P.S. Here's a little puzzle courtesy of New Scientist:
Clever logic should enable you to find the nine-figure number that I have in mind. It consists of the digits 1 to 9 in some order, and in the number each digit is next to another that differs from it by one.
In just one case a digit has both neighbours differing from it by one. Furthermore, the solution is exactly divisible by more than three-quarters of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
> P.S. > Here's a little puzzle courtesy of New Scientist:
> Clever logic should enable you to find the nine-figure number that I have in mind. It consists of the digits 1 to 9 in some order, and in the number each digit is next to another that differs from it by one.
> In just one case a digit has both neighbours differing from it by one. Furthermore, the solution is exactly divisible by more than three-quarters of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
On Fri, Apr 13, 2012 at 11:18 AM, Timothy Perrett <tperr...@gmail.com> wrote: > Doh, that clashes with ScalaDays this year. Next month perhaps! Have fun :-)
> Cheers, Tim
> On 13 Apr 2012, at 11:15, Arwyn wrote:
>> Hi All,
>> Are we on for next Tuesday (17th)?
>> Arwyn
>> P.S. >> Here's a little puzzle courtesy of New Scientist:
>> Clever logic should enable you to find the nine-figure number that I have in mind. It consists of the digits 1 to 9 in some order, and in the number each digit is next to another that differs from it by one.
>> In just one case a digit has both neighbours differing from it by one. Furthermore, the solution is exactly divisible by more than three-quarters of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
> Clever logic should enable you to find the nine-figure number that I > have in mind. It consists of the digits 1 to 9 in some order, and in > the number each digit is next to another that differs from it by one.
> In just one case a digit has both neighbours differing from it by one. > Furthermore, the solution is exactly divisible by more than > three-quarters of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
> What is the nine-figure number?
Totally off topic.
Did anyone solve this? I made some progress, but didn't nail it.
Lemma 1: The solution must be divisible by 4.
Assume the solution is not divisible by 4. It is therefore not divisible by 8 or 12. However, this would mean that the solution is only divisible by at most 3/4th of the numbers between 1 and 12, which contradicts the problem, which states it is divisible by strictly more than 3/4th.
We could apply the same argument to 2 and 3, but we don't need to...
Lemma 2: The solution is divisible by 1, 2, 3, 4, 6, 7, 8, 9, 11 and 12, but not 5 or 10.
All numbers divisible by 10 end in 0. The solution doesn't contain a 0. We know from Lemma 1 that it's divisible by 4. If it is also divisible by 5, it must also end in 0. If the solution were not divisible by any other numbers on the list, it would only be divisible by 3/4ths, which contradicts the problem, so it must be divisible by the remaining numbers.
Lemma 3: The solution must end in 12, 32, 56 or 76
Because the solution is divisible by 4, it's last two digits must be divisible by 4. The two digits must differ by 1. The second digit must be even, since it is divisible by 4, so the first digit must be odd.
This only leaves a few possibilities. Checking them, we find that 12, 32, 56 and 76 fit the bill.
Lemma 4: The solution cannot end in 32.
Assume the number ended with 32. According to the rules, the digit 1 must be next to digit 2. However, digit 2 is at the end, so this is a contradiction.
Where next?
Ending 12 can be split up 3 ways:
- Containing (34 or 43) and (56 or 65) and (789 or 987)
- Containing (34 or 43) and (567 or 765) and (89 or 98)
- Containing (345 or 543) and (67 or 76) and (89 or 98)
= 3*2*2*2*3*2*1=144 possibilities
Ending 56 can only split 1 way:
- Containing (12 or 21) and (34 or43) and (789 or 987)
= 1*2*2*2*3*2*1=48 possibilities
(Note that, ending 456 is not allowed unless the preceding digit is 3, because then we'd have to form a group from 123, or leave 3 on it's own)
Ending 76 can be split up 2 ways:
- Containing (12 or 21) and (345 or 543) and (89 or 98)
- Containing (123 or 321) and (45 or 54) and (89 or 98)
= 2*2*2*2*3*2*1= 96 possibilities
Brute forcible, but the question suggests we shouldn't need to.
Can we exclude more possibilities using other divisors?
2: Not really any help, as we already made use of 4, which is strictly stronger.
3: No help - if you rearrange the digits of a number divisible by 3, you get another number divisible by 3. So we can't exclude any wrong answers, because they will also be divisible by 3.
