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Math Help - fun little problem

  1. #1
    Eater of Worlds
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    fun little problem

    Here is a good problem if anyone wants to give it a go.

    Find the smallest positive integer such that when the first digit is shifted to the end, the result is 3/2 times the original number.

    For instance, 1284 becomes 2841. But 2841 is not 3/2 times 1284.

    Fermat's Little Theorem may come in handy.

    You know, a^{p-1}\equiv 1(mod \;\ p)

    But, feel free to do it however you like.
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  2. #2
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    Quote Originally Posted by galactus View Post
    Here is a good problem if anyone wants to give it a go.

    Find the smallest positive integer such that when the first digit is shifted to the end, the result is 3/2 times the original number.

    For instance, 1284 becomes 2841. But 2841 is not 3/2 times 1284.

    Fermat's Little Theorem may come in handy.

    You know, a^{p-1}\equiv 1(mod \;\ p)

    But, feel free to do it however you like.
    this question was asked on SOS over a year ago and this was my solution: let n=10^{r-1}a_1 + 10^{r-2}a_2 + \cdots + 10a_{r-1} + a_r be the smallest such number. we want to have:

     3(10^{r - 1}a_1 + 10^{r - 2}a_2 + ... + 10a_{r - 1} + a_r) = <br />
2(10^{r - 1}a_2 + 10^{r - 2}a_3 + ... + 10a_r + a_1), which after simplifying gives us (3 \times 10^{r - 1} - 2)a_1 = 17m, where m=10^{r-2}a_2 + 10^{r-3}a_3 + \cdots +10a_{r-1} + a_r.

    so we must have: 3 \times 10^{r - 1} \equiv 2 \mod 17, which after multiplying by 60, gives us: 10^r \equiv 120 \equiv 1 \mod 17. thus r = 16. hence: 17m = (3 \times 10^{15} - 2)a_1. in order to get the smallest number,

    we let a_1 = 1. then m= \frac{3 \times 10^{15} - 2}{17} = 176470588235294. therefore: n=10^{r-1}a_1 +m = 1176470588235294. \ \Box
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