8-digit number and divisibility puzzle

AI Thread Summary
The discussion centers on finding an 8-digit integer P, represented as ABCDEFGH, which uses each digit from 1 to 8 exactly once and meets specific divisibility conditions. The conditions require that pairs of digits (AB, BC, CD, DE, EF, FG, GH) be divisible by 8, 7, 6, 5, 4, 3, and 2 respectively. The process begins with identifying possible values for each pair based on their divisibility rules, followed by eliminating combinations that include the digits 0 or 9 or repeat any digits. Notably, the digit E must be 5, limiting the possible values for EF. Through systematic testing of combinations, the viable sequences are narrowed down. Ultimately, the only configuration that satisfies all conditions is 72185634. This number meets all divisibility requirements and uses each digit from 1 to 8 exactly once.
K Sengupta
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P is a 8-digit base ten positive integer having the form ABCDEFGH that uses each of the nonzero digits from 1 to 8 exactly once, and satisfies all of these conditions:

(i) AB is divisible by 8.
(ii) BC is divisible by 7.
(iii) CD is divisible by 6.
(iv) DE is divisible by 5.
(v) EF is divisible by 4.
(vi) FG is divisible by 3.
(vii) GH is divisible by 2.

Determine all possible value(s) that P can assume.
 
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72185634

Force. Brute and dirty.
 
K Sengupta said:
P is a 8-digit base ten positive integer having the form ABCDEFGH that uses each of the nonzero digits from 1 to 8 exactly once, and satisfies all of these conditions:

(i) AB is divisible by 8.
(ii) BC is divisible by 7.
(iii) CD is divisible by 6.
(iv) DE is divisible by 5.
(v) EF is divisible by 4.
(vi) FG is divisible by 3.
(vii) GH is divisible by 2.

Determine all possible value(s) that P can assume.

Step 1 (list the multiplication table for XY < 90):
AB has to be: 16,24,32,40,48,56,64,72,80,88
BC has to be: 14,21,28,35,42,49,56,63,70,77,84,
CD has to be: 12,18,24,30,36,42,48,54,60,66,72,78,84
DE has to be: 10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85
EF has to be: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88
FG has to be: 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87
GH has to be: 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88

Step 2 Now eliminate any numbers that have 9 in them or 0 in them or numbers with the same digit twice:
AB has to be: 16,24,32,48,56,64,72
BC has to be: 14,21,28,35,42,56,63,84,
CD has to be: 12,18,24,36,42,48,54,72,78,84
DE has to be: 15,25,35,45,65,75,85
EF has to be: 12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84
FG has to be: 12, 15, 18, 21, 24, 27, 36, 42, 45, 48, 51, 54, 57, 63, 72, 75, 78, 81, 84, 87
GH has to be: 12, 14, 16, 18, 24, 26, 28, 32, 34, 36, 38, 42, 46, 48, 52, 54, 56, 58, 62, 64, 68, 72, 74, 76, 78, 82, 84, 86

Step 3:
DE has only endings of 5, so E = 5
Therefore EF is limited to 50 <= EF <= 59

so EF has to be: 52, 56
FG then has to be: 21, 24, 27, 63
GH then has to be: 14, 16, 18, 32, 34, 38, 46, 48, 74, 76

Step 4: Start making connections:

If AB was 16, BC is 63, CD is 36, which can't be possible because of repeating
If AB was 24, BC is 42, so nope
If AB was 32, BC is 21 or 28, then CD is 12,18,or 84. 12 doesn't fit. So try 18 first. DE can then be 85 only. EF is 56, FG is 63. However 3 has shown up already, so try 84. DE is then 45, EF is then 56. Again FG is 63, so doesn't work.

AB = 48, BC = 84, doesn't work
AB = 56 => 63 => 36, doesn't work
64 => 42 => 24 doesn't work.

So AB = 72, so A = 7 B = 2
BC = 21 or 28
72 => 21 => 18 => 85 => 56 => 63 => 34

So the final number is 72185634
 
a=7 b=2 c=1 d=8 e=5 f=6 g=3 h=4
I solved it by looking at the multiples of 8 for AB. Then the last digit of AB must equal the first digit of BC which is a multiple of 7. So I just checked the multiples and crossed out numbers that had either a zero or a digit that occurred twice.
 
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