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How to check if some number, say a, can be expressed as a = b + reverse(b) ? Where b is some natural number.

In how many ways can a number be expressed as a sum of consecutive numbers?How Many Natural Numbers Can Get Expressed as the Sum of Consecutive Natural Numbers in only One Way?What is the significance of the mirror numbers?Are numbers that are already palindromes Lychrel numbers?How to prove that every natural number not of the form $4^n(8m+7)$ can be written as $x^2 + y^2 + z^2$?Number of ways in which a natural number can be expressed as a sum of natural numbers?Natural number which can be expressed as sum of two perfect squares in two different ways?How to prove Prime numbers can be expressed as $6kpm 1$How do I show that any natural number of this expression is a natural linear combination?Proving every natural number can be expressed as $2^{x} cdot {y}$

1

1

$begingroup$

How to check if some number, say $a$, can be expressed as $a = b + mathrm{reverse}(b)$ ? Where $a$ and $b$ are some natural numbers.

Some examples:

1. 
   $22 = 11 + mathrm{rev}(11) = 11 + 11$;
   That is, $22$ can be expressed as sum of "another natural number and its reverse".


2. 
   $121 = 29 + mathrm{rev}(29) = 29 + 92$;
   $9$ can't be expressed as sum of "a number and its reverse".

number-theory elementary-number-theory recreational-mathematics

share|cite|improve this question

edited Mar 13 at 2:01

Brian

758116

asked Mar 12 at 22:42

kapilkapil

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's also not necessary, if we always assume that the reverse of a number takes its smallest decimal representation. e.g. 8 = 4 + rev(4).
  $endgroup$
  – ConMan
  Mar 12 at 22:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonThousand Divisibility by $11$ only works if the number of digits in the seed number is even, or if the seed number is divisible by $11$. $123+321=444$.
  $endgroup$
  – Mark Bennet
  Mar 12 at 22:54
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonThousand It isn't a necessary condition. $444$ is not divisible by $11$
  $endgroup$
  – Mark Bennet
  Mar 12 at 22:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check all natural numbers $0<b<a$ of which there are only a finite number of them.
  $endgroup$
  – Somos
  Mar 12 at 22:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Relevant oeis
  $endgroup$
  – Don Thousand
  Mar 12 at 23:02
  

  
  
  
  

add a comment | 

1

1

$begingroup$

How to check if some number, say $a$, can be expressed as $a = b + mathrm{reverse}(b)$ ? Where $a$ and $b$ are some natural numbers.

Some examples:

1. 
   $22 = 11 + mathrm{rev}(11) = 11 + 11$;
   That is, $22$ can be expressed as sum of "another natural number and its reverse".


2. 
   $121 = 29 + mathrm{rev}(29) = 29 + 92$;
   $9$ can't be expressed as sum of "a number and its reverse".

number-theory elementary-number-theory recreational-mathematics

share|cite|improve this question

edited Mar 13 at 2:01

Brian

758116

asked Mar 12 at 22:42

kapilkapil

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's also not necessary, if we always assume that the reverse of a number takes its smallest decimal representation. e.g. 8 = 4 + rev(4).
  $endgroup$
  – ConMan
  Mar 12 at 22:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonThousand Divisibility by $11$ only works if the number of digits in the seed number is even, or if the seed number is divisible by $11$. $123+321=444$.
  $endgroup$
  – Mark Bennet
  Mar 12 at 22:54
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DonThousand It isn't a necessary condition. $444$ is not divisible by $11$
  $endgroup$
  – Mark Bennet
  Mar 12 at 22:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check all natural numbers $0<b<a$ of which there are only a finite number of them.
  $endgroup$
  – Somos
  Mar 12 at 22:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Relevant oeis
  $endgroup$
  – Don Thousand
  Mar 12 at 23:02
  

  
  
  
  

add a comment | 

$begingroup$

How to check if some number, say $a$, can be expressed as $a = b + mathrm{reverse}(b)$ ? Where $a$ and $b$ are some natural numbers.

Some examples:

1. 
   $22 = 11 + mathrm{rev}(11) = 11 + 11$;
   That is, $22$ can be expressed as sum of "another natural number and its reverse".


