Houses and Telephone #s












7












$begingroup$


There are 16 houses consecutively numbered 1-16. All houses have 7-digit telephone numbers, numbered consecutively, in the same order. For example, if house #3's phone number is 1111111, house #4 will be 1111112. The phone number is always divisible by the house number. House #13's phone number is also divisible by the number #17. What is the phone number that would correspond to house #17, if one was built?






logical-deduction calculation-puzzle algebra







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$endgroup$

















    7












    $begingroup$


    There are 16 houses consecutively numbered 1-16. All houses have 7-digit telephone numbers, numbered consecutively, in the same order. For example, if house #3's phone number is 1111111, house #4 will be 1111112. The phone number is always divisible by the house number. House #13's phone number is also divisible by the number #17. What is the phone number that would correspond to house #17, if one was built?






    logical-deduction calculation-puzzle algebra







    share|improve this question











    $endgroup$















      7












      7








      7





      $begingroup$


      There are 16 houses consecutively numbered 1-16. All houses have 7-digit telephone numbers, numbered consecutively, in the same order. For example, if house #3's phone number is 1111111, house #4 will be 1111112. The phone number is always divisible by the house number. House #13's phone number is also divisible by the number #17. What is the phone number that would correspond to house #17, if one was built?






      logical-deduction calculation-puzzle algebra







      share|improve this question











      $endgroup$




      There are 16 houses consecutively numbered 1-16. All houses have 7-digit telephone numbers, numbered consecutively, in the same order. For example, if house #3's phone number is 1111111, house #4 will be 1111112. The phone number is always divisible by the house number. House #13's phone number is also divisible by the number #17. What is the phone number that would correspond to house #17, 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 18 mins ago









      Rubio♦

      28.9k565178




      28.9k565178










      asked May 28 '18 at 16:38









      user18842sosuser18842sos

      1




      1






















          2 Answers
          2






          active

          oldest

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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



















          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











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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

            6,35011855




            6,35011855








            • 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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