One to Nine
Math brain teasers require computations to solve.
Arrange the digits from 1 to 9 to make a 9-digit number ABCDEFGHI which satisfies the following conditions:
1) AB is divisible by 2;
2) ABC is divisible by 3;
3) ABCD is divisible by 4;
4) ABCDE is divisible by 5;
5) ABCDEF is divisible by 6;
6) ABCDEFG is divisible by 7;
7) ABCDEFGH is divisible by 8;
8) ABCDEFGHI is divisible by 9.
There is only one solution.
HintCondition 8 is the easiest to satisfy just because it is automatically satisfied. But condition 6 is the hardest to satisfy, and should be left last.
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Answer
From condition 4, we know that E equals 5.
From conditions 1, 3, 5 and 7, we know that B, D, F, H are even numbers, therefore A, C, G, I are 1, 3, 7, 9 in some order.
Furthermore, from conditions 3 and 7 we know that CD is divisible by 4 and GH is divisible by 8 (because FGH is divisible by 8 and F is even). Because C and G are odd, D and H must be 2 and 6 in some order.
From conditions 2 and 5, we know that A+B+C, D+E+F, G+H+I are all divisible by 3.
If D=2, then F=8, H=6, B=4. A+4+C is divisible by 3, therefore A and C must be 1 and 7 in some order, G and I must be 3 and 9 in some order. G6 is divisible by 8, therefore G=9. But neither 1472589 nor 7412589 is divisible by 7.
Therefore D=6, and F=4, H=2, B=8. G2 is divisible by 8, therefore G=3 or 7. A+8+C is divisible by 3, therefore one of A and C is chosen from 1 and 7, the other is chosen from 3 and 9.
If G=3, then one of A and C is 9, the other is chosen from 1 and 7. But none of 1896543, 7896543, 9816543 and 9876543 is divisible by 7.
Therefore G=7, one of A and C is 1, the other is chosen from 3 and 9. From 1836547, 1896547, 3816547 and 9816547, only 3816547 is divisible by 7 (the quotient is 545221).
Therefore, the number we are looking for is 381654729.
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Comments
mercenary007   
Apr 12, 2006
| wow... very good teaser, though I thought this to be more of a Logic brainteaser than a Math brainteaser |
tfrangou
Apr 12, 2006
| well there is more than one number. I found the following number that also satisfy the conditions: 222.456.564 |
signmeup74  
Apr 12, 2006
| I think you are only supposed to use each of the digits 1-9 ONCE.
You can't repeat 2, etc.
I agree with the logical nature, too. I didn't do much math.  |
calmsavior 
Apr 12, 2006
| i actually found deciphering the sixth statement as easy, because i can actually mentally determine if a number is divisble by 7, and what remainder it would have after the division (fast) |
shenqiang
Apr 12, 2006
| I meant that, comparing to others, this is the hardest. |
grungy49
Nov 24, 2006
| I tried to make another number - but looks like you were right, the highest I got up to was 1,620,549 without repeating digits. |
javaguru
Jan 02, 2009
| Great teaser! My process of solving it was exactly the way you describe it, down to the order of checking for division by 7.  |
javaguru
Jan 02, 2009
| I do agree that this is more of a logic brain teaser than math...although knowing the rules for divisibility requires some minimal math knowledge. |
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