modems and cas #s












7












$begingroup$


There are 12 cars built. All modems have have 0-number telephone numbers, numbered consecutively, in the same order. For example, if house #3's phone number is 453672888, house 2 will be 523492. The cas number is always multiplied by the modem number. What is the cas number that would correspond to modem 12, if one was built?






logical-deduction calculation-puzzle algebra







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    Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
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    Jun 1 '18 at 22:21










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7












$begingroup$


There are 12 cars built. All modems have have 0-number telephone numbers, numbered consecutively, in the same order. For example, if house #3's phone number is 453672888, house 2 will be 523492. The cas number is always multiplied by the modem number. What is the cas number that would correspond to modem 12, if one was built?






logical-deduction calculation-puzzle algebra







share|improve this question











$endgroup$








  • 1




    $begingroup$
    Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
    $endgroup$
    – Rubio♦
    Jun 1 '18 at 22:21










  • $begingroup$
    Please don't make more work for other people by vandalizing your posts. By posting on the Stack Exchange (SE) network, you've granted a non-revocable right, under the CC BY-SA 3.0 license for SE to distribute that content. By SE policy, any vandalism will be reverted. If you want to know more about deleting a post, consider taking a look at: How does deleting work?
    $endgroup$
    – iBot
    39 mins ago














7












7








7





$begingroup$


There are 12 cars built. All modems have have 0-number telephone numbers, numbered consecutively, in the same order. For example, if house #3's phone number is 453672888, house 2 will be 523492. The cas number is always multiplied by the modem number. What is the cas number that would correspond to modem 12, if one was built?






logical-deduction calculation-puzzle algebra







share|improve this question











$endgroup$




There are 12 cars built. All modems have have 0-number telephone numbers, numbered consecutively, in the same order. For example, if house #3's phone number is 453672888, house 2 will be 523492. The cas number is always multiplied by the modem number. What is the cas number that would correspond to modem 12, if one was built?






logical-deduction calculation-puzzle algebra




logical-deduction calculation-puzzle algebra






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 15 mins ago







user18842sos

















asked May 28 '18 at 16:38









user18842sosuser18842sos

1023




1023








  • 1




    $begingroup$
    Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
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    – Rubio♦
    Jun 1 '18 at 22:21










  • $begingroup$
    Please don't make more work for other people by vandalizing your posts. By posting on the Stack Exchange (SE) network, you've granted a non-revocable right, under the CC BY-SA 3.0 license for SE to distribute that content. By SE policy, any vandalism will be reverted. If you want to know more about deleting a post, consider taking a look at: How does deleting work?
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    – iBot
    39 mins ago














  • 1




    $begingroup$
    Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
    $endgroup$
    – Rubio♦
    Jun 1 '18 at 22:21










  • $begingroup$
    Please don't make more work for other people by vandalizing your posts. By posting on the Stack Exchange (SE) network, you've granted a non-revocable right, under the CC BY-SA 3.0 license for SE to distribute that content. By SE policy, any vandalism will be reverted. If you want to know more about deleting a post, consider taking a look at: How does deleting work?
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    – iBot
    39 mins ago








1




1




$begingroup$
Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
$endgroup$
– Rubio♦
Jun 1 '18 at 22:21




$begingroup$
Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
$endgroup$
– Rubio♦
Jun 1 '18 at 22:21












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Please don't make more work for other people by vandalizing your posts. By posting on the Stack Exchange (SE) network, you've granted a non-revocable right, under the CC BY-SA 3.0 license for SE to distribute that content. By SE policy, any vandalism will be reverted. If you want to know more about deleting a post, consider taking a look at: How does deleting work?
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2 Answers
2






active

oldest

votes


















7












$begingroup$


$7927937$.




