How many $10-$digit numbers are divided by $11.111$ and all the digits are different?
$begingroup$
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
asked Jan 6 at 8:47
ElementaryElementary
360111
$endgroup$
add a comment |
$begingroup$
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
asked Jan 6 at 8:47
ElementaryElementary
360111
$endgroup$
1
$begingroup$
Downvote without any comment?
$endgroup$
– Elementary
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
$endgroup$
– Elementary
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
add a comment |
$begingroup$
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
asked Jan 6 at 8:47
ElementaryElementary
360111
$endgroup$
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
asked Jan 6 at 8:47
ElementaryElementary
360111
asked Jan 6 at 8:47
ElementaryElementary
360111
edited Jan 6 at 9:48
Elementary
asked Jan 6 at 8:47
ElementaryElementary
360111
asked Jan 6 at 8:47
ElementaryElementary
360111
asked Jan 6 at 8:47
ElementaryElementary
360111
360111
1
$begingroup$
Downvote without any comment?
$endgroup$
– Elementary
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
$endgroup$
– Elementary
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
add a comment |
1
$begingroup$
Downvote without any comment?
$endgroup$
– Elementary
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
$endgroup$
– Elementary
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?
$endgroup$
– Elementary
Jan 6 at 8:51
$begingroup$
Downvote without any comment?
$endgroup$
– Elementary
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
$endgroup$
– Elementary
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
$endgroup$
– Elementary
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$2^5cdot 5!-2^4cdot 4! = 3456. $$
edited Jan 6 at 13:53
Moo
5,63131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Elementary
Jan 6 at 9:33
add a comment |
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
});
}
});
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
Required, but never shown
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
$begingroup$
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
$$2^5cdot 5!-2^4cdot 4! = 3456. $$
edited Jan 6 at 13:53
Moo
5,63131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Elementary
Jan 6 at 9:33
add a comment |
$begingroup$
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
$$2^5cdot 5!-2^4cdot 4! = 3456. $$
edited Jan 6 at 13:53
Moo
5,63131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Elementary
Jan 6 at 9:33
add a comment |
$begingroup$
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
$$2^5cdot 5!-2^4cdot 4! = 3456. $$
edited Jan 6 at 13:53
Moo
5,63131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
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
$$2^5cdot 5!-2^4cdot 4! = 3456. $$
edited Jan 6 at 13:53
Moo
5,63131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
282k23272507
edited Jan 6 at 13:53
Moo
5,63131020
edited Jan 6 at 13:53
Moo
5,63131020
edited Jan 6 at 13:53
Moo
5,63131020
5,63131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
282k23272507
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
282k23272507
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
282k23272507
282k23272507
$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Elementary
Jan 6 at 9:33
add a comment |
$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Elementary
Jan 6 at 9:33
$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Elementary
Jan 6 at 9:33
$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Elementary
Jan 6 at 9:33
add a comment |
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.
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
Required, but never shown
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
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
Downvote without any comment?
$endgroup$
– Elementary
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
$endgroup$
– Elementary
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