How many $10-$digit numbers are divided by $11.111$ and all the digits are different?












0












$begingroup$


The Problem:




How many $10-$digit numbers are divided by $11.111$ and all the digits are different?



A) $3250$



B) $3456$



C) $3624$



D) $3842$



E) $4020$




The Problematic point is, "all digits must be different".



I could find only all $10-$ digit numbers.



$$999990000+11.111k≤9.999.999.999$$



$$1≤k≤810.009$$



The problem is, I have no method how can I calculate "all digits are different numbers."






algebra-precalculus contest-math problem-solving natural-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Downvote without any comment?
    $endgroup$
    – Beginner
    Jan 6 at 8:51






  • 1




    $begingroup$
    I don't quite understand the downvote and close vote. It may be because most countries do not use a period to separate the thousands like you've written so it looks like you are asking for 10 digit numbers that divide by a number slightly larger than 11 rather than 11111.
    $endgroup$
    – JessicaK
    Jan 6 at 9:04










  • $begingroup$
    @JessicaK if you understand the mean of question, can you edit ? If there is a mistake in translation into English
    $endgroup$
    – Beginner
    Jan 6 at 9:09










  • $begingroup$
    A clarification on mathematical English grammar: "is/are divided by" is incorrect. Either "can be divided by" or "is divisible by" would work; the latter adjective form is most standard. Also, in English, it's standard to use a period for the decimal point separator and a comma to break out blocks of digits (when we do so).
    $endgroup$
    – jmerry
    Jan 6 at 9:34
















0












$begingroup$


The Problem:




How many $10-$digit numbers are divided by $11.111$ and all the digits are different?



A) $3250$



B) $3456$



C) $3624$



D) $3842$



E) $4020$




The Problematic point is, "all digits must be different".



I could find only all $10-$ digit numbers.



$$999990000+11.111k≤9.999.999.999$$



$$1≤k≤810.009$$



The problem is, I have no method how can I calculate "all digits are different numbers."






algebra-precalculus contest-math problem-solving natural-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Downvote without any comment?
    $endgroup$
    – Beginner
    Jan 6 at 8:51






  • 1




    $begingroup$
    I don't quite understand the downvote and close vote. It may be because most countries do not use a period to separate the thousands like you've written so it looks like you are asking for 10 digit numbers that divide by a number slightly larger than 11 rather than 11111.
    $endgroup$
    – JessicaK
    Jan 6 at 9:04










  • $begingroup$
    @JessicaK if you understand the mean of question, can you edit ? If there is a mistake in translation into English
    $endgroup$
    – Beginner
    Jan 6 at 9:09










  • $begingroup$
    A clarification on mathematical English grammar: "is/are divided by" is incorrect. Either "can be divided by" or "is divisible by" would work; the latter adjective form is most standard. Also, in English, it's standard to use a period for the decimal point separator and a comma to break out blocks of digits (when we do so).
    $endgroup$
    – jmerry
    Jan 6 at 9:34














0












0








0


0



$begingroup$


The Problem:




How many $10-$digit numbers are divided by $11.111$ and all the digits are different?



A) $3250$



B) $3456$



C) $3624$



D) $3842$



E) $4020$




The Problematic point is, "all digits must be different".



I could find only all $10-$ digit numbers.



$$999990000+11.111k≤9.999.999.999$$



$$1≤k≤810.009$$



The problem is, I have no method how can I calculate "all digits are different numbers."






algebra-precalculus contest-math problem-solving natural-numbers







share|cite|improve this question











$endgroup$




The Problem:




How many $10-$digit numbers are divided by $11.111$ and all the digits are different?



A) $3250$



B) $3456$



C) $3624$



D) $3842$



E) $4020$




The Problematic point is, "all digits must be different".



I could find only all $10-$ digit numbers.



$$999990000+11.111k≤9.999.999.999$$



$$1≤k≤810.009$$



The problem is, I have no method how can I calculate "all digits are different numbers."






algebra-precalculus contest-math problem-solving natural-numbers




algebra-precalculus contest-math problem-solving natural-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 9:48







Beginner

















asked Jan 6 at 8:47









BeginnerBeginner

33110




33110








  • 1




    $begingroup$
    Downvote without any comment?
    $endgroup$
    – Beginner
    Jan 6 at 8:51






  • 1




    $begingroup$
    I don't quite understand the downvote and close vote. It may be because most countries do not use a period to separate the thousands like you've written so it looks like you are asking for 10 digit numbers that divide by a number slightly larger than 11 rather than 11111.
    $endgroup$
    – JessicaK
    Jan 6 at 9:04










  • $begingroup$
    @JessicaK if you understand the mean of question, can you edit ? If there is a mistake in translation into English
    $endgroup$
    – Beginner
    Jan 6 at 9:09










  • $begingroup$
    A clarification on mathematical English grammar: "is/are divided by" is incorrect. Either "can be divided by" or "is divisible by" would work; the latter adjective form is most standard. Also, in English, it's standard to use a period for the decimal point separator and a comma to break out blocks of digits (when we do so).
    $endgroup$
    – jmerry
    Jan 6 at 9:34