Explanation:




Let the "$0$th" house have number $x$, i.e., house $i$ has number $x+i$. For each house number $i$ from $1$ to $16$, $i$ divides $x+i$, so $i$ must divide $x$. Therefore, $x$ must be a multiple of the LCM of the first $16$ positive integers, which is $720720$. Say $x=720720k$. We also know that $17$ divides $x+13=720720k+13$, so because $720720equiv5pmod{17}$, $0equiv720720k+13equiv5k+13pmod{17}$. From here, we can quickly test small values to obtain that $k=11$ is the smallest positive integer satisfying that $17$ divides $5k+13=68$, so the $17$th house is $x+17=720720cdot11+17=7927937$. This solution is also unique because $kequiv11pmod{17}$, so the next smallest possible value of $k$ is $11+17=28$, but then $x=720720cdot28=20180160$ is $8$ digits long.







share|improve this answer











$endgroup$









  • 1




    $begingroup$
    Perhaps you can say something about how at first you only know that $kequiv 11pmod{17}$ and that the 7-digit restriction forces you to pick $k=11$.
    $endgroup$
    – Riley
    May 28 '18 at 17:23



















1












$begingroup$


If House 1's number is $abcdefg$, House 2's number must be even, so even-numbered houses have even phone numbers, and odd-numbered ones have odd phone numbers.

Since House 5's number is odd, it can only be 5, so $g=1$.

House 16's number is $9k+7$, $16k$, $5k+1$, $7k+2$, $11k+5$, $13k+3$, $17k+3$, so it can also be written as $13*17k+3=221k+3$.

$221k+3 = 16n$

$13k+3=16n$ (mod 16)

$k=16x+1$

$abcdefg + 16 = 221(16x+1) + 3 = 3536x+224$

$3536x+224 = 11y+5$

$5x+10 = 11y$ (mod 11)

$x = 11y-2$ (mod 11)

$3536(11y-2)+224 = 38896y-6848$

$38896y-6848 = 9z+7$

$7y-6 = 9z$

$y = 9z+6$ (mod 9)

$38896(9z+6)-6848 = 350064z + 226528$

$350064z + 226528 = 7t+2$

$z + 6 = 7t$

$z = 7t+1$ (mod 7)

$350064(7t+1) + 226528 = 2450048t+576592$

It ends with 6, so $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.







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    2 Answers
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    2 Answers
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    active

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    7












    $begingroup$


    $7927937$.




    Explanation:




    Let the "$0$th" house have number $x$, i.e., house $i$ has number $x+i$. For each house number $i$ from $1$ to $16$, $i$ divides $x+i$, so $i$ must divide $x$. Therefore, $x$ must be a multiple of the LCM of the first $16$ positive integers, which is $720720$. Say $x=720720k$. We also know that $17$ divides $x+13=720720k+13$, so because $720720equiv5pmod{17}$, $0equiv720720k+13equiv5k+13pmod{17}$. From here, we can quickly test small values to obtain that $k=11$ is the smallest positive integer satisfying that $17$ divides $5k+13=68$, so the $17$th house is $x+17=720720cdot11+17=7927937$. This solution is also unique because $kequiv11pmod{17}$, so the next smallest possible value of $k$ is $11+17=28$, but then $x=720720cdot28=20180160$ is $8$ digits long.







    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Perhaps you can say something about how at first you only know that $kequiv 11pmod{17}$ and that the 7-digit restriction forces you to pick $k=11$.
      $endgroup$
      – Riley
      May 28 '18 at 17:23
















    7












    $begingroup$


    $7927937$.




    Explanation:




    Let the "$0$th" house have number $x$, i.e., house $i$ has number $x+i$. For each house number $i$ from $1$ to $16$, $i$ divides $x+i$, so $i$ must divide $x$. Therefore, $x$ must be a multiple of the LCM of the first $16$ positive integers, which is $720720$. Say $x=720720k$. We also know that $17$ divides $x+13=720720k+13$, so because $720720equiv5pmod{17}$, $0equiv720720k+13equiv5k+13pmod{17}$. From here, we can quickly test small values to obtain that $k=11$ is the smallest positive integer satisfying that $17$ divides $5k+13=68$, so the $17$th house is $x+17=720720cdot11+17=7927937$. This solution is also unique because $kequiv11pmod{17}$, so the next smallest possible value of $k$ is $11+17=28$, but then $x=720720cdot28=20180160$ is $8$ digits long.







