Usbrth

Why did the IBM 650 use bi-quinary?

Is above average number of years spent on PhD considered a red flag in future academia or industry positions?

When is phishing education going too far?

Models of set theory where not every set can be linearly ordered

How can I make names more distinctive without making them longer?

Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?

Output the ŋarâþ crîþ alphabet song without using (m)any letters

Bonus calculation: Am I making a mountain out of a molehill?

Should I call the interviewer directly, if HR aren't responding?

The logistics of corpse disposal

What's the purpose of writing one's academic bio in 3rd person?

Is there a concise way to say "all of the X, one of each"?

Storing hydrofluoric acid before the invention of plastics

When -s is used with third person singular. What's its use in this context?

What happens to sewage if there is no river near by?

"Seemed to had" is it correct?

3 doors, three guards, one stone

Why is "Consequences inflicted." not a sentence?

Were Kohanim forbidden from serving in King David's army?

Sorting numerically

Using et al. for a last / senior author rather than for a first author

How to find all the available tools in macOS terminal?

Right-skewed distribution with mean equals to mode?

Can inflation occur in a positive-sum game currency system such as the Stack Exchange reputation system?

How many marbles?

0

$begingroup$

One dozen of big marbles and small marbles is 132 gram. If one big marbles is 3 gram heavier than one small marbles, then specify the possibilities of how many are the big marbles and the small marbles.

I got $x+y=12$, $ax+by=132$, and $a=3+b$
with $x$ is how many big marbles, $y$ is small marbles, $a$ and $b$ are their mass. I don't know what to do next.

number-theory diophantine-equations greatest-common-divisor euclidean-algorithm least-common-multiple

share|cite|improve this question

edited Apr 6 at 3:46

Fifi12

asked Apr 6 at 3:08

Fifi12Fifi12

93

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think your third equation should be $a=3+b$
  $endgroup$
  – Strichcoder
  Apr 6 at 3:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Currently, the problem is underdefined. All solutions for $y=4b-32$ are viable.
  $endgroup$
  – Strichcoder
  Apr 6 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh yes I made a mistake and also the question didn't ask for a define answer, but just the possibility of how many they are. I'm sorry, I translated the question on my own from another language, so I left some things.
  $endgroup$
  – Fifi12
  Apr 6 at 3:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can edit your question. Do you still need help?
  $endgroup$
  – Strichcoder
  Apr 6 at 3:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I got it... I found that if $t$ is an integer, then $a=12+t$, $b=9+t$, $x=8-4t$, and $y=4+4t$
  $endgroup$
  – Fifi12
  Apr 6 at 4:12
  

  
  
  
  

 | 
show 10 more comments

0

$begingroup$

One dozen of big marbles and small marbles is 132 gram. If one big marbles is 3 gram heavier than one small marbles, then specify the possibilities of how many are the big marbles and the small marbles.

I got $x+y=12$, $ax+by=132$, and $a=3+b$
with $x$ is how many big marbles, $y$ is small marbles, $a$ and $b$ are their mass. I don't know what to do next.

number-theory diophantine-equations greatest-common-divisor euclidean-algorithm least-common-multiple

share|cite|improve this question

edited Apr 6 at 3:46

Fifi12

asked Apr 6 at 3:08

Fifi12Fifi12

93

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think your third equation should be $a=3+b$
  $endgroup$
  – Strichcoder
  Apr 6 at 3:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Currently, the problem is underdefined. All solutions for $y=4b-32$ are viable.
  $endgroup$
  – Strichcoder
  Apr 6 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh yes I made a mistake and also the question didn't ask for a define answer, but just the possibility of how many they are. I'm sorry, I translated the question on my own from another language, so I left some things.
  $endgroup$
  – Fifi12
  Apr 6 at 3:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can edit your question. Do you still need help?
  $endgroup$
  – Strichcoder
  Apr 6 at 3:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I got it... I found that if $t$ is an integer, then $a=12+t$, $b=9+t$, $x=8-4t$, and $y=4+4t$
  $endgroup$
  – Fifi12
  Apr 6 at 4:12
  

  
  
  
  

 | 
show 10 more comments

$begingroup$

One dozen of big marbles and small marbles is 132 gram. If one big marbles is 3 gram heavier than one small marbles, then specify the possibilities of how many are the big marbles and the small marbles.

I got $x+y=12$, $ax+by=132$, and $a=3+b$
with $x$ is how many big marbles, $y$ is small marbles, $a$ and $b$ are their mass. I don't know what to do next.

number-theory diophantine-equations greatest-common-divisor euclidean-algorithm least-common-multiple

share|cite|improve this question

edited Apr 6 at 3:46

Fifi12

asked Apr 6 at 3:08

Fifi12Fifi12

93

$endgroup$

share|cite|improve this question

edited Apr 6 at 3:46

Fifi12

asked Apr 6 at 3:08

Fifi12Fifi12

93

share|cite|improve this question

edited Apr 6 at 3:46

Fifi12

asked Apr 6 at 3:08

Fifi12Fifi12

93

asked Apr 6 at 3:08

Fifi12Fifi12

93

asked Apr 6 at 3:08

Fifi12Fifi12

93

1 Answer
1

active

oldest

votes

0

$begingroup$

Let's express everything in terms of $b$. We have $a=b+3$. Putting that in the second equation gives $(b+3)x+by=132$, which is $3x+b(x+y)=132$, but from the first equation $x+y=12$, so $3x+12b=132$, so $x=44-4b$. Then $y=12-x=4b-32$.

