Posted 31 Aug 2013 (edited) I have four positive numbers, a, b, c and d (not necessarily integers). Obviously, there are six ways to multiply pairs of them, yielding the products a*b, a*c, a*d, b*c, b*d and c*d. I tell you what five of these products are (but not which product is what): 2, 3, 4, 5 and 6 What's the sixth product? There's a quick and elegant way to answer that question without actually needing to know what the numbers are. Edited 31 Aug 2013 by BMAD 0 Share this post Link to post Share on other sites

0 Posted 1 Sep 2013 If we multiply all 6 pairs we will get (a*b*c*d)^3.Assume that the missing product was a*b, then among the 5 given products we see (a*c)*(b*d)=(a*d)*(b*c), if we look at the numbers obviously 5 is out of the equation because there is nothing to balance it so the only remaining match is 2*6=3*4, so a*b*c*d=2*6=12.So the product of all 6 pairs should be 12^3, the product of the given 5 is 720, so the missing one is 12/5=2.4 0 Share this post Link to post Share on other sites

0 Posted 12 Sep 2013 The four numbers (in no particular order) are either {sqrt(6/5),sqrt(24/5),sqrt(10/3),sqrt(15/2)} or {sqrt(18/5),sqrt(8/5),sqrt(10),sqrt(5/2)}. 0 Share this post Link to post Share on other sites

0 Posted 12 Sep 2013 (edited) The six products listed can obviously be split into 3 pairs of products, each which equal a*b*c*d. So by listing out all possible products that can be found by multiplying any two of the 5 given products, we're guaranteed to find exactly 2 of these products that equal a*b*c*d. If one tries doing this for 2,3,4,5,6 only one end product is repeated twice (12) ensuring that 12 = a*b*c*d. Once we have this, finding the last product is easy. the two pairs of procucts that multiple to 12 are (2,6) and (3,4). Thus the last pair must be (5,2.4), giving the sixth product as 2.4 The product of the 6 products (as mentioned in the first sentence) is a^{3}b^{3}c^{3}d^{3} = (a*b*c*d)^{3}. (a*b*c*d)^{3} = 2*3*4*5*6*x where x is the sixth product. 12^{3} = 720x x = 1728/270 = 2.4 Edited 12 Sep 2013 by vinay.singh84 0 Share this post Link to post Share on other sites

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