    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Perhaps you can say something about how at first you only know that $kequiv 11pmod{17}$ and that the 7-digit restriction forces you to pick $k=11$.
      $endgroup$
      – Riley
      May 28 '18 at 17:23














    7












    7








    7





    $begingroup$


    $7927937$.




    Explanation:




    Let the "$0$th" house have number $x$, i.e., house $i$ has number $x+i$. For each house number $i$ from $1$ to $16$, $i$ divides $x+i$, so $i$ must divide $x$. Therefore, $x$ must be a multiple of the LCM of the first $16$ positive integers, which is $720720$. Say $x=720720k$. We also know that $17$ divides $x+13=720720k+13$, so because $720720equiv5pmod{17}$, $0equiv720720k+13equiv5k+13pmod{17}$. From here, we can quickly test small values to obtain that $k=11$ is the smallest positive integer satisfying that $17$ divides $5k+13=68$, so the $17$th house is $x+17=720720cdot11+17=7927937$. This solution is also unique because $kequiv11pmod{17}$, so the next smallest possible value of $k$ is $11+17=28$, but then $x=720720cdot28=20180160$ is $8$ digits long.







    share|improve this answer











    $endgroup$




    $7927937$.




    Explanation:




    Let the "$0$th" house have number $x$, i.e., house $i$ has number $x+i$. For each house number $i$ from $1$ to $16$, $i$ divides $x+i$, so $i$ must divide $x$. Therefore, $x$ must be a multiple of the LCM of the first $16$ positive integers, which is $720720$. Say $x=720720k$. We also know that $17$ divides $x+13=720720k+13$, so because $720720equiv5pmod{17}$, $0equiv720720k+13equiv5k+13pmod{17}$. From here, we can quickly test small values to obtain that $k=11$ is the smallest positive integer satisfying that $17$ divides $5k+13=68$, so the $17$th house is $x+17=720720cdot11+17=7927937$. This solution is also unique because $kequiv11pmod{17}$, so the next smallest possible value of $k$ is $11+17=28$, but then $x=720720cdot28=20180160$ is $8$ digits long.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited May 28 '18 at 17:31

























    answered May 28 '18 at 16:43









    noednenoedne

    6,34011855




    6,34011855








    • 1




      $begingroup$
      Perhaps you can say something about how at first you only know that $kequiv 11pmod{17}$ and that the 7-digit restriction forces you to pick $k=11$.
      $endgroup$
      – Riley
      May 28 '18 at 17:23














    • 1




      $begingroup$
      Perhaps you can say something about how at first you only know that $kequiv 11pmod{17}$ and that the 7-digit restriction forces you to pick $k=11$.
      $endgroup$
      – Riley
      May 28 '18 at 17:23








    1




    1




    $begingroup$
    Perhaps you can say something about how at first you only know that $kequiv 11pmod{17}$ and that the 7-digit restriction forces you to pick $k=11$.
    $endgroup$
    – Riley
    May 28 '18 at 17:23




    $begingroup$
    Perhaps you can say something about how at first you only know that $kequiv 11pmod{17}$ and that the 7-digit restriction forces you to pick $k=11$.
    $endgroup$
    – Riley
    May 28 '18 at 17:23











    1












    $begingroup$


    If House 1's number is $abcdefg$, House 2's number must be even, so even-numbered houses have even phone numbers, and odd-numbered ones have odd phone numbers.

    Since House 5's number is odd, it can only be 5, so $g=1$.

    House 16's number is $9k+7$, $16k$, $5k+1$, $7k+2$, $11k+5$, $13k+3$, $17k+3$, so it can also be written as $13*17k+3=221k+3$.

    $221k+3 = 16n$

    $13k+3=16n$ (mod 16)

    $k=16x+1$

    $abcdefg + 16 = 221(16x+1) + 3 = 3536x+224$

    $3536x+224 = 11y+5$

    $5x+10 = 11y$ (mod 11)

    $x = 11y-2$ (mod 11)

    $3536(11y-2)+224 = 38896y-6848$

    $38896y-6848 = 9z+7$

    $7y-6 = 9z$

    $y = 9z+6$ (mod 9)

    $38896(9z+6)-6848 = 350064z + 226528$

    $350064z + 226528 = 7t+2$

    $z + 6 = 7t$

    $z = 7t+1$ (mod 7)

    $350064(7t+1) + 226528 = 2450048t+576592$

    It ends with 6, so $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.







    share|improve this answer











    $endgroup$


















      1












      $begingroup$


      If House 1's number is $abcdefg$, House 2's number must be even, so even-numbered houses have even phone numbers, and odd-numbered ones have odd phone numbers.

