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
342110
$endgroup$
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
342110
$endgroup$
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
342110
$endgroup$
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
342110
asked Jan 6 at 8:47
ElementaryElementary
342110
edited Jan 6 at 9:48
Elementary
asked Jan 6 at 8:47
ElementaryElementary
342110
asked Jan 6 at 8:47
ElementaryElementary
342110
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
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
5,61131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
$begingroup$
(+) Thank you very much. I'm trying to understand your answer.
$endgroup$
– Elementary
Jan 6 at 9:33
add a comment |
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$begingroup$
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
5,61131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
$begingroup$
(+) 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
5,61131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
$begingroup$
(+) 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
5,61131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
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
5,61131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
279k23271502
edited Jan 6 at 13:53
Moo
5,61131020
edited Jan 6 at 13:53
Moo
5,61131020
edited Jan 6 at 13:53
Moo
5,61131020
5,61131020
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
279k23271502
answered Jan 6 at 9:23
Hagen von EitzenHagen von Eitzen
279k23271502
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
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 |
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$begingroup$
Downvote without any comment?
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– 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