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he and she - er und sie

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6

$begingroup$

The formula $HE=sqrtSHE$ translates in German to $ER=sqrtSIE$.
Find the solution for both, where each letter represents a digit.

(These are two separate puzzles: digits represented by S and E don't have to be the same between languages.)

pattern

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edited 7 hours ago

Bass

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asked 8 hours ago

ThomasLThomasL

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$endgroup$

add a comment | 

6

$begingroup$

The formula $HE=sqrtSHE$ translates in German to $ER=sqrtSIE$.
Find the solution for both, where each letter represents a digit.

(These are two separate puzzles: digits represented by S and E don't have to be the same between languages.)

pattern

share|improve this question

edited 7 hours ago

Bass

33.9k44 gold badges8181 silver badges199199 bronze badges

asked 8 hours ago

ThomasLThomasL

21311 silver badge88 bronze badges

$endgroup$

add a comment | 

$begingroup$

The formula $HE=sqrtSHE$ translates in German to $ER=sqrtSIE$.
Find the solution for both, where each letter represents a digit.

(These are two separate puzzles: digits represented by S and E don't have to be the same between languages.)

pattern

share|improve this question

edited 7 hours ago

Bass

33.9k44 gold badges8181 silver badges199199 bronze badges

asked 8 hours ago

ThomasLThomasL

21311 silver badge88 bronze badges

$endgroup$

share|improve this question

edited 7 hours ago

Bass

33.9k44 gold badges8181 silver badges199199 bronze badges

asked 8 hours ago

ThomasLThomasL

21311 silver badge88 bronze badges

share|improve this question

edited 7 hours ago

Bass

33.9k44 gold badges8181 silver badges199199 bronze badges

asked 8 hours ago

ThomasLThomasL

21311 silver badge88 bronze badges

edited 7 hours ago

Bass

33.9k44 gold badges8181 silver badges199199 bronze badges

edited 7 hours ago

Bass

33.9k44 gold badges8181 silver badges199199 bronze badges

asked 8 hours ago

ThomasLThomasL

21311 silver badge88 bronze badges

asked 8 hours ago

ThomasLThomasL

21311 silver badge88 bronze badges

1 Answer
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7

$begingroup$

If $HE^2=SHE$ then

$HE$ (call it $x$ for convenience) has the property that $x=x^2$ mod 100. Hence $x$ is either 0 or 1 mod each of 4,25. We can't have either 0,0 or 1,1 because then we actually have $x^2=x$ which would mean $S=0$ (and also $H=0$). 0,1 yields $HE=76$ whose square is clearly too big. $1,0$ yields $HE=25$ which works ($SHE=625$).

If $ER^2=SIE$ then

clearly $0<Eleq3$ (else the square has too many digits) and $E$ is the last digit of a square so $E=1$. Then $R$ must be 1 (no!) or 9, leading to $ER=19$ and $SIE=361$.

share|improve this answer

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

75.4k33 gold badges189189 silver badges291291 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And here I thought they used the same rule...
  $endgroup$
  – AxiomaticSystem
  8 hours ago
  

  
  
  
  

add a comment | 

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7

$begingroup$

If $HE^2=SHE$ then

$HE$ (call it $x$ for convenience) has the property that $x=x^2$ mod 100. Hence $x$ is either 0 or 1 mod each of 4,25. We can't have either 0,0 or 1,1 because then we actually have $x^2=x$ which would mean $S=0$ (and also $H=0$). 0,1 yields $HE=76$ whose square is clearly too big. $1,0$ yields $HE=25$ which works ($SHE=625$).

If $ER^2=SIE$ then

clearly $0<Eleq3$ (else the square has too many digits) and $E$ is the last digit of a square so $E=1$. Then $R$ must be 1 (no!) or 9, leading to $ER=19$ and $SIE=361$.

share|improve this answer

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

75.4k33 gold badges189189 silver badges291291 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And here I thought they used the same rule...
  $endgroup$
  – AxiomaticSystem
  8 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

If $HE^2=SHE$ then

$HE$ (call it $x$ for convenience) has the property that $x=x^2$ mod 100. Hence $x$ is either 0 or 1 mod each of 4,25. We can't have either 0,0 or 1,1 because then we actually have $x^2=x$ which would mean $S=0$ (and also $H=0$). 0,1 yields $HE=76$ whose square is clearly too big. $1,0$ yields $HE=25$ which works ($SHE=625$).

If $ER^2=SIE$ then

clearly $0<Eleq3$ (else the square has too many digits) and $E$ is the last digit of a square so $E=1$. Then $R$ must be 1 (no!) or 9, leading to $ER=19$ and $SIE=361$.

share|improve this answer

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

75.4k33 gold badges189189 silver badges291291 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And here I thought they used the same rule...
  $endgroup$
  – AxiomaticSystem
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If $HE^2=SHE$ then

$HE$ (call it $x$ for convenience) has the property that $x=x^2$ mod 100. Hence $x$ is either 0 or 1 mod each of 4,25. We can't have either 0,0 or 1,1 because then we actually have $x^2=x$ which would mean $S=0$ (and also $H=0$). 0,1 yields $HE=76$ whose square is clearly too big. $1,0$ yields $HE=25$ which works ($SHE=625$).

If $ER^2=SIE$ then

clearly $0<Eleq3$ (else the square has too many digits) and $E$ is the last digit of a square so $E=1$. Then $R$ must be 1 (no!) or 9, leading to $ER=19$ and $SIE=361$.

share|improve this answer

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

75.4k33 gold badges189189 silver badges291291 bronze badges

$endgroup$

share|improve this answer

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

75.4k33 gold badges189189 silver badges291291 bronze badges

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

75.4k33 gold badges189189 silver badges291291 bronze badges

answered 8 hours ago

Gareth McCaughan♦Gareth McCaughan

75.4k33 gold badges189189 silver badges291291 bronze badges