      Since House 5's number is odd, it can only be 5, so $g=1$.

      House 16's number is $9k+7$, $16k$, $5k+1$, $7k+2$, $11k+5$, $13k+3$, $17k+3$, so it can also be written as $13*17k+3=221k+3$.

      $221k+3 = 16n$

      $13k+3=16n$ (mod 16)

      $k=16x+1$

      $abcdefg + 16 = 221(16x+1) + 3 = 3536x+224$

      $3536x+224 = 11y+5$

      $5x+10 = 11y$ (mod 11)

      $x = 11y-2$ (mod 11)

      $3536(11y-2)+224 = 38896y-6848$

      $38896y-6848 = 9z+7$

      $7y-6 = 9z$

      $y = 9z+6$ (mod 9)

      $38896(9z+6)-6848 = 350064z + 226528$

      $350064z + 226528 = 7t+2$

      $z + 6 = 7t$

      $z = 7t+1$ (mod 7)

      $350064(7t+1) + 226528 = 2450048t+576592$

      It ends with 6, so $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.







      share|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$


        If House 1's number is $abcdefg$, House 2's number must be even, so even-numbered houses have even phone numbers, and odd-numbered ones have odd phone numbers.

        Since House 5's number is odd, it can only be 5, so $g=1$.

        House 16's number is $9k+7$, $16k$, $5k+1$, $7k+2$, $11k+5$, $13k+3$, $17k+3$, so it can also be written as $13*17k+3=221k+3$.

        $221k+3 = 16n$

        $13k+3=16n$ (mod 16)

        $k=16x+1$

        $abcdefg + 16 = 221(16x+1) + 3 = 3536x+224$

        $3536x+224 = 11y+5$

        $5x+10 = 11y$ (mod 11)

        $x = 11y-2$ (mod 11)

        $3536(11y-2)+224 = 38896y-6848$

        $38896y-6848 = 9z+7$

        $7y-6 = 9z$

        $y = 9z+6$ (mod 9)

        $38896(9z+6)-6848 = 350064z + 226528$

        $350064z + 226528 = 7t+2$

        $z + 6 = 7t$

        $z = 7t+1$ (mod 7)

        $350064(7t+1) + 226528 = 2450048t+576592$

        It ends with 6, so $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.







        share|improve this answer











        $endgroup$




        If House 1's number is $abcdefg$, House 2's number must be even, so even-numbered houses have even phone numbers, and odd-numbered ones have odd phone numbers.

        Since House 5's number is odd, it can only be 5, so $g=1$.

        House 16's number is $9k+7$, $16k$, $5k+1$, $7k+2$, $11k+5$, $13k+3$, $17k+3$, so it can also be written as $13*17k+3=221k+3$.

        $221k+3 = 16n$

        $13k+3=16n$ (mod 16)

        $k=16x+1$

        $abcdefg + 16 = 221(16x+1) + 3 = 3536x+224$

        $3536x+224 = 11y+5$

        $5x+10 = 11y$ (mod 11)

        $x = 11y-2$ (mod 11)

        $3536(11y-2)+224 = 38896y-6848$

        $38896y-6848 = 9z+7$

        $7y-6 = 9z$

        $y = 9z+6$ (mod 9)

        $38896(9z+6)-6848 = 350064z + 226528$

        $350064z + 226528 = 7t+2$

        $z + 6 = 7t$

        $z = 7t+1$ (mod 7)

        $350064(7t+1) + 226528 = 2450048t+576592$

        It ends with 6, so $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited May 28 '18 at 19:00

























        answered May 28 '18 at 18:52









        NautilusNautilus

        4,048525




        4,048525






























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            Tabula Rosettana

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            Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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