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Enigma 1692 - Key factors
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Chappy
May 16, 2012, 8:52:31 PM5/16/12
to
Enigma 1692 - Key factors
New Scientist magazine, 07 April 2012.
By Susan Denham.
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?
Ciao,
Chappy.
New Scientist magazine, 07 April 2012.
By Susan Denham.
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?
Ciao,
Chappy.
Ilan Mayer
May 16, 2012, 10:09:28 PM5/16/12
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The number cannot be divisible by 10 as there is no 0. It cannot be
divisible by 5 either since then too few divisors in the range 1-12
are left.
Therefore the number must be divisible by 1, 2, 3, 4, 6, 7, 8, 9, 11
and 12.
Since it is divisible by 8, it must end in 12, 32, 56 or 76.
Using a cryptarithm solver finds only one solution:
436578912
Please reply to ilan dot mayer at hotmail dot com
__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||
John Jones
May 17, 2012, 1:10:29 PM5/17/12
to
In article <5946558.1127.1337215951817.JavaMail.geo-discussion-
forums@pbqm7>, petergreg...@hotmail.com says...
> Spoiler
3/4 of 12 is 9 so we need to cast only two of those divisors out.
obviously 10 and 5 then.
3 and 9 are given (digit-sum is 45 div by 3 and 9).
1 too.
If div by 2 must end with an even digit, and div by 6 then a given.
Probably far enough. The lcm of the divisors (less 10 and 5) is 5544, so
the end result must be divisible by 5544 which is a useful test.
Since there is a special number adjacent to both its neighbours we can
divide the numbers into pairs and one triplet. Each pair has difference
of one.
The triplet is a run of three.
We know the last digit is even, but it could be the triplet or a pair.
TBQH I have not yet analysed the former. But carrying on regardless:-
suppose the end is a pair.
Then in some order we have pair pair triplet + pair.
lets write x' x for a pair and t'' t' t for the triplet
then the number is something like
y' y x' x t'' t' t e' e
(e is even)
Now a number is divisible by 11 if the two alternating sums of digits
are equal modulus 11.
so
-y' + (y-x') +(x-t'') + (t'-t) +(e'-e) is the value that should
be zero.
rearranging brackets we get
(-y'+y) +(-x'+x) + (-t'' + t') + -t + (e'-e)
Each bracketted item is +/-1, so to be divisible by 11 we must have
t = 1 : 1 : 1 : 1 where each ":" is replaced by + or - in all possible
ways.
so t = -4 -2 0 2 or 4 mod 11
ie t = 2 4 7 or 9 (no zero)
Since 234 432 456 leave an odd number of lower digits we couldnt then
pair the rest, so t= 7 or 9 and either way the triple is 789 or 987.
The obvious thing to do next is to attack divisibility by 7, which isnt
as easy.
Consider the number as groups of digits: V1 V2 V3 V4.
Let the triple come third, V3, (this is the only case I tried).
Think of dividing by 7, using short division. You end up with a
remainder coming into V4 from the left. Since V4 is a pair, if R3 is the
remainder coming in, we must have 100*R3+V4 divisible by 7, so
R3 = -V4 / 100 mod 7 = 3 * V4 mod 7
Now V3 has a remainder R2 coming in, and is a triple, so to produce R3
we need
1000*R2 + V3 = R3 mod 7
R2 = V3 - R3 mod 7
Finally V2 has R1 coming from the division of V1, R1=V1 mod 7, so we can
deduce
100*V1 + V2 = R2 mod 7
V1 = 4*(R2 - V2) mod 7.
Doesnt look hopeful, but allows a little table to be drawn up. I will
illustrate the case V4 = 5, corresponding to V4 digits = 12. The only
other cases are 34 and 56 and the tables are equally short for those but
have no solutions. V3 and V2 show all possible values, and V1 is
calculated from the above formulae (as are R2 R3).
line V1 V2 R2 V3 R3 V4 number
1 2 0 4 5 1 5
2 5 1 4 5 1 5
3 1 2 4 5 1 5 436578912
4 3 5 4 5 1 5
5 6 6 4 5 1 5
6 3 0 6 0 1 5
7 6 1 6 0 1 5
8 2 2 6 0 1 5
9 4 5 6 0 1 5
10 0 6 6 0 1 5 563498712
V3 can only be 789(0 mod 7) or 987 (5 mod 7).
V2 can be any of 21(0) 56(0) 43(1) 65(2) 12(5) 34(6)
V1 must be any of 21(0) 56(0) 43(1) 65(2) 12(5) 34(6)
where the number in brackets is mod 7 of the two digit number before it.
