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smallest n digit number which is multiple of 7

Finding the smallest baseA big “smallest” numberWhat's the smallest number with first digit 1 that triples when this digit is moved to the end?Find the smallest natural number $n$Determine the smallest prime which does not divide any five digit number…Smallest multiple containing only 1's and 0'sSmallest number starting with N divisble by every non zero digit of NBy which convention 04 (for example) is no valid 2 digit number?What is the smallest number $n>2$ of this kind?What's the smallest three digit number that satisfies the following system of congruences?

2

2

$begingroup$

I want to find the smallest n digit number which is a multiple of $7$. For example, when $n=1$ then $7$ is the answer, if $n=2$ then $14$ is the answer, if $n=3$ then $105$ is the answer. What will be the answer for $n$th term?

sequences-and-series elementary-number-theory

share|cite|improve this question

edited Apr 29 at 15:48

tarit goswami

2,5291522

asked Apr 29 at 15:10

sr123sr123

374

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with modular arithmetic?
  $endgroup$
  – Arthur
  Apr 29 at 15:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes i am familiar.
  $endgroup$
  – sr123
  Apr 29 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $!bmod 7!:, 10^large n-1!+k,equiv, 0iff k,equiv,ldots $
  $endgroup$
  – Bill Dubuque
  Apr 29 at 15:16
  
  

  
  
  
  

add a comment | 

2

2

$begingroup$

I want to find the smallest n digit number which is a multiple of $7$. For example, when $n=1$ then $7$ is the answer, if $n=2$ then $14$ is the answer, if $n=3$ then $105$ is the answer. What will be the answer for $n$th term?

sequences-and-series elementary-number-theory

share|cite|improve this question

edited Apr 29 at 15:48

tarit goswami

2,5291522

asked Apr 29 at 15:10

sr123sr123

374

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with modular arithmetic?
  $endgroup$
  – Arthur
  Apr 29 at 15:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes i am familiar.
  $endgroup$
  – sr123
  Apr 29 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $!bmod 7!:, 10^large n-1!+k,equiv, 0iff k,equiv,ldots $
  $endgroup$
  – Bill Dubuque
  Apr 29 at 15:16
  
  

  
  
  
  

add a comment | 

$begingroup$

I want to find the smallest n digit number which is a multiple of $7$. For example, when $n=1$ then $7$ is the answer, if $n=2$ then $14$ is the answer, if $n=3$ then $105$ is the answer. What will be the answer for $n$th term?

sequences-and-series elementary-number-theory

share|cite|improve this question

edited Apr 29 at 15:48

tarit goswami

2,5291522

asked Apr 29 at 15:10

sr123sr123

374

$endgroup$

share|cite|improve this question

edited Apr 29 at 15:48

tarit goswami

2,5291522

asked Apr 29 at 15:10

sr123sr123

374

share|cite|improve this question

edited Apr 29 at 15:48

tarit goswami

2,5291522

asked Apr 29 at 15:10

sr123sr123

374

edited Apr 29 at 15:48

tarit goswami

2,5291522

edited Apr 29 at 15:48

tarit goswami

2,5291522

asked Apr 29 at 15:10

sr123sr123

374

asked Apr 29 at 15:10

sr123sr123

374

3 Answers
3

active

oldest

votes

4

$begingroup$

The $n$ th term of the sequence will be $$10^n-1+big(7-(10^n-1pmod7)big)$$

Because to get a $n$ digit number you will need at least $10^n-1$, now suppose the number is of the form $10^n-1+k$ for some natural $k$, for divisibility we need $(10^n-1+k)equiv 0 pmod7$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^n-1pmod7in [0,6]$, hence the result:
$$a_n=10^n-1+big(7-(10^n-1pmod7)big)$$

share|cite|improve this answer

edited Apr 29 at 15:36

answered Apr 29 at 15:20

tarit goswamitarit goswami

2,5291522

$endgroup$

add a comment | 

3

$begingroup$

Hint The function $k mapsto m leftlceil frackm rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.

For an integer $n > 0$ the smallest $n$-digit number is $10^n - 1$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac10^n - 17rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^n - 1 equiv 3^n - 1 pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^n - 1 + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.

share|cite|improve this answer

edited Apr 29 at 16:36

answered Apr 29 at 15:22

TravisTravis

65.1k769152

$endgroup$

add a comment | 

2

$begingroup$

Solve

$$(10^n+c)bmod 7=0$$ for the smallest positive $c$.

