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    Find the sum of those numbers between 1000 and 6000 every one of whose digits is one of the numbers 0, 2, 5 or 7, giving your answer as a product of primes.

    i started getting somewhere but i still end up having to sum too many numbers. i'm sure i must be missing something...
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    (Original post by toad)
    Find the sum of those numbers between 1000 and 6000 every one of whose digits is one of the numbers 0, 2, 5 or 7, giving your answer as a product of primes.

    i started getting somewhere but i still end up having to sum too many numbers. i'm sure i must be missing something...

    Don't try writing them all down - if you think of the number as ABCD then A can be 2 or 5 and B,C,D can be independently 0,2,5,7. Why does this mean there are 128 numbers in all?

    Then to sum them break the numbers into their constituent parts: thousands, hundreds, tens, units. Why, for example, do all the hundreds add to (0+2+5+7) x 32 x 100?

    I got the answer 2^6 x 7 x 11 x 101.
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    Find the sum of those numbers between 1000 and 6000 every one of whose digits is one of the numbers 0, 2, 5 or 7, giving your answer as a product of primes.
    (Original post by toad)

    i started getting somewhere but i still end up having to sum too many numbers. i'm sure i must be missing something...
    Collecting the digits ABCD is a clever approach.

    Now you know that only 2 and 5 can take the A positions since any of the others will be >6000 or <2000. So for the first position we have 2C1, choices.

    For the last three digits BCD we have three possibilities.
    1)If B does not equalC, B does not equal D and C does not equal D (the last three numbers are all different) the number of ways for this to occur is:

    2C1*4C3 * (3!) = 48

    2)If B=C or C=D or B=D (two of three last numbers are the same.)

    Then, the number of ways in which this can occur is:

    2C1* 4C2* (3! /2!) * 2 = 72

    3)If B=C=D (the three last digits are the same)

    Then the ways in which this can occur is:

    2C1* 4C1* (3! /3!) = 8

    This gives the 128 ways.

    Now we know there are 64 numbers in the 2000, and 64 numbers in the 5000.

    So for 2000 we have 64
    The first four lines look like this
    2000 2002 2005 2007 = 2000*4+ 16
    2020 2022 2025 2027 = 2000*4+ 94
    2050 2052 2055 2057 = 2000*4+ 214
    2070 2072 2075 2077 = 2000*4+ 294

    These first four lines sums to: 2000*16+616
    The next three lines give:
    2200*16+616
    2500*16+616
    2700*16+616
    This gives:
    2000*64+1400*16+616*4=152864


    Do the same for the 5000 and you will get:
    5000*64+1400*16+616*4= 344864

    Hence the final total is 497728. The prime factor of this number is:

    2^6*7*11*101
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    I don't think there's so much need to do so much direct calculating.

    For example, as A can be 2 or 5, and B,C,D can be 0,2,5,7 entirely _independently_ then there are

    2 x 4 x 4 x 4 = 128 such numbers. [No need to treat cases]

    Now if we break these numbers into thousands, hundreds, tens, units, and sum them, we get:

    (2+5) x 64 x 1000 [from the thousands - 64 each starting with 2 or 5]

    + (0+2+5+7) x 32 x 100 [from the hundreds - 32 with second digit 0 etc.]

    +(0+2+5+7) x 32 x 10 [from tens - 32 with third digit 0 etc.]

    +(0+2+5+7) x 32 x 1 [from units - 32 with fourth digit 0 etc.]

    = [grouping]

    32 x (14000 + 1400 + 140 + 14) =

    2^6 x 7 x (1000+100+10+1) =

    2^6 x 7 x 1111 =

    2^6 x 7 x 11 x 101.
 
 
 
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