  • 1




    $begingroup$
    Downvote without any comment?
    $endgroup$
    – Beginner
    Jan 6 at 8:51






  • 1




    $begingroup$
    I don't quite understand the downvote and close vote. It may be because most countries do not use a period to separate the thousands like you've written so it looks like you are asking for 10 digit numbers that divide by a number slightly larger than 11 rather than 11111.
    $endgroup$
    – JessicaK
    Jan 6 at 9:04










  • $begingroup$
    @JessicaK if you understand the mean of question, can you edit ? If there is a mistake in translation into English
    $endgroup$
    – Beginner
    Jan 6 at 9:09










  • $begingroup$
    A clarification on mathematical English grammar: "is/are divided by" is incorrect. Either "can be divided by" or "is divisible by" would work; the latter adjective form is most standard. Also, in English, it's standard to use a period for the decimal point separator and a comma to break out blocks of digits (when we do so).
    $endgroup$
    – jmerry
    Jan 6 at 9:34








1




1




$begingroup$
Downvote without any comment?
$endgroup$
– Beginner
Jan 6 at 8:51




$begingroup$
Downvote without any comment?
$endgroup$
– Beginner
Jan 6 at 8:51




1




1




$begingroup$
I don't quite understand the downvote and close vote. It may be because most countries do not use a period to separate the thousands like you've written so it looks like you are asking for 10 digit numbers that divide by a number slightly larger than 11 rather than 11111.
$endgroup$
– JessicaK
Jan 6 at 9:04




$begingroup$
I don't quite understand the downvote and close vote. It may be because most countries do not use a period to separate the thousands like you've written so it looks like you are asking for 10 digit numbers that divide by a number slightly larger than 11 rather than 11111.
$endgroup$
– JessicaK
Jan 6 at 9:04












$begingroup$
@JessicaK if you understand the mean of question, can you edit ? If there is a mistake in translation into English
$endgroup$
– Beginner
Jan 6 at 9:09




$begingroup$
@JessicaK if you understand the mean of question, can you edit ? If there is a mistake in translation into English
$endgroup$
– Beginner
Jan 6 at 9:09












$begingroup$
A clarification on mathematical English grammar: "is/are divided by" is incorrect. Either "can be divided by" or "is divisible by" would work; the latter adjective form is most standard. Also, in English, it's standard to use a period for the decimal point separator and a comma to break out blocks of digits (when we do so).
$endgroup$
– jmerry
Jan 6 at 9:34




$begingroup$
A clarification on mathematical English grammar: "is/are divided by" is incorrect. Either "can be divided by" or "is divisible by" would work; the latter adjective form is most standard. Also, in English, it's standard to use a period for the decimal point separator and a comma to break out blocks of digits (when we do so).
$endgroup$
– jmerry
Jan 6 at 9:34










1 Answer
1






active

oldest

votes


















3












$begingroup$

If all digits are different, they must be all ten digits. In particular, the digit sum is $45$ and hence our number is a multiple of $9$. Thus we are in fact looking for certain multiples of $99999$. This reduces your $k$ range down to about $90000$ possibilities - still unfeasible do work out by hand.



If the ten digit number is $abcdefghij$, then after subtracting $99999cdot abcde$ we still have a multiple of $99999$, namely $fghij+abcde$. As this sum is certainly $>0$ and $<99999+99999$, we conclude that $$fghij+abcde=99999.$$
In particular, $j+e=9$ without carry. Then also $i+d=9$ without carry, and so on. Thus the digit pairs ${a,f},{b,g},{c,h},{d,i},{e,j}$ must be the pairs ${0,9},{1,8},{2,7},{3,6},{4,5}$ in some order. There are $5!$ such permutations and then for each pair there are $2$ ways to match. This gives us $2^5cdot 5!$ numbers of the desired form. However, among these are $2^4cdot 4!$ where we attempt to set $a=0$ (and $f=9$), i.e., that are not really ten-digit numbers. Hence the final answer is
$$2^5cdot 5!-2^4cdot 4! = 3456. $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+) Thank you very much. I'm trying to understand your answer.
    $endgroup$
    – Beginner
    Jan 6 at 9:33













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063620%2fhow-many-10-digit-numbers-are-divided-by-11-111-and-all-the-digits-are-diffe%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

If all digits are different, they must be all ten digits. In particular, the digit sum is $45$ and hence our number is a multiple of $9$. Thus we are in fact looking for certain multiples of $99999$. This reduces your $k$ range down to about $90000$ possibilities - still unfeasible do work out by hand.