4: Already used
9: No help for the same reasons as 3.
8 can be used to restrict the 3rd from last digit, but it only excludes a handful of possibilities (it doesn't even uniquely determine the 3rd digit). Depending on whether the last 2 digits are divisible by 8, the second to last digit must be either even or odd. Some possibilities are excluded because the result in illegal combinations of digits)
My gut feeling tells me that 12 isn't very useful - we already made use of one it's factors (4) and we know that the other one (3) provides no useful information, so I don't think 12 tells us anything. (But I'm not sure I have a watertight argument, so I could be wrong here)
So I'm pretty sure we need to make use of 7 and 11 - but I can't see an easy way or doing so (without doing a tonne of algebra, which I don't think the question is looking for)
That's an impressive effort! You certainly got much further than I did with logic.
I must admit that I suggested this as a programming problem, so I was kind of ignoring the "Clever logic..." bit.
My original aim was to write something that reads like a statement of the problem and also solves it. In a language like Prolog I think that might not be too difficult. However, working in Haskell I have found it quite challenging. I did a filter-based solution which essentially brute-forces the whole thing (9! perms). However, I would like to do something a bit more clever, by which I mean something that tests fewer candidates.
You have already cut the space down considerably: 144+48+96=288 << 9! I think we can cut it down a bit further.
E.g. taking:
>> Ending 12 can be split up 3 ways: >> - Containing (34 or 43) and (56 or 65) and (789 or 987) >> - Containing (34 or 43) and (567 or 765) and (89 or 98) >> - Containing (345 or 543) and (67 or 76) and (89 or 98) >> = 3*2*2*2*3*2*1=144 possibilities
Some of the permutations can be excluded by the rule: "In just one case a digit has both neighbours differing from it by one." E.g. the first permutation enumerated would be 345678912 which has 5 digits which have both neighbours differing by 1. Your three lines above already give the three cases for which digit has 2 neighbours differing by one: 8, 6 and 4 respectively. So we just have to exclude all the incidental ones; I'll call them "collisions".
We still have 3x2x1 perms of the arrangement of groups, however all of them result in at least 1 possible collision and 2 result in 2. If there is 1 possible collision, that reduces the number of perms of the ordering within the groups from 8 to 6 (3 for the collided groups x 2 for the other one). If there are 2 possible collisions, that reduces the number of such perms to 5 because 1 of those 6 is excluded.
What this means is that instead of each line being 6x8 = 48 permutations we have 4x6+2x5 = 34. By three lines gives 102.
Using similar logic we get 4x6+2x8 = 40 perms for Ending 56 and 80 for Ending 76.
So now we're down to 102+40+80 = 222 possibilities. Can we do any better?
On Thursday, 26 April 2012 19:32:49 UTC+1, Jimbo F wrote:
> On 2012-04-13 11:15, Arwyn wrote:
> > Clever logic should enable you to find the nine-figure number that I > > have in mind. It consists of the digits 1 to 9 in some order, and in > > the number each digit is next to another that differs from it by one.
> > In just one case a digit has both neighbours differing from it by one. > > Furthermore, the solution is exactly divisible by more than > > three-quarters of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
> > What is the nine-figure number?
> Totally off topic.
> Did anyone solve this? I made some progress, but didn't nail it.
> Lemma 1: The solution must be divisible by 4.
> Assume the solution is not divisible by 4. It is therefore not > divisible by 8 or 12. However, this would mean that the solution is > only divisible by at most 3/4th of the numbers between 1 and 12, which > contradicts the problem, which states it is divisible by strictly more > than 3/4th.
> We could apply the same argument to 2 and 3, but we don't need to...
> Lemma 2: The solution is divisible by 1, 2, 3, 4, 6, 7, 8, 9, 11 and 12, > but not 5 or 10.
> All numbers divisible by 10 end in 0. The solution doesn't contain a > 0. We know from Lemma 1 that it's divisible by 4. If it is also > divisible by 5, it must also end in 0. If the solution were not > divisible by any other numbers on the list, it would only be divisible > by 3/4ths, which contradicts the problem, so it must be divisible by the > remaining numbers.
> Lemma 3: The solution must end in 12, 32, 56 or 76
> Because the solution is divisible by 4, it's last two digits must be > divisible by 4. The two digits must differ by 1. The second digit must > be even, since it is divisible by 4, so the first digit must be odd. > This only leaves a few possibilities. Checking them, we find that 12, > 32, 56 and 76 fit the bill.
