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?




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 $720720\equiv5\pmod{17}$, $0\equiv720720k+13\equiv5k+13\pmod{17}$. 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 $x+17=720720\cdot11+17=7927937$. This solution is also unique because $k\equiv11\pmod{17}$, so the next smallest possible value of $k$ is $11+17=28$, but then $x=720720\cdot28=20180160$ is $8$ digits long.

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

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)


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

$3536x+224 = 11y+5$

$5x+10 = 11y$ (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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