You have

$$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$

so that $$c=7-10^nbmod 7.$$

share|cite|improve this answer

edited Apr 29 at 15:27

answered Apr 29 at 15:22

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

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4

$begingroup$

The $n$ th term of the sequence will be $$10^n-1+big(7-(10^n-1pmod7)big)$$

Because to get a $n$ digit number you will need at least $10^n-1$, now suppose the number is of the form $10^n-1+k$ for some natural $k$, for divisibility we need $(10^n-1+k)equiv 0 pmod7$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^n-1pmod7in [0,6]$, hence the result:
$$a_n=10^n-1+big(7-(10^n-1pmod7)big)$$

share|cite|improve this answer

edited Apr 29 at 15:36

answered Apr 29 at 15:20

tarit goswamitarit goswami

2,5291522

$endgroup$

add a comment | 

4

$begingroup$

The $n$ th term of the sequence will be $$10^n-1+big(7-(10^n-1pmod7)big)$$

Because to get a $n$ digit number you will need at least $10^n-1$, now suppose the number is of the form $10^n-1+k$ for some natural $k$, for divisibility we need $(10^n-1+k)equiv 0 pmod7$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^n-1pmod7in [0,6]$, hence the result:
$$a_n=10^n-1+big(7-(10^n-1pmod7)big)$$

share|cite|improve this answer

edited Apr 29 at 15:36

answered Apr 29 at 15:20

tarit goswamitarit goswami

2,5291522

$endgroup$

add a comment | 

$begingroup$

The $n$ th term of the sequence will be $$10^n-1+big(7-(10^n-1pmod7)big)$$

Because to get a $n$ digit number you will need at least $10^n-1$, now suppose the number is of the form $10^n-1+k$ for some natural $k$, for divisibility we need $(10^n-1+k)equiv 0 pmod7$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^n-1pmod7in [0,6]$, hence the result:
$$a_n=10^n-1+big(7-(10^n-1pmod7)big)$$

share|cite|improve this answer

edited Apr 29 at 15:36

answered Apr 29 at 15:20

tarit goswamitarit goswami

2,5291522

$endgroup$

share|cite|improve this answer

edited Apr 29 at 15:36

answered Apr 29 at 15:20

tarit goswamitarit goswami

2,5291522

answered Apr 29 at 15:20

tarit goswamitarit goswami

2,5291522

answered Apr 29 at 15:20

tarit goswamitarit goswami

2,5291522

3

$begingroup$

Hint The function $k mapsto m leftlceil frackm rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.

For an integer $n > 0$ the smallest $n$-digit number is $10^n - 1$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac10^n - 17rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^n - 1 equiv 3^n - 1 pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^n - 1 + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.

share|cite|improve this answer

edited Apr 29 at 16:36

answered Apr 29 at 15:22

TravisTravis

65.1k769152

$endgroup$

add a comment | 

3

$begingroup$

Hint The function $k mapsto m leftlceil frackm rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.

For an integer $n > 0$ the smallest $n$-digit number is $10^n - 1$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac10^n - 17rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^n - 1 equiv 3^n - 1 pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^n - 1 + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.

share|cite|improve this answer

edited Apr 29 at 16:36

answered Apr 29 at 15:22

TravisTravis

65.1k769152

$endgroup$

add a comment | 

$begingroup$

Hint The function $k mapsto m leftlceil frackm rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.

For an integer $n > 0$ the smallest $n$-digit number is $10^n - 1$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac10^n - 17rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^n - 1 equiv 3^n - 1 pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^n - 1 + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.

share|cite|improve this answer

edited Apr 29 at 16:36

answered Apr 29 at 15:22

TravisTravis

65.1k769152

$endgroup$

share|cite|improve this answer

edited Apr 29 at 16:36

answered Apr 29 at 15:22

TravisTravis

65.1k769152

answered Apr 29 at 15:22

TravisTravis

65.1k769152

answered Apr 29 at 15:22

TravisTravis

65.1k769152

2

$begingroup$

Solve

$$(10^n+c)bmod 7=0$$ for the smallest positive $c$.

You have

$$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$

so that $$c=7-10^nbmod 7.$$

share|cite|improve this answer

edited Apr 29 at 15:27

answered Apr 29 at 15:22

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

2

$begingroup$

Solve

$$(10^n+c)bmod 7=0$$ for the smallest positive $c$.

You have

$$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$

so that $$c=7-10^nbmod 7.$$

share|cite|improve this answer

edited Apr 29 at 15:27

answered Apr 29 at 15:22

Yves DaoustYves Daoust

135k676233

$endgroup$

add a comment | 

$begingroup$

Solve

$$(10^n+c)bmod 7=0$$ for the smallest positive $c$.

You have

$$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$

so that $$c=7-10^nbmod 7.$$

share|cite|improve this answer

edited Apr 29 at 15:27

answered Apr 29 at 15:22

Yves DaoustYves Daoust

135k676233

$endgroup$

share|cite|improve this answer

edited Apr 29 at 15:27

answered Apr 29 at 15:22

Yves DaoustYves Daoust

135k676233

answered Apr 29 at 15:22

Yves DaoustYves Daoust

135k676233

answered Apr 29 at 15:22

Yves DaoustYves Daoust

135k676233