Ttdfjt

Why did Robert F. Kennedy loathe Lyndon B. Johnson?

Can a landlord force all residents to use the landlord's in-house debit card accounts?

Why does Trump want a citizenship question on the census?

Is there a formal/better word than "skyrocket" for the given context?

Can one block with a protection from color creature?

Interpretation of non-significant results as "trends"

What was the profession 芸者 (female entertainer) called in Russia?

Is "wissen" the only verb in German to have an irregular present tense?

Moving millions of files to a different directory with specfic name patterns

How should I ask for a "pint" in countries that use metric?

Write a function

Four ships at the ocean with the same distance

Why won't the U.S. sign a peace treaty with North Korea?

Was it ever illegal to name a pig "Napoleon" in France?

What do you call a situation where you have choices but no good choice?

Strong Password Detection in Python

When do flights get cancelled due to fog?

Sense of humor in your sci-fi stories

Is it possible for a character at any level to cast all 44 Cantrips in one week without Magic Items?

Why is a mixture of two normally distributed variables only bimodal if their means differ by at least two times the common standard deviation?

How to evaluate the performance of open source solver?

My professor has told me he will be the corresponding author. Will it hurt my future career?

Passwordless authentication - how and when to invalidate a login code

Blocks from @ jafe

he and she - er und sie

Oh no! Yet another number series… - Find the pattern in the two examples and provide an extensionAnother Company of ThirteenName ProgressionA basic calculator, a simple game. What was I playing at?The London safes and their mysterious combinationsColumn-complete latin square problem: everyone rotates through all seats, never sitting at the same seat as another more than onceAn enigmatic pilgrimageHow to tell if a Mathematical Rubiks puzzle is solvableHotel Combination & Room Number: Pattern?Five Letter Boxed puzzles with special solutions

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

7

$begingroup$

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 Jun 28 at 21:54

Bass

34.3k44 gold badges8282 silver badges201201 bronze badges

asked Jun 28 at 21:16

ThomasLThomasL

34311 silver badge1111 bronze badges

$endgroup$

add a comment | 

7

$begingroup$

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 Jun 28 at 21:54

Bass

34.3k44 gold badges8282 silver badges201201 bronze badges

asked Jun 28 at 21:16

ThomasLThomasL

34311 silver badge1111 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 Jun 28 at 21:54

Bass

34.3k44 gold badges8282 silver badges201201 bronze badges

asked Jun 28 at 21:16

ThomasLThomasL

34311 silver badge1111 bronze badges

$endgroup$

share|improve this question

edited Jun 28 at 21:54

Bass

34.3k44 gold badges8282 silver badges201201 bronze badges

asked Jun 28 at 21:16

ThomasLThomasL

34311 silver badge1111 bronze badges

share|improve this question

edited Jun 28 at 21:54

Bass

34.3k44 gold badges8282 silver badges201201 bronze badges

asked Jun 28 at 21:16

ThomasLThomasL

34311 silver badge1111 bronze badges

edited Jun 28 at 21:54

Bass

34.3k44 gold badges8282 silver badges201201 bronze badges

edited Jun 28 at 21:54

Bass

34.3k44 gold badges8282 silver badges201201 bronze badges

asked Jun 28 at 21:16

ThomasLThomasL

34311 silver badge1111 bronze badges

asked Jun 28 at 21:16

ThomasLThomasL

34311 silver badge1111 bronze badges

1 Answer
1

active

oldest

votes

8

$begingroup$

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 Jun 28 at 21:34

Gareth McCaughan♦Gareth McCaughan

76k33 gold badges189189 silver badges292292 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And here I thought they used the same rule...
  $endgroup$
  – AxiomaticSystem
  Jun 28 at 21:37
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "559"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f85658%2fhe-and-she-er-und-sie%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

8

$begingroup$

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 Jun 28 at 21:34

Gareth McCaughan♦Gareth McCaughan

76k33 gold badges189189 silver badges292292 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And here I thought they used the same rule...
  $endgroup$
  – AxiomaticSystem
  Jun 28 at 21:37
  

  
  
  
  

add a comment | 

8

$begingroup$

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 Jun 28 at 21:34

Gareth McCaughan♦Gareth McCaughan

76k33 gold badges189189 silver badges292292 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And here I thought they used the same rule...
  $endgroup$
  – AxiomaticSystem
  Jun 28 at 21:37
  

  
  
  
  

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 Jun 28 at 21:34

Gareth McCaughan♦Gareth McCaughan

76k33 gold badges189189 silver badges292292 bronze badges

$endgroup$

share|improve this answer

answered Jun 28 at 21:34

Gareth McCaughan♦Gareth McCaughan

76k33 gold badges189189 silver badges292292 bronze badges

answered Jun 28 at 21:34

Gareth McCaughan♦Gareth McCaughan

76k33 gold badges189189 silver badges292292 bronze badges

answered Jun 28 at 21:34

Gareth McCaughan♦Gareth McCaughan

76k33 gold badges189189 silver badges292292 bronze badges