> Lemma 4: The solution cannot end in 32.
> Assume the number ended with 32. According to the rules, the digit 1 > must be next to digit 2. However, digit 2 is at the end, so this is a > contradiction.
> Where next?
> Ending 12 can be split up 3 ways: > - Containing (34 or 43) and (56 or 65) and (789 or 987) > - Containing (34 or 43) and (567 or 765) and (89 or 98) > - Containing (345 or 543) and (67 or 76) and (89 or 98) > = 3*2*2*2*3*2*1=144 possibilities
> Ending 56 can only split 1 way: > - Containing (12 or 21) and (34 or43) and (789 or 987) > = 1*2*2*2*3*2*1=48 possibilities
> (Note that, ending 456 is not allowed unless the preceding digit is 3, > because then we'd have to form a group from 123, or leave 3 on it's own)
> Ending 76 can be split up 2 ways: > - Containing (12 or 21) and (345 or 543) and (89 or 98) > - Containing (123 or 321) and (45 or 54) and (89 or 98) > = 2*2*2*2*3*2*1= 96 possibilities
> Brute forcible, but the question suggests we shouldn't need to.
> Can we exclude more possibilities using other divisors?
> 2: Not really any help, as we already made use of 4, which is strictly > stronger. > 3: No help - if you rearrange the digits of a number divisible by 3, you > get another number divisible by 3. So we can't exclude any wrong > answers, because they will also be divisible by 3. > 4: Already used > 9: No help for the same reasons as 3.
> 8 can be used to restrict the 3rd from last digit, but it only excludes > a handful of possibilities (it doesn't even uniquely determine the 3rd > digit). Depending on whether the last 2 digits are divisible by 8, the > second to last digit must be either even or odd. Some possibilities are > excluded because the result in illegal combinations of digits)
> My gut feeling tells me that 12 isn't very useful - we already made use > of one it's factors (4) and we know that the other one (3) provides no > useful information, so I don't think 12 tells us anything. (But I'm > not sure I have a watertight argument, so I could be wrong here)
> So I'm pretty sure we need to make use of 7 and 11 - but I can't see an > easy way or doing so (without doing a tonne of algebra, which I don't > think the question is looking for)
> So now we're down to 102+40+80 = 222 possibilities.
> Can we do any better?
We do seem to be able to do a bit better. After seeing some pretty
exciting code and demos at EuroClojure, I tried this out using
core.logic which is a logic programming library for Clojure.
core.logic is similar in its capabilities to prolog, but is just a
regular clojure lib so you can mix logic programming with functional
very easily.
Jim's analysis is really excellent, and all I had to do was write it
out in the format that core.logic expects. I added to Jim's lemmas
1-4 a further constraint that the first 2 digits must differ by one.
The result (which takes my laptop about a minute and a half) is a
list of 30 candidates, from which the correct one can easily be
selected by eye. 436578912.
Can anyone constrain the solution list further? To one item,
ideally ;)
Matthew
On Apr 27, 10:57 am, Arwyn <arwyn.robe...@gmail.com> wrote:
> That's an impressive effort!
> You certainly got much further than I did with logic.
> I must admit that I suggested this as a programming problem, so I was kind
> of ignoring the "Clever logic..." bit.
> My original aim was to write something that reads like a statement of the
> problem and also solves it.
> In a language like Prolog I think that might not be too difficult.
> However, working in Haskell I have found it quite challenging.
> I did a filter-based solution which essentially brute-forces the whole
> thing (9! perms).
> However, I would like to do something a bit more clever, by which I mean
> something that tests fewer candidates.
> You have already cut the space down considerably: 144+48+96=288 << 9!
> I think we can cut it down a bit further.
> E.g. taking:
> >> Ending 12 can be split up 3 ways:
> >> - Containing (34 or 43) and (56 or 65) and (789 or 987)
> >> - Containing (34 or 43) and (567 or 765) and (89 or 98)
> >> - Containing (345 or 543) and (67 or 76) and (89 or 98)
> >> = 3*2*2*2*3*2*1=144 possibilities
> Some of the permutations can be excluded by the rule: "In just one case a
> digit has both neighbours differing from it by one."
