 5 \times (1 + 2 + 3 + \hdots + 9) \:=\:5(45) \:=\:225

    Since this happens again in the 100's, 200's and 300's,
    . . the ten-digits total:  4 \times 225 \:=\:\boxed{900}


    Now, the hundred-digits.
    . . They are all 0 in the first 99 numbers.
    . . They are all 1 in the next hundred numbers.
    . . They are all 2 in the next hundred numbers.
    . . They are all 3 in the next hundred numbers.
    . . Finally, the "400" has a 4.
    Hence, the hundred-digits add up to: 100 + 200 + 300 + 4 \:=\:\boxed{604}


    Therefore, the total of all the digits is: 800 + 900 + 604 \:=\:\boxed{2304}


    Hope I got it right this time . . .
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  8. #8
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    Quote Originally Posted by jarny View Post
    Soroban, you went up to 200 but it would be up to four hundred right? so it would be 0 as the third digit in the first 49 a 0, next 50 a 1, third 50 a 2, fourth 50 a 3, then last one 4 right? could you reexplain the work then?
    Hello,

    you are right but I presume it's too late now, but nevertheless...:

    Do the same method as Soroban has explained.
    You'll get:
    1. Sum of all one-digits: 40 * 20 = 800
    2. Sum of all ten-digits: 4 * 225 = 900
    3. Sum of all hundred-digits: 100 * (1+2+3) +4

    Total: 2104

    EB
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