Place the numbers 1 through 25 in the grid below:

   1 2 3 4 5
A | | | | | |
B | | | | | |
C | | | | | |
D | | | | | |
E | | | | | |

1. The sum of each column is odd
   
2. The sum of each row, except C, is even
   
3. The sum of row A is not greater than the sum of any other row
   
4. The sum of the diagonal E1 to A5 is less than that of the diagonal A1 to E5
   
5. A4 + B4 > C4 + D4 + E4
   
6. A1 + B1 = D1 + E1
   
7. A1 > E1
   
8. A1, A3 and B1 are all prime numbers
   
9. (A3 + E3) is a prime number
   
10. A5, D1, D3 and E1 are all squares
    
11. B2, C2 and D2 are ascending consecutive numbers
    
12. B3, C3 and D3 are ascending consecutive numbers
    
13. B5 + D5 = A5 + C5
    
14. (C1)² + (C5)² = (E3)²
    
15. C5 is a two digit number
    
16. D5 is a multiple of E5
    
17. E1 + E3 = E2 + E4 + E5

^{Let's start with (14) : (C1)² + (C5)² = (E3)².}

^{From (1) and (6) we conclude that C1 is odd, and from (15) we know that C5 has 2 digits.}

^{The possibilities are : (C1, C5, E3) = (5,12,13) or (7,24,25) or (9,12,15) or (15,20,25).} 

^{In all of them, E3 is odd. Since, by (9), (A3+B3) is prime, so A3 must be even. And since by (8), A3 is a prime, the unique possibility is A3=2. But, since (A3+E3)=(2+E3) is prime, the unique possility of the four above is A3=2, C1=9, C5=12 and E3=15.}

 

^{So, we arrive at :}

 

^{A1,A2,02,A4,A5}

^{B1,B2,B3,B4,B5}

^{09,C2,C3,C4,12}

^{D1,D2,D3,D4,D5}

^{E1,E2,15,E4,E5}

 

^{Since by (10), D3 is a square, and by (12), (B3,C3,D3) are consecutive crescent numbers, the unique value for D3 is 25, so B3=23 and C3=24.}

 

^{We have until now :}

 

^{A1,A2,02,A4,A5}

^{B1,B2,23,B4,B5}

^{09,C2,24,C4,12}

^{D1,D2,25,D4,D5}

^{E1,E2,15,E4,E5}

 

^{By (10), D1 and E1 are squares, so they must be 1 or 4 or 16, since 9 and 25 are already placed.}

 

^{Since by (6), (A1+B1)=(D1+E1), and by (7), A1 > E1, B1 is smaller than D1. So, D1 can't be 1 (because there are no zero).}

 

^{We have now two alternatives : D1=4 or D1=16.}

 

 ^{a)First hypothesis : D1 = 4.}

^{A1,A2,2,A4,A5}

^{B1,B2,23,B4,B5}

^{09,C2,24,C4,12}

^{04,D2,25,D4,D5}

^{E1,E2,15,E4,E5}

 

^{E1 (square) would then be or 1 or 16. But E1 can't be 1, since (D1+E1) would result in 5, and since (A1+B1)=(E1+D1), there are no values for A1 and B1, since 2 is already placed.}

^{So, E1=16, and A5 (by (10) is a square will be 1.}

^{A1,A2,02,A4,01}

^{B1,B2,23,B4,B5}

^{09,C2,24,C4,12}

^{04,D2,25,D4,D5}

^{16,E2,15,E4,E5}

 

^{By (8), A1 and B1 are primes, and by (6) their sum is 20 (4+16), so (A1,B1) would be (3,17) or (17,3)or (7,13) or (13,7). But, since by (7) A1 is greater than E1, the unique possibility is (A1,B1)=(17,3).}

 

^{17,A2,02,A4,01}

^{03,B2,23,B4,B5}

^{09,C2,24,C4,12}

^{04,D2,25,D4,D5}

^{16,E2,15,E4,E5}

 

^{By (13), (B5+D5)=(A5+C5), so (B5+D5)=(1+12)=13. The alternatives are (B5,D5)=(5,8) or (8,5) or (6,7) or (7,6). Since, by (16), D5 is a multiple of E5, none of the alternatives above will serve, for the first pair, E5 would have to be 1 or 2 or 4, all already placed; for the second and third pairs, E5 would be 1 (already placed) and for the last pair, E5 would have to be 1 or 2 or 3 (all already placed).}

 

^{SO, THE HYPOTHESIS D1=4 IS FALSE, AND D1= 16.}

 

^{We have, then :}

^{A1,A2,02,A4,A5}

^{B1,B2,23,B4,B5}

^{09,C2,24,C4,12}

^{16,D2,25,D4,D5}

^{E1,E2,15,E4,E5}

 