2. 
   $121 = 29 + mathrm{rev}(29) = 29 + 92$;
   $9$ can't be expressed as sum of "a number and its reverse".

number-theory elementary-number-theory recreational-mathematics

share|cite|improve this question

edited Mar 13 at 2:01

Brian

758116

asked Mar 12 at 22:42

kapilkapil

112

$endgroup$

share|cite|improve this question

edited Mar 13 at 2:01

Brian

758116

asked Mar 12 at 22:42

kapilkapil

112

share|cite|improve this question

edited Mar 13 at 2:01

Brian

758116

asked Mar 12 at 22:42

kapilkapil

112

edited Mar 13 at 2:01

Brian

758116

edited Mar 13 at 2:01

Brian

758116

asked Mar 12 at 22:42

kapilkapil

112

asked Mar 12 at 22:42

kapilkapil

112

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $b = sumlimits_{k=0}^n b_k 10^k$

$b + rev(b) = sumlimits_{k=0}^{n}(b_k + b_{n-k})10^k < 2*10^{k+1}$.

Suppose $a$ is an $n+1$ digit number.

Case 1: The first digit of $a$ is not equal to $1$. Then if $a = b + rev(b)$ then $b$ is also an digit number and $b_0 + b_k < 10$. So the first digit and last digit of $a$ are are $b_0 + b_k$ and equal.

If the first digit and last digit of $a$ are not equal then $a $ is not is not $b+ rev(b)$. If the first digit and last digit or then remove the first digit and last digit, divide by $10$ and repeat the process.

Case 1a: The first and last digit are $1$. To see if $b_0 + b_k = 1$ and $a = b + rev(b)$, revert to case 1. If it fails, continue to case 2.

Case 2: The first digit of $a$ is $1$. If $b_0 + b_k ge 10+ $ last digit of $a = m$ and $a = b + rev (b)$ then subtract $m$ and $m*10^{n-1}$ from $a$. Divide by $10$. Repeat the process.

Either the process will end at a single digit and $a = b+ rev(b)$ or the process will fail and $a ne b + rev (b)$ for any $b$.

Example: $59406 + 60495=119901$ so

so subtract $100001$ to get $19900$ and divide by $10$ to get $1990$. Subtract $10$ and $1000$ to get $980$ and divide by $10$ to get $98$ and that was a failure.

So go back to $119901$ and subtract $11$ and $110000$ so get $9890$ and divide by $10$ to get $989$. Subtract $909$ and get $80$ and divid by $10$ to get $8$. That's a success.

share|cite|improve this answer

answered Mar 12 at 23:46

fleabloodfleablood

72.8k22788

$endgroup$

add a comment | 

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1

$begingroup$

Let $b = sumlimits_{k=0}^n b_k 10^k$

$b + rev(b) = sumlimits_{k=0}^{n}(b_k + b_{n-k})10^k < 2*10^{k+1}$.

Suppose $a$ is an $n+1$ digit number.

Case 1: The first digit of $a$ is not equal to $1$. Then if $a = b + rev(b)$ then $b$ is also an digit number and $b_0 + b_k < 10$. So the first digit and last digit of $a$ are are $b_0 + b_k$ and equal.

If the first digit and last digit of $a$ are not equal then $a $ is not is not $b+ rev(b)$. If the first digit and last digit or then remove the first digit and last digit, divide by $10$ and repeat the process.

Case 1a: The first and last digit are $1$. To see if $b_0 + b_k = 1$ and $a = b + rev(b)$, revert to case 1. If it fails, continue to case 2.

Case 2: The first digit of $a$ is $1$. If $b_0 + b_k ge 10+ $ last digit of $a = m$ and $a = b + rev (b)$ then subtract $m$ and $m*10^{n-1}$ from $a$. Divide by $10$. Repeat the process.

Either the process will end at a single digit and $a = b+ rev(b)$ or the process will fail and $a ne b + rev (b)$ for any $b$.

Example: $59406 + 60495=119901$ so

so subtract $100001$ to get $19900$ and divide by $10$ to get $1990$. Subtract $10$ and $1000$ to get $980$ and divide by $10$ to get $98$ and that was a failure.