Now, we need $xge0$, so $ble11$, and $yge0$, so $bge8$. So, $8le ble11$. Further, $x$ and $y$ need to be whole numbers, so $4b$ must be a whole number, so $b$ must be one of the 13 numbers $8,33/4,17/2,dots,11$. For each such $b$, you can use the formulas already derived to work out $x$, $y$, and $a$.

EDIT: The upshot is that $x$ can be any whole number between zero and twelve, inclusive. Then $y=12-x$, $b=11-(x/4)$, and $a=14-(x/4)$.

MORE EDIT: If you read the question to imply there must be at least one marble of each of the two types, then $x$ can be any whole number between one and eleven, inclusive (and $y,b,a$ are given by the formula above).

share|cite|improve this answer

edited Apr 12 at 0:14

answered Apr 9 at 2:50

Gerry MyersonGerry Myerson

148k8152306

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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%2fmath.stackexchange.com%2fquestions%2f3176628%2fhow-many-marbles%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Let's express everything in terms of $b$. We have $a=b+3$. Putting that in the second equation gives $(b+3)x+by=132$, which is $3x+b(x+y)=132$, but from the first equation $x+y=12$, so $3x+12b=132$, so $x=44-4b$. Then $y=12-x=4b-32$.

Now, we need $xge0$, so $ble11$, and $yge0$, so $bge8$. So, $8le ble11$. Further, $x$ and $y$ need to be whole numbers, so $4b$ must be a whole number, so $b$ must be one of the 13 numbers $8,33/4,17/2,dots,11$. For each such $b$, you can use the formulas already derived to work out $x$, $y$, and $a$.

EDIT: The upshot is that $x$ can be any whole number between zero and twelve, inclusive. Then $y=12-x$, $b=11-(x/4)$, and $a=14-(x/4)$.

MORE EDIT: If you read the question to imply there must be at least one marble of each of the two types, then $x$ can be any whole number between one and eleven, inclusive (and $y,b,a$ are given by the formula above).

share|cite|improve this answer

edited Apr 12 at 0:14

answered Apr 9 at 2:50

Gerry MyersonGerry Myerson

148k8152306

$endgroup$

add a comment | 

0

$begingroup$

Let's express everything in terms of $b$. We have $a=b+3$. Putting that in the second equation gives $(b+3)x+by=132$, which is $3x+b(x+y)=132$, but from the first equation $x+y=12$, so $3x+12b=132$, so $x=44-4b$. Then $y=12-x=4b-32$.

Now, we need $xge0$, so $ble11$, and $yge0$, so $bge8$. So, $8le ble11$. Further, $x$ and $y$ need to be whole numbers, so $4b$ must be a whole number, so $b$ must be one of the 13 numbers $8,33/4,17/2,dots,11$. For each such $b$, you can use the formulas already derived to work out $x$, $y$, and $a$.

EDIT: The upshot is that $x$ can be any whole number between zero and twelve, inclusive. Then $y=12-x$, $b=11-(x/4)$, and $a=14-(x/4)$.

MORE EDIT: If you read the question to imply there must be at least one marble of each of the two types, then $x$ can be any whole number between one and eleven, inclusive (and $y,b,a$ are given by the formula above).

share|cite|improve this answer

edited Apr 12 at 0:14

answered Apr 9 at 2:50

Gerry MyersonGerry Myerson

148k8152306

$endgroup$

add a comment | 

$begingroup$

Let's express everything in terms of $b$. We have $a=b+3$. Putting that in the second equation gives $(b+3)x+by=132$, which is $3x+b(x+y)=132$, but from the first equation $x+y=12$, so $3x+12b=132$, so $x=44-4b$. Then $y=12-x=4b-32$.

Now, we need $xge0$, so $ble11$, and $yge0$, so $bge8$. So, $8le ble11$. Further, $x$ and $y$ need to be whole numbers, so $4b$ must be a whole number, so $b$ must be one of the 13 numbers $8,33/4,17/2,dots,11$. For each such $b$, you can use the formulas already derived to work out $x$, $y$, and $a$.

EDIT: The upshot is that $x$ can be any whole number between zero and twelve, inclusive. Then $y=12-x$, $b=11-(x/4)$, and $a=14-(x/4)$.

MORE EDIT: If you read the question to imply there must be at least one marble of each of the two types, then $x$ can be any whole number between one and eleven, inclusive (and $y,b,a$ are given by the formula above).

share|cite|improve this answer

edited Apr 12 at 0:14

answered Apr 9 at 2:50

Gerry MyersonGerry Myerson

148k8152306

$endgroup$

share|cite|improve this answer

edited Apr 12 at 0:14

answered Apr 9 at 2:50

Gerry MyersonGerry Myerson

148k8152306

answered Apr 9 at 2:50

Gerry MyersonGerry Myerson

148k8152306

answered Apr 9 at 2:50

Gerry MyersonGerry Myerson

148k8152306