Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11

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Question:
Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7

edited 2 hours ago

boboquack

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asked 3 hours ago

DeNel

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Should I add no-computer tag?
â€“Â Omega Krypton
8 mins ago

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up vote
1
down vote

favorite

Question:
Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7

edited 2 hours ago

boboquack

14.5k144112

asked 3 hours ago

DeNel

New contributor

Should I add no-computer tag?
â€“Â Omega Krypton
8 mins ago

add a commentÂ |Â

up vote
1
down vote

favorite

Question:
Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7

edited 2 hours ago

boboquack

14.5k144112

asked 3 hours ago

DeNel

New contributor

Question:
Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7

formation-of-numbers number-theory number-property

edited 2 hours ago

boboquack

14.5k144112

asked 3 hours ago

DeNel

New contributor

edited 2 hours ago

boboquack

14.5k144112

asked 3 hours ago

DeNel

New contributor

edited 2 hours ago

boboquack

14.5k144112

edited 2 hours ago

boboquack

14.5k144112

edited 2 hours ago

boboquack

14.5k144112

asked 3 hours ago

DeNel

New contributor

asked 3 hours ago

DeNel

asked 3 hours ago

DeNel

New contributor

DeNel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Should I add no-computer tag?
â€“Â Omega Krypton
8 mins ago

add a commentÂ |Â

Should I add no-computer tag?
â€“Â Omega Krypton
8 mins ago

Should I add no-computer tag?
â€“Â Omega Krypton
8 mins ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

Alright. First of all,

3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.

Now,

the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.

Now, let's look at the second-last and third-last digits.

The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.

Now

all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.

edited 2 hours ago

answered 2 hours ago

Excited Raichu

2,149227

I have 1385679240 & 7165932840 neither divisible by 7
â€“Â DeNel
2 hours ago

My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that.
â€“Â DeNel
2 hours ago

"The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4.
â€“Â elias
33 mins ago

add a commentÂ |Â

up vote
4
down vote

This is a good problem to attack by computer:

It turns out that there are 7344 solutions, including 1056 to the bonus.

So I thought, how far can we go?

And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.

Code used (Python 3 IDLE):


>>> import itertools
>>> normal=
>>> bonus=
>>> extra=
>>> for i in itertools.permutations('123456789'):
... n=int(''.join(i+('0',)))
... if 0==n%8==n%9==n%11:
... normal.append(n)
... if 0==n%7:
... bonus.append(n)
... if 0==n%13==n%16==n%17:
... extra.append(n)
...
>>> len(normal)
7344
>>> len(bonus)
1056
>>> extra
[2438195760, 3785942160, 4753869120, 4876391520]
>>> [i%19 for i in extra]
[12, 13, 17, 5]

edited 30 mins ago

answered 2 hours ago

boboquack

14.5k144112

Oh my goodness, that is amazing. IÃ¢Â€Â™m totally Computer illiterate. you have to totally teach me how to do this problem on a computer.
â€“Â DeNel
2 hours ago

@DeNel I used Python 3, I'll attach some code that I used in a few moments.
â€“Â boboquack
37 mins ago

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

Alright. First of all,

3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.

Now,

the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.

Now, let's look at the second-last and third-last digits.

The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.

Now

all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.

edited 2 hours ago

answered 2 hours ago

Excited Raichu

2,149227

I have 1385679240 & 7165932840 neither divisible by 7
â€“Â DeNel
2 hours ago

My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that.
â€“Â DeNel
2 hours ago

"The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4.
â€“Â elias
33 mins ago

add a commentÂ |Â

up vote
2
down vote

accepted

Alright. First of all,

3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.

Now,

the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.

Now, let's look at the second-last and third-last digits.

The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.

Now

all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.

edited 2 hours ago

answered 2 hours ago

Excited Raichu

2,149227

I have 1385679240 & 7165932840 neither divisible by 7
â€“Â DeNel
2 hours ago

My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that.
â€“Â DeNel
2 hours ago

"The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4.
â€“Â elias
33 mins ago

add a commentÂ |Â

up vote
2
down vote

accepted

Alright. First of all,

3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.

Now,

the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.

Now, let's look at the second-last and third-last digits.

The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.

Now

all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.

edited 2 hours ago

answered 2 hours ago

Excited Raichu

2,149227

Alright. First of all,

3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.

Now,

the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.

Now, let's look at the second-last and third-last digits.

The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.

Now

all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.

edited 2 hours ago

answered 2 hours ago

Excited Raichu

2,149227

edited 2 hours ago

answered 2 hours ago

Excited Raichu

2,149227

answered 2 hours ago

Excited Raichu

2,149227

answered 2 hours ago

Excited Raichu

2,149227

I have 1385679240 & 7165932840 neither divisible by 7
â€“Â DeNel
2 hours ago

My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that.
â€“Â DeNel
2 hours ago

"The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4.
â€“Â elias
33 mins ago

add a commentÂ |Â

I have 1385679240 & 7165932840 neither divisible by 7
â€“Â DeNel
2 hours ago

My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that.
â€“Â DeNel
2 hours ago

"The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4.
â€“Â elias
33 mins ago

I have 1385679240 & 7165932840 neither divisible by 7
â€“Â DeNel
2 hours ago

My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that.
â€“Â DeNel
2 hours ago

"The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4.
â€“Â elias
33 mins ago

add a commentÂ |Â

up vote
4
down vote

This is a good problem to attack by computer:

It turns out that there are 7344 solutions, including 1056 to the bonus.