> E.g. the first permutation enumerated would be 345678912 which has 5 digits
> which have both neighbours differing by 1.
> Your three lines above already give the three cases for which digit has 2
> neighbours differing by one: 8, 6 and 4 respectively.
> So we just have to exclude all the incidental ones; I'll call them
> "collisions".
> We still have 3x2x1 perms of the arrangement of groups, however all of them
> result in at least 1 possible collision and 2 result in 2.
> If there is 1 possible collision, that reduces the number of perms of the
> ordering within the groups from 8 to 6 (3 for the collided groups x 2 for
> the other one).
> If there are 2 possible collisions, that reduces the number of such perms
> to 5 because 1 of those 6 is excluded.
> What this means is that instead of each line being 6x8 = 48 permutations we
> have 4x6+2x5 = 34.
> By three lines gives 102.
> Using similar logic we get 4x6+2x8 = 40 perms for Ending 56 and 80 for
> Ending 76.
> So now we're down to 102+40+80 = 222 possibilities.
> Can we do any better?
> Cheers,
> Arwyn
> On Thursday, 26 April 2012 19:32:49 UTC+1, Jimbo F wrote:
> > On 2012-04-13 11:15, Arwyn wrote:
> > > Clever logic should enable you to find the nine-figure number that I
> > > have in mind. It consists of the digits 1 to 9 in some order, and in
> > > the number each digit is next to another that differs from it by one.
> > > In just one case a digit has both neighbours differing from it by one.
> > > Furthermore, the solution is exactly divisible by more than
> > > three-quarters of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
> > > What is the nine-figure number?
> > Totally off topic.
> > Did anyone solve this? I made some progress, but didn't nail it.
> > Lemma 1: The solution must be divisible by 4.
> > Assume the solution is not divisible by 4. It is therefore not
> > divisible by 8 or 12. However, this would mean that the solution is
> > only divisible by at most 3/4th of the numbers between 1 and 12, which
> > contradicts the problem, which states it is divisible by strictly more
> > than 3/4th.
> > We could apply the same argument to 2 and 3, but we don't need to...
> > Lemma 2: The solution is divisible by 1, 2, 3, 4, 6, 7, 8, 9, 11 and 12,
> > but not 5 or 10.
> > All numbers divisible by 10 end in 0. The solution doesn't contain a
> > 0. We know from Lemma 1 that it's divisible by 4. If it is also
> > divisible by 5, it must also end in 0. If the solution were not
> > divisible by any other numbers on the list, it would only be divisible
> > by 3/4ths, which contradicts the problem, so it must be divisible by the
> > remaining numbers.
> > Lemma 3: The solution must end in 12, 32, 56 or 76
> > Because the solution is divisible by 4, it's last two digits must be
> > divisible by 4. The two digits must differ by 1. The second digit must
> > be even, since it is divisible by 4, so the first digit must be odd.
> > This only leaves a few possibilities. Checking them, we find that 12,
> > 32, 56 and 76 fit the bill.
> > Lemma 4: The solution cannot end in 32.
> > Assume the number ended with 32. According to the rules, the digit 1
> > must be next to digit 2. However, digit 2 is at the end, so this is a
> > contradiction.
> > Where next?
> > Ending 12 can be split up 3 ways:
> > - Containing (34 or 43) and (56 or 65) and (789 or 987)
> > - Containing (34 or 43) and (567 or 765) and (89 or 98)
> > - Containing (345 or 543) and (67 or 76) and (89 or 98)
> > = 3*2*2*2*3*2*1=144 possibilities
> > Ending 56 can only split 1 way:
> > - Containing (12 or 21) and (34 or43) and (789 or 987)
> > = 1*2*2*2*3*2*1=48 possibilities
> > (Note that, ending 456 is not allowed unless the preceding digit is 3,
> > because then we'd have to form a group from 123, or leave 3 on it's own)
> > Ending 76 can be split up 2 ways:
> > - Containing (12 or 21) and (345 or 543) and (89 or 98)
> > - Containing (123 or 321) and (45 or 54) and (89 or 98)
> > = 2*2*2*2*3*2*1= 96 possibilities
> > Brute forcible, but the question suggests we shouldn't need to.
> > Can we exclude more possibilities using other divisors?
> > 2: Not really any help, as we already made use of 4, which is strictly
> > stronger.