V1 is calculated, as above. Note that V1 does not have
3 or 4 as possibles and so lines 4 6 and 9 can be struck off.
line 1 requires V1 to be 2, corresponding to 65 and V2 to be 0
corresponding to 21 or 56. Since V4 is 12 neither of these work.
Similar considerations cast out all but lines 3 and 10. Line 10
corresponds to 563498712 which satisfies everything save it is not
divisible by 11, and so is not divisible by "more than" 3/4 of the
numbers 1 to 12.
Line 3 is the solution already posted.
Is there a simpler way? Susan Denham is rather clever at these aint she?
HTH
JJ
forums@pbqm7>, petergreg...@hotmail.com says...
> Spoiler
obviously 10 and 5 then.
3 and 9 are given (digit-sum is 45 div by 3 and 9).
1 too.
If div by 2 must end with an even digit, and div by 6 then a given.
Probably far enough. The lcm of the divisors (less 10 and 5) is 5544, so
the end result must be divisible by 5544 which is a useful test.
Since there is a special number adjacent to both its neighbours we can
divide the numbers into pairs and one triplet. Each pair has difference
of one.
The triplet is a run of three.
We know the last digit is even, but it could be the triplet or a pair.
TBQH I have not yet analysed the former. But carrying on regardless:-
suppose the end is a pair.
Then in some order we have pair pair triplet + pair.
lets write x' x for a pair and t'' t' t for the triplet
then the number is something like
y' y x' x t'' t' t e' e
(e is even)
Now a number is divisible by 11 if the two alternating sums of digits
are equal modulus 11.
so
-y' + (y-x') +(x-t'') + (t'-t) +(e'-e) is the value that should
be zero.
rearranging brackets we get
(-y'+y) +(-x'+x) + (-t'' + t') + -t + (e'-e)
Each bracketted item is +/-1, so to be divisible by 11 we must have
t = 1 : 1 : 1 : 1 where each ":" is replaced by + or - in all possible
ways.
so t = -4 -2 0 2 or 4 mod 11
ie t = 2 4 7 or 9 (no zero)
Since 234 432 456 leave an odd number of lower digits we couldnt then
pair the rest, so t= 7 or 9 and either way the triple is 789 or 987.
The obvious thing to do next is to attack divisibility by 7, which isnt
as easy.
Consider the number as groups of digits: V1 V2 V3 V4.
Let the triple come third, V3, (this is the only case I tried).
Think of dividing by 7, using short division. You end up with a
remainder coming into V4 from the left. Since V4 is a pair, if R3 is the
remainder coming in, we must have 100*R3+V4 divisible by 7, so
R3 = -V4 / 100 mod 7 = 3 * V4 mod 7
Now V3 has a remainder R2 coming in, and is a triple, so to produce R3
we need
1000*R2 + V3 = R3 mod 7
R2 = V3 - R3 mod 7
Finally V2 has R1 coming from the division of V1, R1=V1 mod 7, so we can
deduce
100*V1 + V2 = R2 mod 7
V1 = 4*(R2 - V2) mod 7.
Doesnt look hopeful, but allows a little table to be drawn up. I will
illustrate the case V4 = 5, corresponding to V4 digits = 12. The only
other cases are 34 and 56 and the tables are equally short for those but
have no solutions. V3 and V2 show all possible values, and V1 is
calculated from the above formulae (as are R2 R3).
line V1 V2 R2 V3 R3 V4 number
1 2 0 4 5 1 5
2 5 1 4 5 1 5
3 1 2 4 5 1 5 436578912
4 3 5 4 5 1 5
5 6 6 4 5 1 5
6 3 0 6 0 1 5
7 6 1 6 0 1 5
8 2 2 6 0 1 5
9 4 5 6 0 1 5
10 0 6 6 0 1 5 563498712
V3 can only be 789(0 mod 7) or 987 (5 mod 7).
V2 can be any of 21(0) 56(0) 43(1) 65(2) 12(5) 34(6)
V1 must be any of 21(0) 56(0) 43(1) 65(2) 12(5) 34(6)
where the number in brackets is mod 7 of the two digit number before it.
V1 is calculated, as above. Note that V1 does not have
3 or 4 as possibles and so lines 4 6 and 9 can be struck off.
line 1 requires V1 to be 2, corresponding to 65 and V2 to be 0
corresponding to 21 or 56. Since V4 is 12 neither of these work.
Similar considerations cast out all but lines 3 and 10. Line 10
corresponds to 563498712 which satisfies everything save it is not
divisible by 11, and so is not divisible by "more than" 3/4 of the
numbers 1 to 12.
Line 3 is the solution already posted.
Is there a simpler way? Susan Denham is rather clever at these aint she?
HTH
JJ
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