modems and cas #s
$begingroup$
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
asked May 28 '18 at 16:38
user18842sosuser18842sos
1023
$endgroup$
add a comment |
$begingroup$
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
asked May 28 '18 at 16:38
user18842sosuser18842sos
1023
$endgroup$
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
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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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– iBot
27 mins ago
add a comment |
$begingroup$
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
asked May 28 '18 at 16:38
user18842sosuser18842sos
1023
$endgroup$
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
asked May 28 '18 at 16:38
user18842sosuser18842sos
1023
asked May 28 '18 at 16:38
user18842sosuser18842sos
1023
edited 3 mins ago
user18842sos
asked May 28 '18 at 16:38
user18842sosuser18842sos
1023
asked May 28 '18 at 16:38
user18842sosuser18842sos
1023
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 :)
$endgroup$
– 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
27 mins ago
add a comment |
1
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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
27 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
$begingroup$
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
27 mins ago
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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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– iBot
27 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$7927937$.
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.
answered May 28 '18 at 16:43
noednenoedne
6,34011855
$endgroup$
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
add a comment |
$begingroup$
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 $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.
answered May 28 '18 at 18:52
NautilusNautilus
4,048525
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$7927937$.
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.
answered May 28 '18 at 16:43
noednenoedne
6,34011855
$endgroup$
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
add a comment |
$begingroup$
$7927937$.
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.
answered May 28 '18 at 16:43
noednenoedne
6,34011855
$endgroup$
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
add a comment |
$begingroup$
$7927937$.
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.
answered May 28 '18 at 16:43
noednenoedne
6,34011855
$endgroup$
$7927937$.
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.
answered May 28 '18 at 16:43
noednenoedne
6,34011855
edited May 28 '18 at 17:31
answered May 28 '18 at 16:43
noednenoedne
6,34011855
answered May 28 '18 at 16:43
noednenoedne
6,34011855
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
add a comment |
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
add a comment |
$begingroup$
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 $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.
answered May 28 '18 at 18:52
NautilusNautilus
4,048525
$endgroup$
add a comment |
$begingroup$
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 $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.
answered May 28 '18 at 18:52
NautilusNautilus
4,048525
$endgroup$
add a comment |
$begingroup$
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 $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.
answered May 28 '18 at 18:52
NautilusNautilus
4,048525
$endgroup$
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 $t=3 => abcdefg+17 = 7927936 + 1 = 7927937$.
answered May 28 '18 at 18:52
NautilusNautilus
4,048525
edited May 28 '18 at 19:00
answered May 28 '18 at 18:52
NautilusNautilus
4,048525
answered May 28 '18 at 18:52
NautilusNautilus
4,048525
answered May 28 '18 at 18:52
NautilusNautilus
4,048525
4,048525
add a comment |
add a comment |
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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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– 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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– iBot
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