For the first round, it looks tricky and I found one way by making 11 without 3 and 100, which is possible since 11 = 2 x 7 - 3: 50 x 7 = 350 350 - 75 = 275 275 / 25 = 11 3 x 100 = 300 300 + 11 = 311 Actually, Adam's solution is just (100 + 3) x 3 + 2 = 311. For the second round, I got the solution 254 = 127 x 2 immediately: 100 + 25 + 2 = 127 6 - 4 = 2 127 x 2 = 254 Besides, there is another simple solution: 25 x 6 = 150 150 + 100 + 4 = 254 For the third round, I think Rachel's solution is the only possible one. The last round is easy. I got the same solution as Adam's once the target was announced.
In the first round, Adam scooped my solution. But there is another similar way of getting 311 with those numbers: 100 x 75 = 7500 50 x 7 = 350 7500 + 350 = 7850 7850 / 25 = 314 314 - 3 = 311 And I also had these solutions: 75 x 3 = 225 225 + 100 = 325 50 / 25 = 2 7 x 2 = 14 325 - 14 = 311 100 x 7 = 700 700 - 75 - 3 = 622 50 / 25 = 2 622 / 2 = 311 You can also go into the tens of thousands by multiplying by 25 before dividing by 50, but there's no real point to it. In the second round, I saw both contestants' solutions, but also had these: 25 x 6 = 150 150 + 100 + 4 = 254 100 + 25 + 2 = 127 6 - 4 = 2 127 x 2 = 254 25 - 4 = 21 21 x 6 = 126 126 + 1 = 127 127 x 2 = 254 In the third round, I saw Rachel's solution because 100 x 25 / 3, 100 x 50 / 6, and 100 x 75 / 9 are all ways of getting to solutions in the vicinity of 833. However, there are no other solutions. In the fourth round, I saw both contestants' solutions and also had these alternatives: 100 + 25 + 5 = 130 130 - 8 = 122 122 x 5 = 610 610 + 7 = 617 100 - 25 = 75 75 x 8 = 600 600 + 7 + 5 + 5 = 617
After seeing Adam's solution for the first round, I thought he could get the 821 as well, but it seems that he's just a specialist of the four-large. Here's my solution, found in time unlike Rachel ;) 50 x 100 = 5000 25 x 3 = 75 5000 - 75 = 4925 4925 + 1 = 4926 *4926 / 6 = 821*
Colin loves Countdown so much. I'm so glad he's presenting now.
Such an exciting ending. Thanks for keeping in the conundrum and contestants reflections.
3:25 - Oh Rachel lol. Love this reaction.
75*3=225, +100=325, 50/25=2, *7=14, 325-14=311
Exactly the way I saw it.
1. 3x75+100-(7x50/25)=311
2. 6x25+100+4=254
3. (100+50)x6-(25+1)x3=822
4. 25x5x5-8=617
5:58
((50x100)-(3x25)+1)/6 = 821
For the first round, it looks tricky and I found one way by making 11 without 3 and 100, which is possible since 11 = 2 x 7 - 3:
50 x 7 = 350
350 - 75 = 275
275 / 25 = 11
3 x 100 = 300
300 + 11 = 311
Actually, Adam's solution is just (100 + 3) x 3 + 2 = 311.
For the second round, I got the solution 254 = 127 x 2 immediately:
100 + 25 + 2 = 127
6 - 4 = 2
127 x 2 = 254
Besides, there is another simple solution:
25 x 6 = 150
150 + 100 + 4 = 254
For the third round, I think Rachel's solution is the only possible one.
The last round is easy. I got the same solution as Adam's once the target was announced.
254:
100 / 2 = 50 (or 25 * 2 = 50)
6 - 1 = 5
50 * 5 = 250
250 + 4 = 254
25 * 6 = 150
150 + 100 = 250
250 + 4 = 254
6 + 1 = 7
7 * 2 = 14
25 * 14 = 350
350 - 100 = 250
250 + 4 = 254
100 + 25 + 2 = 127
6 - 4 = 2
127 * 2 = 254
100 + 1 = 101
101 * 4 = 404
25 * 6 = 150
404 - 150 = 254
617:
100 - 25 = 75
75 * 8 = 600
5 + 5 + 7 = 17
600 + 17 = 617
5 / 5 = 1
7 - 1 = 6
6 * 100 = 600
25 - 8 = 17
600 + 17 = 617
100 + 25 = 125
8 - 5 = 3
125 - 3 = 122
122 * 5 = 610
610 + 7 = 617
Shame that Richard didn't host at least 1 round of the numbers round.
Would love Richard to take over. Way better than Colin.
In the first round, Adam scooped my solution. But there is another similar way of getting 311 with those numbers:
100 x 75 = 7500
50 x 7 = 350
7500 + 350 = 7850
7850 / 25 = 314
314 - 3 = 311
And I also had these solutions:
75 x 3 = 225
225 + 100 = 325
50 / 25 = 2
7 x 2 = 14
325 - 14 = 311
100 x 7 = 700
700 - 75 - 3 = 622
50 / 25 = 2
622 / 2 = 311
You can also go into the tens of thousands by multiplying by 25 before dividing by 50, but there's no real point to it.
In the second round, I saw both contestants' solutions, but also had these:
25 x 6 = 150
150 + 100 + 4 = 254
100 + 25 + 2 = 127
6 - 4 = 2
127 x 2 = 254
25 - 4 = 21
21 x 6 = 126
126 + 1 = 127
127 x 2 = 254
In the third round, I saw Rachel's solution because 100 x 25 / 3, 100 x 50 / 6, and 100 x 75 / 9 are all ways of getting to solutions in the vicinity of 833. However, there are no other solutions.
In the fourth round, I saw both contestants' solutions and also had these alternatives:
100 + 25 + 5 = 130
130 - 8 = 122
122 x 5 = 610
610 + 7 = 617
100 - 25 = 75
75 x 8 = 600
600 + 7 + 5 + 5 = 617
for the final round:
5/5=1
7-1=6
6*100=600
600+25-8=617
617
100 x 5 = 500
25 x 5 = 125
500 + 125 = 625
625 - 8 = 617
[(100 × 7) - 75 - 3] / (50 / 25) = 311
After seeing Adam's solution for the first round, I thought he could get the 821 as well, but it seems that he's just a specialist of the four-large. Here's my solution, found in time unlike Rachel ;)
50 x 100 = 5000
25 x 3 = 75
5000 - 75 = 4925
4925 + 1 = 4926
*4926 / 6 = 821*
254=100+25×6+4=(100+25+2)×(6-4)
254
100 + 25 = 125
125 x 2 = 250
250 + 4 = 254
311:
100 * 3 = 300
50 / 25 = 2
75 + 2 = 77
77 / 7 = 11
300 + 11 = 311
100 * 3 = 300
7 * 50 = 350
350 - 75 = 275
275 / 25 = 11
300 + 11 = 311
75 * 3 = 225
225 + 100 = 325
50 / 25 = 2
7 * 2 = 14
325 - 14 = 311
75 * 7 = 525
525 + 100 = 625
625 - 3 = 622
50 / 25 = 2
622 / 2 = 311
Oh no! Tom has lost!
:(