In the first numbers round, I saw Paul's solution and also had these solutions: 50 + 5 = 55 10 - 3 - 1 = 6 55 x 6 = 330 330 + 9 = 339 5 + 1 = 6 10 x 6 = 60 60 + 50 = 110 110 x 3 = 330 330 + 9 = 339 5 - 3 = 2 50 / 2 = 25 25 + 9 = 34 34 x 10 = 340 340 - 1 = 339 In the second round, I again saw Paul's solution but also had these: 75 x 2 = 150 150 / 3 = 50 50 + 5 + 1 = 56 56 x 7 = 392 75 - 7 = 68 68 x 2 = 136 136 - 5 = 131 131 x 3 = 393 393 - 1 = 392 In the third round, I was surprised that neither contestant got this solution: 10 x 9 = 90 4 x 3 = 12 90 + 12 = 102 102 x 6 = 612 I also factored like Rachel did for these solutions: 10 x 3 = 30 30 + 6 = 36 4 x 2 = 8 9 + 8 = 17 36 x 17 = 612 10 x 3 = 30 30 + 4 = 34 34 x 9 x 2 = 612 And here I had an alternative method of getting to 68 10 x 6 = 60 4 x 2 = 8 60 + 8 = 68 68 x 9 = 612 With the 10 present and with such a factorizeable number, this six-small was not too difficult. In the final round, I yet again saw Paul's solution, but I also had this solution because I always try to solve low-target 1-large games as "five small": 7 + 6 = 13 13 x 9 = 117 117 + 4 = 121 And I also wanted to do something with the fact that this is a perfect square, and so I found this solution: 75 - 9 = 66 66 / 6 = 11 7 + 4 = 11 11 x 11 = 121
For the first round I did: 10-1-3=6 50+5=55 55*6=330 330+9 =339 For the second round I did like the challenger: 75*5=375 7*2=15 375+14+3=392 For the third round I did: 9*6=54 54+3+4=61 61*10=610 610+2=612 For the last round I did: 7*6=42 75+42+4=121
1. (1+5)x10+50=110x3+9=339 or (50+5)x(10-1-3)+9=339 or 50/5+1=11x10x3+9=339 2. (75+2)x5+7=392 or 75+(3x7)+3=98x(1+3)=392 or 75/3=25x2+1+5=56x7=392 3. (6x9-2)x3x4=612 or (9x10+3x4)x6=612 or ((6+9)x10+3)x4=612 3. 75+(4x9)+7+9-6=612 4. 75+9+9+(4x7)=121
612 was really pretty easy for nobody to get too close. Rachel's way was at best the 3rd easiest way to do it. Also could have done: ((10x9)+(4x3))x6 = 612 [this is just 6x102] or ((9x6)+(4+3))x10+2=612 [this is just 61x10+2]
Colin even hinted at a 3rd easy solution. 612 is an easy multiple of 6, 6x102. You can easily get 102 while saving the 6 by doing (9x10)+(4x3). Oddly easy miss for a player as strong as Paul.
Do we have to use all the numbers? Because I can get 392 on the second one without using the 7 or the 1 75,2,3,5,7,1 75+3=78 78*5=390 390+2=392 7 and 1 not used