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azlink10

  • 3 years ago

how many positive odd integers less than 70,000 can be represented using the digits 2,3,4,5,and 6

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  1. azlink10
    • 3 years ago
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    greaaaaat so nobody knows:(

  2. Andras
    • 3 years ago
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    just looked at it now, thinking

  3. dhatraditya
    • 3 years ago
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    you mean as a arithmetic combination of these numbers or does it mean like you can use only 2,3,4,5,6 to represent the number? like 23 is a valid number but 37 is not because it uses 7?

  4. 3pwood
    • 3 years ago
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    I'm sure there's some fancy statistics way to do this, but thinking about it the long way... For one-digit numbers: only 3 works so 1 For two digit numbers: 3 still has to be the second digit, but any of the 5 numbers can be first, so 5 For three digit numbers: 25, by similar logic for four: 125 for five: 625 for six: 3125 so 1 + 5 + 5^2 + 5^3 + 5^4 + 5^5 = 3906 I'm not totally sure of my method. That make good sense?

  5. 3pwood
    • 3 years ago
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    Oh woops, went one exponent too far. Six-digit numbers don't qualify. so 1 + 5 + 5^2 + 5^3 + 5^4 = 781

  6. Andras
    • 3 years ago
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    but 5 is an odd number too :)

  7. 3pwood
    • 3 years ago
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    Right...somehow didn't even see the 5. Apologies. What would that do?...it would mean that both 3 and 5 could be the last digit, and make for twice as many possibilities, right?

  8. Andras
    • 3 years ago
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    not sure, but I think not

  9. Andras
    • 3 years ago
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    I give up, it is too late here :) tomorrow I will give it another go if nobody solved it till then

  10. 3pwood
    • 3 years ago
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    I'm relatively confident that it's 781*2 = 1562, but I know my work is pretty sloppy.

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