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Re: official quant thread for CAT 2009 -
22-07-2009, 02:50 AM
(m)*A1+ (m-1)*A2+(m-2)*A3+... Am =99
Where, A1 to Am are either 3 or 7.
If all, were 7, then for m=5, Sum =105
and if all were 3 then, for m=7, sum=84
This shows that m must be seleceted out of 5,6 and 7
When we replace 7 with 3 or 3 with 7 there will be a change of 4 units, hence m=5 can never give sum as 99 because (105-99) is not of the form 4k.
For m=7, and all 7s, we have sum = 196. Since (196-99) is not 4k hence m=7 is not possible.
For, m=6, and all 7s, we have sum = 147. Since (147-99) is of the form 4k, and the difference is 48, we can have thje following combinations:
3,7,3,7,3,7
7,3,3,3,7,7
3,3,7,7,7,3
3,7,7,3 ,3,3
7,3,3,7,3,3
Hence 5 solutions.
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