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a Maths problem - urgent please


hgupta

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consider A to I. The values of A to I are equal to 1 to 9, not necessarily in sequential manner. And A=4. Now

A+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

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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

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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.

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