Let's first make a few starting observations regarding the left and right numbers for a given number (result) and the number it is created from by a magical operation (operand):
1) If the left number of the result is 1 then the left and right numbers of the operand must either sum to 10 or more, or sum to 9 with a 1 carried from the addition of any middle numbers.
2) If the left number of the result is 1 more than the right number, then the left and right numbers of the operand must sum to the right number of the result, with a 1 carried from the addition of any middle numbers.
3) If the left and right numbers of the result are equal, then the left and right numbers of operand must sum to one of them, and there is no carried 1 from the addition of any middle numbers.
Any combination of left and right numbers for the result that don't meet the above criteria can't be formed by any operand.
Starting with the sixth number as 17347
, then we can say that the fifth number must be...
a 4-digit number, since there is no 5-digit number that can produce the sixth number after a magical operation. The only two single-digit numbers that can produce 17 are 8 and 9, so the fifth number has to be
9ab8
. The 9 has to be on the left so the numbers that sum to 8 on the right (from the prior magical operation on the fourth number) can be incremented by a carried 1. We can also determine thata
andb
have to sum to 3, and that the carried one that creates the 9 will also be added toa
, makinga
one larger thanb
. The only solution isa=2
andb=1
.
This gives us the fifth number:9218
which also has to be a 4-digit number
abcd
. We know thata
andd
have to add to 8, andb
andc
have to add to 11, and are therefore different numbers. Let's first consider the possible values fora
andd
. The combinations2bc6
,3bc5
,5bc3
,6bc2
, and7bc1
are immediately out since there are no combination of numbers in a 3-digit or 4-digit third number that could make those. If it's4bc4
, then a 4-digit third number would have to be either1gh3
,2gh2
, or3gh1
, and there are no combinations ofg
orh
that could produce two different values forb
andc
(since there can be no carried 1).
The only option left for the fourth number is1bc7
. This would be created by a third numberxyz
wherex
andz
would have to add to create a7
on the right and a1
on the left, and we know from above this would mean it's9y8
. The value forb
could be either7
ify+y+1
is less than 10, or8
ify+y+1
is greater than 10. Sinceb+c
must be 11, the only combination that works isb=8
,c=3
, andy=6
.
This gives us the fourth number:1837
As well as the third number:968
which hashave to be a 3-digit number
a8c
, wherea+c
equals 8 and the middle number is 8 to produce a carried 1. As above, the options286
,385
,583
,682
, and781
are out since there are no 2-digit or 3-digit first numbers that can result in those. We're left with484
or187
.
This gives us two possibilities for a second number:187
and484
which forFor a second number
484
the first number must be a 3-digit numbera4c
, wherea+c
is 4 anda
must be strictly less than (not equal to)c
. Onlya=1
andc=3
satisfy this.
For a second number187
, only a first number89
works, since98
has a left number greater than the right.
This gives us two possibilities for a first number:89
and143