^{E1 (square) will be 1 or 4. For E1=1, by (6), (A1+B1)=(16+1)=17, and doesn't exist a pair of primes (by (8), A1 and B1 are primes) that sum to 17. So E1=4 and by (10), A5=1.}

 

^{A1,A2,02,A4,01}

^{B1,B2,23,B4,B5}

^{09,C2,24,C4,12}

^{16,D2,25,D4,D5}

^{04,E2,15,E4,E5.}

 

^{By (8), A1 and B1 are primes and by (6) their sum is 20. So the alternatives are (A1,B1)=(3,17) or (17,3) or (7,13) or (13,7). And since by (7) A1 is greater than E1, we can eliminate the alternative (3,17).}

 

^{We have now to analyse each one of these 3 alternatives.}

 

^{FIRST ONE : (A1,B1)=(17,3).}

 

^{17,A2,02,A4,01}

^{03,B2,23,B4,B5}

^{09,C2,24,C4,12}

^{16,D2,25,D4,D5}

^{04,E2,15,E4,E5}

 

^{By (13), (B5+D5)=(A5+C5)=13. So, (B5,D5)=(5,8) or (8,5) or (6,7) or (7,6). Since by (16), D5 is a multiple of E5, none of these pairs serve. For the first, E5 would have to be 1 or 2 or 4 (all already placed), for the second and third, E5 would have to be 1 (already placed) and for the fourth, E5 would have to be 1 or 2 or 3 (all already placed).}

^{SO THE FIRST HYPOTHESIS (A1,B1)=(17,3), IS FALSE.}

 

^{SECOND HYPOTHESIS : (A1,B1)=(7,13).}

 

^{07,A2,02,A4,01}

^{13,B2,23,B4,B5}

^{09,C2,24,C4,12}

^{16,D2,25,D4,D5}

^{04,E2,15,E4,E5}

 

^{For (13), (B5+D5)=(A5+C5)=13. So the alternatives are (B5,D5)=(3,10) or (10,3) or (5,8) or (8,5).}

^{By (16), D5 is a multiple of E5, so the last three pairs don't serve, as showed before. So, B5=3 and D5=10 and since D5 is a multiple of E5, and can't be neither 1 or 2, E5=5.}

 

^{07,A2,02,A4,01}

^{13,B2,23,B4,03}

^{09,C2,24,C4,12}

^{16,D2,25,D4,10}

^{04,E2,15,E4,05}

 

^{By (17), (E1+E3)=(E2+E4+E5), so (E2+E4)=14, that leads to (E2,E4) = (6,8) or (8,6), and results in 38 for the sum of line E.}

^{By (3), the sum of line A is not greater than any other line. Since line E sums to 38, (A2+A4) will have to be less or equal to 28.}

^{The numbers not placed are : 11,14,17,18,19,20,21 and 22.}

^{So (A2,A4) could only be (11,14) or (14,11) or (11,17) or (17,11).}

^{By (11), (B2,C2,D2) are consecutive crescent numbers, and the possibilities are (17,18,19) or (18,19,20) or (19,20,21) or (20,21,22).}

 

^{a) For (B2,C2,D2)=(17,18,19) or (19,20,21), since E2 is equal to 6 or 8 (an even number), and by (1), the sum of each column is odd, A2 must be even. The pairs possibles are (A2,A4)=(11,14),(11,17),(17,11).}

 

^{The sumof line A, by (2) is even. So, (A2+A4) should be even, and rests the pairs (A2,A4)=(11,17) or (17,11).}

 

^{So, (B2,C2,D2) should only be (19,20,21).}

 

^{We have, then two possibilities now:}

^{07,11,02,17,01}

^{13,19,23,B4,03}

^{09,20,24,C4,12}

^{16,21,25,D4,10}

^{04,E2,15,E4,05}

 

^OR

 

^{07,17,02,11,01}

^{13,19,23,B4,03}

^{09,20,24,C4,12}

^{16,21,25,D4,10}

^{04,E2,15,E4,05}

 

^{In the two alternatives, the numbers for B4,C4,D4 are 14, 18, 22, not necessarily in this order. And, any one of these numbers, for D4, will make the sum of line C odd, wich confirms (2).}

^{By (4), (A1+B2+D4+E5) is greater than (E1+D2+B4+A5), that is, in both cases (7+19+D4+5) > (4+21+B4+1).}

^{So, D4 + 5 > B4. And also, by (5), (A4+B4) > (C4+D4+E4).}

 

^{In the first case, A4=17, so B4 > (C4+D4+E4-17).}

 

^{If B4=14, then (C4+D4)=(18+22)=40 that leads to the impossibility 14 > E4 + 23.}

 