If the ten digit number is $abcdefghij$, then after subtracting $99999cdot abcde$ we still have a multiple of $99999$, namely $fghij+abcde$. As this sum is certainly $>0$ and $<99999+99999$, we conclude that $$fghij+abcde=99999.$$
In particular, $j+e=9$ without carry. Then also $i+d=9$ without carry, and so on. Thus the digit pairs ${a,f},{b,g},{c,h},{d,i},{e,j}$ must be the pairs ${0,9},{1,8},{2,7},{3,6},{4,5}$ in some order. There are $5!$ such permutations and then for each pair there are $2$ ways to match. This gives us $2^5cdot 5!$ numbers of the desired form. However, among these are $2^4cdot 4!$ where we attempt to set $a=0$ (and $f=9$), i.e., that are not really ten-digit numbers. Hence the final answer is
$$2^5cdot 5!-2^4cdot 4! = 3456. $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+) Thank you very much. I'm trying to understand your answer.
    $endgroup$
    – Beginner
    Jan 6 at 9:33


















3












$begingroup$

If all digits are different, they must be all ten digits. In particular, the digit sum is $45$ and hence our number is a multiple of $9$. Thus we are in fact looking for certain multiples of $99999$. This reduces your $k$ range down to about $90000$ possibilities - still unfeasible do work out by hand.



If the ten digit number is $abcdefghij$, then after subtracting $99999cdot abcde$ we still have a multiple of $99999$, namely $fghij+abcde$. As this sum is certainly $>0$ and $<99999+99999$, we conclude that $$fghij+abcde=99999.$$
In particular, $j+e=9$ without carry. Then also $i+d=9$ without carry, and so on. Thus the digit pairs ${a,f},{b,g},{c,h},{d,i},{e,j}$ must be the pairs ${0,9},{1,8},{2,7},{3,6},{4,5}$ in some order. There are $5!$ such permutations and then for each pair there are $2$ ways to match. This gives us $2^5cdot 5!$ numbers of the desired form. However, among these are $2^4cdot 4!$ where we attempt to set $a=0$ (and $f=9$), i.e., that are not really ten-digit numbers. Hence the final answer is
$$2^5cdot 5!-2^4cdot 4! = 3456. $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+) Thank you very much. I'm trying to understand your answer.
    $endgroup$
    – Beginner
    Jan 6 at 9:33
















3












3








3





$begingroup$

If all digits are different, they must be all ten digits. In particular, the digit sum is $45$ and hence our number is a multiple of $9$. Thus we are in fact looking for certain multiples of $99999$. This reduces your $k$ range down to about $90000$ possibilities - still unfeasible do work out by hand.



If the ten digit number is $abcdefghij$, then after subtracting $99999cdot abcde$ we still have a multiple of $99999$, namely $fghij+abcde$. As this sum is certainly $>0$ and $<99999+99999$, we conclude that $$fghij+abcde=99999.$$
In particular, $j+e=9$ without carry. Then also $i+d=9$ without carry, and so on. Thus the digit pairs ${a,f},{b,g},{c,h},{d,i},{e,j}$ must be the pairs ${0,9},{1,8},{2,7},{3,6},{4,5}$ in some order. There are $5!$ such permutations and then for each pair there are $2$ ways to match. This gives us $2^5cdot 5!$ numbers of the desired form. However, among these are $2^4cdot 4!$ where we attempt to set $a=0$ (and $f=9$), i.e., that are not really ten-digit numbers. Hence the final answer is
$$2^5cdot 5!-2^4cdot 4! = 3456. $$






share|cite|improve this answer











$endgroup$



If all digits are different, they must be all ten digits. In particular, the digit sum is $45$ and hence our number is a multiple of $9$. Thus we are in fact looking for certain multiples of $99999$. This reduces your $k$ range down to about $90000$ possibilities - still unfeasible do work out by hand.



If the ten digit number is $abcdefghij$, then after subtracting $99999cdot abcde$ we still have a multiple of $99999$, namely $fghij+abcde$. As this sum is certainly $>0$ and $<99999+99999$, we conclude that $$fghij+abcde=99999.$$
In particular, $j+e=9$ without carry. Then also $i+d=9$ without carry, and so on. Thus the digit pairs ${a,f},{b,g},{c,h},{d,i},{e,j}$ must be the pairs ${0,9},{1,8},{2,7},{3,6},{4,5}$ in some order. There are $5!$ such permutations and then for each pair there are $2$ ways to match. This gives us $2^5cdot 5!$ numbers of the desired form. However, among these are $2^4cdot 4!$ where we attempt to set $a=0$ (and $f=9$), i.e., that are not really ten-digit numbers. Hence the final answer is
$$2^5cdot 5!-2^4cdot 4! = 3456. $$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 6 at 13:53









Moo

5,53131020




5,53131020










answered Jan 6 at 9:23









Hagen von EitzenHagen von Eitzen

277k22269496




277k22269496












  • $begingroup$
    (+) Thank you very much. I'm trying to understand your answer.
    $endgroup$
    – Beginner
    Jan 6 at 9:33




















  • $begingroup$
    (+) Thank you very much. I'm trying to understand your answer.
    $endgroup$
    – Beginner
    Jan 6 at 9:33


















$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Beginner
Jan 6 at 9:33






$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Beginner
Jan 6 at 9:33




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063620%2fhow-many-10-digit-numbers-are-divided-by-11-111-and-all-the-digits-are-diffe%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

Display a custom attribute below product name in the front-end Magento 1.9.3.8

Image
0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
Read more