So I thought, how far can we go?

And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.

Code used (Python 3 IDLE):


>>> import itertools
>>> normal=
>>> bonus=
>>> extra=
>>> for i in itertools.permutations('123456789'):
... n=int(''.join(i+('0',)))
... if 0==n%8==n%9==n%11:
... normal.append(n)
... if 0==n%7:
... bonus.append(n)
... if 0==n%13==n%16==n%17:
... extra.append(n)
...
>>> len(normal)
7344
>>> len(bonus)
1056
>>> extra
[2438195760, 3785942160, 4753869120, 4876391520]
>>> [i%19 for i in extra]
[12, 13, 17, 5]

edited 30 mins ago

answered 2 hours ago

boboquack

14.5k144112

Oh my goodness, that is amazing. IÃ¢Â€Â™m totally Computer illiterate. you have to totally teach me how to do this problem on a computer.
â€“Â DeNel
2 hours ago

@DeNel I used Python 3, I'll attach some code that I used in a few moments.
â€“Â boboquack
37 mins ago

add a commentÂ |Â

up vote
4
down vote

This is a good problem to attack by computer:

It turns out that there are 7344 solutions, including 1056 to the bonus.

So I thought, how far can we go?

And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.

Code used (Python 3 IDLE):


>>> import itertools
>>> normal=
>>> bonus=
>>> extra=
>>> for i in itertools.permutations('123456789'):
... n=int(''.join(i+('0',)))
... if 0==n%8==n%9==n%11:
... normal.append(n)
... if 0==n%7:
... bonus.append(n)
... if 0==n%13==n%16==n%17:
... extra.append(n)
...
>>> len(normal)
7344
>>> len(bonus)
1056
>>> extra
[2438195760, 3785942160, 4753869120, 4876391520]
>>> [i%19 for i in extra]
[12, 13, 17, 5]

edited 30 mins ago

answered 2 hours ago

boboquack

14.5k144112

Oh my goodness, that is amazing. IÃ¢Â€Â™m totally Computer illiterate. you have to totally teach me how to do this problem on a computer.
â€“Â DeNel
2 hours ago

@DeNel I used Python 3, I'll attach some code that I used in a few moments.
â€“Â boboquack
37 mins ago

add a commentÂ |Â

up vote
4
down vote

This is a good problem to attack by computer:

It turns out that there are 7344 solutions, including 1056 to the bonus.

So I thought, how far can we go?

And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.

Code used (Python 3 IDLE):


>>> import itertools
>>> normal=
>>> bonus=
>>> extra=
>>> for i in itertools.permutations('123456789'):
... n=int(''.join(i+('0',)))
... if 0==n%8==n%9==n%11:
... normal.append(n)
... if 0==n%7:
... bonus.append(n)
... if 0==n%13==n%16==n%17:
... extra.append(n)
...
>>> len(normal)
7344
>>> len(bonus)
1056
>>> extra
[2438195760, 3785942160, 4753869120, 4876391520]
>>> [i%19 for i in extra]
[12, 13, 17, 5]

edited 30 mins ago

answered 2 hours ago

boboquack

14.5k144112

This is a good problem to attack by computer:

It turns out that there are 7344 solutions, including 1056 to the bonus.

So I thought, how far can we go?

And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.

Code used (Python 3 IDLE):


>>> import itertools
>>> normal=
>>> bonus=
>>> extra=
>>> for i in itertools.permutations('123456789'):
... n=int(''.join(i+('0',)))
... if 0==n%8==n%9==n%11:
... normal.append(n)
... if 0==n%7:
... bonus.append(n)
... if 0==n%13==n%16==n%17:
... extra.append(n)
...
>>> len(normal)
7344
>>> len(bonus)
1056
>>> extra
[2438195760, 3785942160, 4753869120, 4876391520]
>>> [i%19 for i in extra]
[12, 13, 17, 5]

edited 30 mins ago

answered 2 hours ago

boboquack

14.5k144112

edited 30 mins ago

answered 2 hours ago

boboquack

14.5k144112

answered 2 hours ago

boboquack

14.5k144112

answered 2 hours ago

boboquack

14.5k144112

Oh my goodness, that is amazing. IÃ¢Â€Â™m totally Computer illiterate. you have to totally teach me how to do this problem on a computer.
â€“Â DeNel
2 hours ago

@DeNel I used Python 3, I'll attach some code that I used in a few moments.
â€“Â boboquack
37 mins ago

add a commentÂ |Â

Oh my goodness, that is amazing. IÃ¢Â€Â™m totally Computer illiterate. you have to totally teach me how to do this problem on a computer.
â€“Â DeNel
2 hours ago

@DeNel I used Python 3, I'll attach some code that I used in a few moments.
â€“Â boboquack
37 mins ago

Oh my goodness, that is amazing. IÃ¢Â€Â™m totally Computer illiterate. you have to totally teach me how to do this problem on a computer.
â€“Â DeNel
2 hours ago

@DeNel I used Python 3, I'll attach some code that I used in a few moments.
â€“Â boboquack
37 mins ago

add a commentÂ |Â

DeNel is a new contributor. Be nice, and check out our Code of Conduct.

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