^{If B4=18, then 18 > E4+19, wich is impossible too.}

 

^{If B4=22, then 22 > E4+15, and the unique value that confirms this inequation is E4=6 (we already knew that E4 is either 6 or 8). And since B4 < D4 + 5, we have 22 < D4 + 5, wich implies D4=18, and by consequence, C4=14 and E2=8.}

 

^{And we obtain a possible solution :}

 

^{07,11,02,17,01}

^{13,19,23,22,03}

^{09,20,24,14,12}

^{16,21,25,18,10}

^{04,08,15,06,10.}

 

^{In the second case, A4=11. So, B4 > (C4+D4+E4-11).}

^{If B4=14, then (C4+D4)=(18+22)=40 and we'll have 14 > E4 +29, wich is imposible.}

 

^{If B4=18, we'll have 18 > E4+25, wich is impossible too.}

 

^{If B4=22, we'll have 22 > E4+21, wich is impossible too.}

 

^{b) For (B2,C2,D2)=(18,19,20) or (20,21,22), since E2=6 or 8, and by  (1), the sum of each column is odd, A2 must be even. The unique pair is (A2,A4)=(14,11), that sums 25 and leads to the sum of line A odd, wich contradicts (2).} <o:p> </o:p><o:p> </o:p>

<o:p></o:p> 

<o:p>SO, THE HYPOTHESIS (A1,B1)=(7,13) LEADS TO A SOLUTION (showed above).</o:p>

<o:p></o:p> 

<o:p></o:p> 

<o:p>We will analyse the third hypothesis, to verify if there is another solution (there isn't).</o:p>

<o:p></o:p> 

<o:p>THIRD HYPOTHESIS : (A1,B1)=(13,7)</o:p>

<o:p></o:p> 

<o:p>13,A2,02,A4,01</o:p>

<o:p>07,B2,23,B4,B5</o:p>

<o:p>09,C2,24,C4,12</o:p>

<o:p>16,D2,25,D4,D5</o:p>

<o:p>04,E2,15,E4,E5</o:p>

<o:p></o:p> 

<o:p>By (13), (B5+D5)=(A5+C5)=13. So, (B5,D5)=(3,10) or (10,3) or (5,8) or (8,5).</o:p>

<o:p></o:p> 

<o:p>By (16), D5 is a multiple of E5. The last three pairs above doens't serve, as showed before. So B5=3 and D5=10, and since D5 (a multiple of E5) can't be 1 or 2, E5=5.</o:p><o:p><o:p> </o:p>

<o:p> </o:p>

13,A2,02,A4,01

07,B2,23,B4,03

09,C2,24,C4,12

16,D2,25,D4,10

04,E2,15,E4,05

 

By (17), (E1+E3)=(E2+E4+E5) wich leads to (E2+E4)=14, so (E2,E4)=(6,8) or (8,6) and the sum of line E is 38.

 

By (3) the sum of line A is not greater than any other line. And since line E sums 38, (A2+A4) would have to be less or equal 22.

 

The numbers that rests to be placed are 11,14,17,18,19,20,21 and 22.

 

And there is no pairs for (A2,A4) that sums less than or equal to 22.

<o:p> </o:p>

<o:p></o:p> 

<o:p>SO THE SOLUTION FOUNDED IS UNIQUE :</o:p>

<o:p></o:p> 

<o:p>07,11,02,17,01</o:p>

<o:p>13,19,23,22,03</o:p>

<o:p>09,20,24,14,12</o:p>

<o:p>16,21,25,18,10</o:p>

<o:p>04,08,15,06,05 </o:p>

</o:p> 

<o:p></o:p> 

<o:p></o:p> 

<o:p></o:p> 

<o:p></o:p> 

<o:p></o:p> 

<o:p></o:p> 

<o:p></o:p> 

<o:p></o:p> 

<o:p>THIRD HYPOTHESIS : (A1,B1)=(13,7).</o:p>

<o:p> </o:p>

 

 

<o:p><o:p><o:p> </o:p><o:p> </o:p><o:p> </o:p>

<o:p> </o:p>

<o:p></o:p> 

SOLUÇÃO POSSÍVEL !!<o:p></o:p>

</o:p> 

</o:p>

<o:p> </o:p>

<o:p> </o:p>

 

<o:p> </o:p>

<o:p> </o:p>

<o:p> </o:p>

 

<o:p> </o:p>

  <o:p> </o:p>

 

 

 

 

 

 <o:p></o:p>

 

 

 

<o:p> </o:p>

<o:p> </o:p>

 

 

<o:p> </o:p>

<o:p> </o:p>

 

 

<o:p> 

</o:p>

<o:p> </o:p>

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Posted by ARLEKIM on 2005-02-10 18:26:14