> > 3: No help - if you rearrange the digits of a number divisible by 3, you
> > get another number divisible by 3. So we can't exclude any wrong
> > answers, because they will also be divisible by 3.
> > 4: Already used
> > 9: No help for the same reasons as 3.
> > 8 can be used to restrict the 3rd from last digit, but it only excludes
> > a handful of possibilities (it doesn't even uniquely determine the 3rd
> > digit). Depending on whether the last 2 digits are divisible by 8, the
> > second to last digit must be either even or odd. Some possibilities are
> > excluded because the result in illegal combinations of digits)
> > My gut feeling tells me that 12 isn't very useful - we already made use
> > of one it's factors (4) and we know that the other one (3) provides no
> > useful information, so I don't think 12 tells us anything. (But I'm
> > not sure I have a watertight argument, so I could be wrong here)
> > So I'm pretty sure we need to make use of 7 and 11 - but I can't see an
> > easy way or doing so (without doing a tonne of algebra, which I don't
> > think the question is looking for)
On Sat, May 26, 2012 at 10:20 PM, Matthew G <matthew.gilli...@gmail.com> wrote:
>> So now we're down to 102+40+80 = 222 possibilities.
>> Can we do any better?
> We do seem to be able to do a bit better. After seeing some pretty
> exciting code and demos at EuroClojure, I tried this out using
> core.logic which is a logic programming library for Clojure.
> core.logic is similar in its capabilities to prolog, but is just a
> regular clojure lib so you can mix logic programming with functional
> very easily.
> Jim's analysis is really excellent, and all I had to do was write it
> out in the format that core.logic expects. I added to Jim's lemmas
> 1-4 a further constraint that the first 2 digits must differ by one.
> The result (which takes my laptop about a minute and a half) is a
> list of 30 candidates, from which the correct one can easily be
> selected by eye. 436578912.
> Can anyone constrain the solution list further? To one item,
> ideally ;)
> Matthew
> On Apr 27, 10:57 am, Arwyn <arwyn.robe...@gmail.com> wrote:
>> Hi Jim,
>> That's an impressive effort!
>> You certainly got much further than I did with logic.
>> I must admit that I suggested this as a programming problem, so I was kind
>> of ignoring the "Clever logic..." bit.
>> My original aim was to write something that reads like a statement of the
>> problem and also solves it.
>> In a language like Prolog I think that might not be too difficult.
>> However, working in Haskell I have found it quite challenging.
>> I did a filter-based solution which essentially brute-forces the whole
>> thing (9! perms).
>> However, I would like to do something a bit more clever, by which I mean
>> something that tests fewer candidates.
>> You have already cut the space down considerably: 144+48+96=288 << 9!
>> I think we can cut it down a bit further.
>> E.g. taking:
>> >> Ending 12 can be split up 3 ways:
>> >> - Containing (34 or 43) and (56 or 65) and (789 or 987)
>> >> - Containing (34 or 43) and (567 or 765) and (89 or 98)
>> >> - Containing (345 or 543) and (67 or 76) and (89 or 98)
>> >> = 3*2*2*2*3*2*1=144 possibilities
>> Some of the permutations can be excluded by the rule: "In just one case a
>> digit has both neighbours differing from it by one."
>> E.g. the first permutation enumerated would be 345678912 which has 5 digits
>> which have both neighbours differing by 1.
>> Your three lines above already give the three cases for which digit has 2
>> neighbours differing by one: 8, 6 and 4 respectively.
>> So we just have to exclude all the incidental ones; I'll call them
>> "collisions".
>> We still have 3x2x1 perms of the arrangement of groups, however all of them
>> result in at least 1 possible collision and 2 result in 2.
>> If there is 1 possible collision, that reduces the number of perms of the
>> ordering within the groups from 8 to 6 (3 for the collided groups x 2 for
>> the other one).
>> If there are 2 possible collisions, that reduces the number of such perms
>> to 5 because 1 of those 6 is excluded.
>> What this means is that instead of each line being 6x8 = 48 permutations we
>> have 4x6+2x5 = 34.
>> By three lines gives 102.
>> Using similar logic we get 4x6+2x8 = 40 perms for Ending 56 and 80 for
>> Ending 76.
>> So now we're down to 102+40+80 = 222 possibilities.
>> Can we do any better?
