Using a definitional approach, 6 and 9 are both multiples of 3, so it is clear that in order to form any large number that is not a multiple of 3, you must use 20. A single use of 20 will create the ability to make a number which when divided by 3 will leave a remainder of 2. A second use of 20 will create the ability to make a number which when divided by 3 will leave a remainder of 1. There is no need to ever use 20 for a third time because a remainder of 0 is achieveable with 6 and 9 alone. Therefore, the largest number that you can get from the factors 6, 9 and 20 must necessarily be a number that when divided by 3 will leave a remainder of 1, but is yet small enough to make it impossible to use 20 twice followed by a combination of 6 and 9. Obviously 40 itself is achieveable with 2 sets of 20, so we can go larger. Looking at numbers larger that 40 which when divided by 3 will give remainders of 1, we have 43, 46, 49. Out of these three, only 43 is unachieveable because 3 < 6.
Posted on: Thu, 01 Aug 2013 20:13:17 +0000