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Thread: Sum of 24 Integers

  1. #1
    Super Member harpazo's Avatar
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    Sum of 24 Integers

    The addition problem below shows 4 of the 24 different integers that can be formed by using each of the digits 1, 2, 3, and 4 exactly once in each integer. What is the sum of these 24 integers?

    1,234 + 1,243 + 1,324 + x,xxx + x,xxx + 4,321 =

    I understand that every x posted must be replaced by one of the given integers. Afterward, I must add all four-digits to obtain a sum. Any ideas?
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  2. #2
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    Re: Sum of 24 Integers

    In these 24 integers, each of "1", "2", "3", "4" is the leading digit in 24/4= 6 of the numbers so the sum will include 1000*6+ 2000*6+ 3000*6+ 4000*6= 60000. Similarly, each will occur in the "100's place" 6 times so we will also have 100*6+ 200*6+ 300*6+ 400*6= 6000. Do the same for the "10's place" and "one's place".
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    Re: Sum of 24 Integers

    Quote Originally Posted by harpazo View Post
    The addition problem below shows 4 of the 24 different integers that can be formed by using each of the digits 1, 2, 3, and 4 exactly once in each integer. What is the sum of these 24 integers?
    1,234 + 1,243 + 1,324 + x,xxx + x,xxx + 4,321 =
    I understand that every x posted must be replaced by one of the given integers. Afterward, I must add all four-digits to obtain a sum. Any ideas?
    If one makes a column of these twenty-four numbers we have 24 rows & 4 columns. In each column 1 appears six times, 2 appears six times, 3 appears six times, and 4 appears six times. Add any column to get $6+12+18+24=60$. What can you do with that hint?
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    Re: Sum of 24 Integers

    a=1, b=2, c=3, d=4
    Since each of the 24 numbers can be represented by (as example) 1000a+100b+10c + d, then:
    6[1000(a+b+c+d)] + 6[100(a+b+c+d)] + 6[10(a+b+c+d] + 6[1(a+b+c+d)]
    = 6666(a+b+c+d) = 6666(10) = 66660
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    Re: Sum of 24 Integers

    This is a simple question of 'permutations and combinations'.

    Consider four blanks _ _ _ _
    Where no. 1-4 has to be filled.
    You can place any of those no. at any blank and each no. will appear there for 3! times.

    Eg- if u place 4 at the last then it will appear there for 3! times (6) and since at the last blank thus its value is unity . If at the third blank then its value is tens and so on.

    Thus, to calculate the value of the expression
    3!(4+40+400+4000)+3!(3+30+300+3000)+3!(2+20+200+20 00)+3!(1+10+100+1000)

    which will equal to 66,660.
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  6. #6
    Super Member harpazo's Avatar
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    Re: Sum of 24 Integers

    Thank you everyone.
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    Re: Sum of 24 Integers

    Quote Originally Posted by DenisB View Post
    a=1, b=2, c=3, d=4
    Since each of the 24 numbers can be represented by (as example) 1000a+100b+10c + d, then:
    6[1000(a+b+c+d)] + 6[100(a+b+c+d)] + 6[10(a+b+c+d] + 6[1(a+b+c+d)]
    = 6666(a+b+c+d) = 6666(10) = 66660
    Denis, surely you know that 66660 is a beast of a number.
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  8. #8
    Super Member harpazo's Avatar
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    Re: Sum of 24 Integers

    Good study notes.
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