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?
    $endgroup$
    – 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
















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?
    $endgroup$
    – 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














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







Elementary

















asked Jan 6 at 8:47









ElementaryElementary

342110




342110








  • 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




    $begingroup$
    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










1 Answer
1






active

oldest

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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.
    $endgroup$
    – Elementary
    Jan 6 at 9:33













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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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.
    $endgroup$
    – Elementary
    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.
    $endgroup$
    – Elementary
    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

5,61131020




5,61131020










answered Jan 6 at 9:23









Hagen von EitzenHagen von Eitzen

279k23271502




279k23271502












  • $begingroup$
    (+) 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






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




















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