hgupta Posted July 15, 2006 Posted July 15, 2006 consider A to I. The values of A to I are equal to 1 to 9, not necessarily in sequential manner. And A=4. NowA+B+C+D = D+E+F+G = G+H+I = 17. What is the value of D,G. could somebody please tell me how to do this?? i am totally lost on this one. hgupta
hgupta Posted July 16, 2006 Author Posted July 16, 2006 got the solution, posting here if someone else might want to know: Since A to I are 1 to 9, then: A+B+C+D+E+F+G+H+I = 45 (eq_1) We know that: A+B+C+D + D+E+F+G + G+H+I = 51 (17 * 3) (eq_2) Therefore the difference between the eq_1 and eq_2 is D+G, so D+G = 51-45 = 6 3 pairs of numbers can add together to form 6: 1 & 5, or 2 & 4, or 3 & 3 The correct pair must be 1 and 5 because we know A = 4, and repetition of 3. If we guess that D=1 and G=5, then 4+B+C+1 = 17 and 5+H+I = 17 This means that B+C = 17-4-1 = 12, and H+I = 17-5 = 12. Three pairs of numbers add together to make 12: 3+9, 4+8, and 5+7. Of these, it can't be 4+8 because A=4, and it can't be 5+7 because we know that either D or G = 5, so the correct pair must be 9+3. But, we have two pairs of numbers that need to add up to 12: B+C and H+I, and only one pair of numbers to acheive it with, therefore our original assumption that D=1 and G=5 must be wrong. Now assume that D = 5 and G = 1: We now have 4+B+C+5 = 17, and 1+H+I = 17. So, B+C = 17-4-5 = 8, and H+I = 17-1 = 16. Only one pair of numbers add up to make 16: 7+9, so H+ I = 7+9 Of the remaining numbers one pair adds up to 8: 2+6. Therefore B+C = 2+6 We are only left with one other pair: 3+8, so E+F must = 3+8. If you now plug all these numebers back into the original equations: 4 + 2 + 6 + 5 = 17 5 + 3 + 8 + 1 = 17 1 + 7 + 9 = 17. Therefore D = 5, and G =1
the tree Posted July 17, 2006 Posted July 17, 2006 i dint understand.pls.explainIs there a specific bit that you don't understand?
hgupta Posted July 17, 2006 Author Posted July 17, 2006 i dint understand.pls.explain umm, i think that's the whole explanation. well if you could not understand a/some specific step(s), post back and i will try to redo it again in a different manner.
chitrangda Posted July 18, 2006 Posted July 18, 2006 i went through it again i got a better understanding of it now thnx 4 taking pains, pls post more questions like this.they r very intresting!
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