Brain Teasers
Divisorama
Arrange the numerals "123456789" to form a 9-digit number (referred to as "ABCDEFGHI") whose 2-digit subsets meet the following criteria:
1) AB is divisible by 2
2) BC is divisible by 3
3) CD is divisible by 4
4) DE is divisible by 5
5) EF is divisible by 6
6) FG is divisible by 7
7) GH is divisible by 8
8) HI is divisible by 9
There are two solutions.
1) AB is divisible by 2
2) BC is divisible by 3
3) CD is divisible by 4
4) DE is divisible by 5
5) EF is divisible by 6
6) FG is divisible by 7
7) GH is divisible by 8
8) HI is divisible by 9
There are two solutions.
Answer
781254963 and 187254963AB, CD, EF, and GH must be divisible by even divisors. Therefore B, D, F, and H must be even numerals. This means that A, C, E, G, and I are odd numerals. DE must be divisible by 5, therefore E is 5. EF must be divisible by 6, therefore EF is 54. FG must be divisible by 7, therefore FG is 49. GH must be divisible by 8, therefore GH is 96. GH must be divisible by 9, therefore GH is 63. The last five digits have now been determined to be 54963. To be divisible by 4, CD must end in 2. This makes B equal 8. The digits are now ?8?254963. The remaining digits, 1 and 7, can be placed into either remaining location.
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Comments
Um.. this should have a logic grid instead of a calculator..
Yeah, this is a Logic-Grid, that's why I didn't understand it well. (because I needed that Grid page with all the stuff)
try the number 123456789!!!!
Therulerofthem, you aren't understanding the question. 123456789 doesn't work because 23 is NOT divisible by 3, 34 is NOT divisible by 4, 56 is NOT divisible by 6, etc.
i don't understand at all...i'm not even gonna try!
Jan 01, 2005
Great puzzle. I had fun working it out.
stupid me... i multiplied my two digits together and got 123456789!
Not that hard once you start at the obvious best starting place - E=5. Then go forwards, realizing the 7 has to be 49, not 42. Then head backwards, see that D is automatically 2 (B, D, F, H are even, A, C, E, G, I are odd, if D = 8, then C must be even, but it can;t be, this also explains the 7 =49 part (though the next line would yield 24, giving 5424 at some point in the number). Easy after that. Well, done after that, pretty much.
I started with deciding the even digits, but once 5 is assigned to E, all the numbers above DE are forced. This leaves 1,2,7,8, which forces D=2, forcing the last even B=8.
Looked like it might be difficult, but actually was trivial. Very cool teaser anyway!
Looked like it might be difficult, but actually was trivial. Very cool teaser anyway!
Fun and challenging! Cool that something so systematic actually works out to only 2 possible answers, without being something overtly significant (e.g., 987654321). Took me about 5 minutes. At first I thought I was going to get to "Correct a Teaser" when I found two answers. But then found that you listed both >_<
Good teaser!
Good teaser!
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