So go back to $119901$ and subtract $11$ and $110000$ so get $9890$ and divide by $10$ to get $989$. Subtract $909$ and get $80$ and divid by $10$ to get $8$. That's a success.

share|cite|improve this answer

answered Mar 12 at 23:46

fleabloodfleablood

72.8k22788

$endgroup$

add a comment | 

1

$begingroup$

Let $b = sumlimits_{k=0}^n b_k 10^k$

$b + rev(b) = sumlimits_{k=0}^{n}(b_k + b_{n-k})10^k < 2*10^{k+1}$.

Suppose $a$ is an $n+1$ digit number.

Case 1: The first digit of $a$ is not equal to $1$. Then if $a = b + rev(b)$ then $b$ is also an digit number and $b_0 + b_k < 10$. So the first digit and last digit of $a$ are are $b_0 + b_k$ and equal.

If the first digit and last digit of $a$ are not equal then $a $ is not is not $b+ rev(b)$. If the first digit and last digit or then remove the first digit and last digit, divide by $10$ and repeat the process.

Case 1a: The first and last digit are $1$. To see if $b_0 + b_k = 1$ and $a = b + rev(b)$, revert to case 1. If it fails, continue to case 2.

Case 2: The first digit of $a$ is $1$. If $b_0 + b_k ge 10+ $ last digit of $a = m$ and $a = b + rev (b)$ then subtract $m$ and $m*10^{n-1}$ from $a$. Divide by $10$. Repeat the process.

Either the process will end at a single digit and $a = b+ rev(b)$ or the process will fail and $a ne b + rev (b)$ for any $b$.

Example: $59406 + 60495=119901$ so

so subtract $100001$ to get $19900$ and divide by $10$ to get $1990$. Subtract $10$ and $1000$ to get $980$ and divide by $10$ to get $98$ and that was a failure.

So go back to $119901$ and subtract $11$ and $110000$ so get $9890$ and divide by $10$ to get $989$. Subtract $909$ and get $80$ and divid by $10$ to get $8$. That's a success.

share|cite|improve this answer

answered Mar 12 at 23:46

fleabloodfleablood

72.8k22788

$endgroup$

add a comment | 

$begingroup$

Let $b = sumlimits_{k=0}^n b_k 10^k$

$b + rev(b) = sumlimits_{k=0}^{n}(b_k + b_{n-k})10^k < 2*10^{k+1}$.

Suppose $a$ is an $n+1$ digit number.

Case 1: The first digit of $a$ is not equal to $1$. Then if $a = b + rev(b)$ then $b$ is also an digit number and $b_0 + b_k < 10$. So the first digit and last digit of $a$ are are $b_0 + b_k$ and equal.

If the first digit and last digit of $a$ are not equal then $a $ is not is not $b+ rev(b)$. If the first digit and last digit or then remove the first digit and last digit, divide by $10$ and repeat the process.

Case 1a: The first and last digit are $1$. To see if $b_0 + b_k = 1$ and $a = b + rev(b)$, revert to case 1. If it fails, continue to case 2.

Case 2: The first digit of $a$ is $1$. If $b_0 + b_k ge 10+ $ last digit of $a = m$ and $a = b + rev (b)$ then subtract $m$ and $m*10^{n-1}$ from $a$. Divide by $10$. Repeat the process.

Either the process will end at a single digit and $a = b+ rev(b)$ or the process will fail and $a ne b + rev (b)$ for any $b$.

Example: $59406 + 60495=119901$ so

so subtract $100001$ to get $19900$ and divide by $10$ to get $1990$. Subtract $10$ and $1000$ to get $980$ and divide by $10$ to get $98$ and that was a failure.

So go back to $119901$ and subtract $11$ and $110000$ so get $9890$ and divide by $10$ to get $989$. Subtract $909$ and get $80$ and divid by $10$ to get $8$. That's a success.

share|cite|improve this answer

answered Mar 12 at 23:46

fleabloodfleablood

72.8k22788

$endgroup$

share|cite|improve this answer

answered Mar 12 at 23:46

fleabloodfleablood

72.8k22788

answered Mar 12 at 23:46

fleabloodfleablood

72.8k22788

answered Mar 12 at 23:46

fleabloodfleablood

72.8k22788