>> Cheers,
>> Arwyn
>> On Thursday, 26 April 2012 19:32:49 UTC+1, Jimbo F wrote:
>> > On 2012-04-13 11:15, Arwyn wrote:
>> > > Clever logic should enable you to find the nine-figure number that I
>> > > have in mind. It consists of the digits 1 to 9 in some order, and in
>> > > the number each digit is next to another that differs from it by one.
>> > > In just one case a digit has both neighbours differing from it by one.
>> > > Furthermore, the solution is exactly divisible by more than
>> > > three-quarters of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
>> > > What is the nine-figure number?
>> > Totally off topic.
>> > Did anyone solve this? I made some progress, but didn't nail it.
>> > Lemma 1: The solution must be divisible by 4.
>> > Assume the solution is not divisible by 4. It is therefore not
>> > divisible by 8 or 12. However, this would mean that the solution is
>> > only divisible by at most 3/4th of the numbers between 1 and 12, which
>> > contradicts the problem, which states it is divisible by strictly more
>> > than 3/4th.
>> > We could apply the same argument to 2 and 3, but we don't need to...
>> > Lemma 2: The solution is divisible by 1, 2, 3, 4, 6, 7, 8, 9, 11 and 12,
>> > but not 5 or 10.
>> > All numbers divisible by 10 end in 0. The solution doesn't contain a
>> > 0. We know from Lemma 1 that it's divisible by 4. If it is also
>> > divisible by 5, it must also end in 0. If the solution were not
>> > divisible by any other numbers on the list, it would only be divisible
>> > by 3/4ths, which contradicts the problem, so it must be divisible by the
>> > remaining numbers.
>> > Lemma 3: The solution must end in 12, 32, 56 or 76
>> > Because the solution is divisible by 4, it's last two digits must be
>> > divisible by 4. The two digits must differ by 1. The second digit must
>> > be even, since it is divisible by 4, so the first digit must be odd.
>> > This only leaves a few possibilities. Checking them, we find that 12,
>> > 32, 56 and 76 fit the bill.
>> > Lemma 4: The solution cannot end in 32.
>> > Assume the number ended with 32. According to the rules, the digit 1
>> > must be next to digit 2. However, digit 2 is at the end, so this is a
>> > contradiction.
>> > Where next?
>> > Ending 12 can be split up 3 ways:
>> > - Containing (34 or 43) and (56 or 65) and (789 or 987)
>> > - Containing (34 or 43) and (567 or 765) and (89 or 98)
>> > - Containing (345 or 543) and (67 or 76) and (89 or 98)
>> > = 3*2*2*2*3*2*1=144 possibilities
>> > Ending 56 can only split 1 way:
>> > - Containing (12 or 21) and (34 or43) and (789 or 987)
>> > = 1*2*2*2*3*2*1=48 possibilities
>> > (Note that, ending 456 is not allowed unless the preceding digit is 3,
>> > because then we'd have to form a group from 123, or leave 3 on it's own)
>> > Ending 76 can be split up 2 ways:
>> > - Containing (12 or 21) and (345 or 543) and (89 or 98)
>> > - Containing (123 or 321) and (45 or 54) and (89 or 98)
>> > = 2*2*2*2*3*2*1= 96 possibilities
>> > Brute forcible, but the question suggests we shouldn't need to.
>> > Can we exclude more possibilities using other divisors?
>> > 2: Not really any help, as we already made use of 4, which is strictly
>> > stronger.
>> > 3: No help - if you rearrange the digits of a number divisible by 3, you
>> > get another number divisible by 3. So we can't exclude any wrong
>> > answers, because they will also be divisible by 3.
>> > 4: Already used
>> > 9: No help for the same reasons as 3.
>> > 8 can be used to restrict the 3rd from last digit, but it only excludes
>> > a handful of possibilities (it doesn't even uniquely determine the 3rd
>> > digit). Depending on whether the last 2 digits are divisible by 8, the
>> > second to last digit must be either even or odd. Some possibilities are
>> > excluded because the result in illegal combinations of digits)
>> > My gut feeling tells me that 12 isn't very useful - we already made use
>> > of one it's factors (4) and we know that the other one (3) provides no
>> > useful information, so I don't think 12 tells us anything. (But I'm
>> > not sure I have a watertight argument, so I could be wrong here)
>> > So I'm pretty sure we need to make use of 7 and 11 - but I can't see an
>> > easy way or doing so (without doing a tonne of algebra, which I don't
>> > think the question is looking for)
