@@peterwilliamskelhorn6675 Have you ever paused it and gotten a calculator, do that for a while and you will be better at it........ Do daredevils start with the hardest stunt, no helmet or net??????? Why should you?
Round 1: (321) 6x9=54+7+3=64x5=320+1=321 (This one had a ton of solutions for a 6 small) Round 2: (352) 50x7=350+2=352 no sense going for a silly solution when it's an ABC answer. Round 3: (329) 8+1=9x9=81x4=324+3+2+=329 (I went 9x8 for 72x4 at first but that came in well too low (288), so 9x9 was my next starting point, took maybe 10 seconds to realiize 80x4=320, another less than baffling 6 small, but one I might have missed if I were on the show with nerves) Round 4: (587) I butchered this one on the clock. Started by thinking the 9 and 3 would be needed for 27 at some point. In retrospect that makes no sense given the last numbers were 87, not 27 or 77, then panicked and went 75+9+50+25=159x3=477+100=577. I'm bad at the 4 large.
I’m surprised that Susie mispronounced the conundrum word! I’ve always heard it pronounced the way it is in its original language 🤷♀️ (ie with the emphasis on the second syllable).
In the first round, I saw Leo's solution, and had a solution similar to Florence's, except that she included a redundant step since you can do it this way: 7 + 5 = 12 12 x 9 = 108 108 - 1 = 107 107 x 3 = 321 I also had this solution: 7 - 1 = 6 9 x 6 x 6 = 324 324 - 3 = 321 In the second numbers round, it was so easy that I tried to use the 30 seconds to come up with a solution that used all the numbers and found this based on the factors (11 x 2^5 = 352): 6 x 3 = 18 50 - 18 = 32 7 x 2 = 14 25 - 14 = 11 32 x 11 = 352 In the third numbers round, I had a solution similar to Richard's, except I got to 330 this way: 8 + 4 = 12 12 x 9 = 108 108 + 2 = 110 110 x 3 = 330 330 - 1 = 329 Since this is the same starting step as my solution for the first problem, if Florence had spotted that 12 x 9 = 108 she might have solved this game as well. And I had this solution from factoring (47 x 7 = 329): 9 + 2 = 11 11 x 4 = 44 44 + 3 = 47 8 - 1 = 7 47 x 7 = 329 In the final round, I saw Florence's solution-it's one of the main techniques for solving four-large puzzles with target numbers in the range of 625-but there are no other solutions.
For the first round, it is immediately noticed that 321 = 3 x 107, so one of the ways is to make 107. Apart from Florence's solution, there is another way to make 107: 7 + 5 = 12 9 x 12 = 108 108 - 1 = 107 107 x 3 = 321 Besides, there is one alternative way as well, apart from Leo's solution that I also got: 5 + 1 = 6 9 x 6 x 6 = 324 324 - 3 = 321 The second round is an easy round. I got the constestants' solution when Rachel declares the target. Besides, there is one alternative way for the target: 50 x 3 = 150 150 + 25 + 7 - 6 = 176 176 x 2 = 352 The third round is similar to the first round, one way is to make 108 + 1 and then multiply by 3: 8 + 4 = 12 12 x 9 = 108 108 + 1 = 109 3 x 109 = 327 327 + 2 = 329 Besides, you may also obtain the target from 328: 3 + 2 = 5 4 x 9 = 36 36 + 5 = 41 41 x 8 = 328 328 + 1 = 329 3 + 2 = 5 5 x 9 = 45 45 - 4 = 41 41 x 8 = 328 328 + 1 = 329 I also got the way by Richard as well. For the final round, I got the same as Florence's solution, but I cannot find any alternative for this one.
Thank you so much Bjoern, and best wishes for this new year to all🎆🌠
For the first time in 2023, Numbers up!
These champion rounds really can't be played much at home. I stink at 6 small, but it's fascinating to see them get the solutions.
+Litigious Society I'm the same with 6 small numbers(I've tried but can't)
@@peterwilliamskelhorn6675 Have you ever paused it and gotten a calculator, do that for a while and you will be better at it........
Do daredevils start with the hardest stunt, no helmet or net??????? Why should you?
For the third round I was trying to get 324 + 5 but timed out before I could get it. It's (8+1)×9×4+2+3.
Round 1: (321) 6x9=54+7+3=64x5=320+1=321 (This one had a ton of solutions for a 6 small)
Round 2: (352) 50x7=350+2=352 no sense going for a silly solution when it's an ABC answer.
Round 3: (329) 8+1=9x9=81x4=324+3+2+=329 (I went 9x8 for 72x4 at first but that came in well too low (288), so 9x9 was my next starting point, took maybe 10 seconds to realiize 80x4=320, another less than baffling 6 small, but one I might have missed if I were on the show with nerves)
Round 4: (587) I butchered this one on the clock. Started by thinking the 9 and 3 would be needed for 27 at some point. In retrospect that makes no sense given the last numbers were 87, not 27 or 77, then panicked and went 75+9+50+25=159x3=477+100=577. I'm bad at the 4 large.
The description still says Vick Hope is in Dictionary Corner but it's Richard Osman this week
50+7-25 = 32
6+3+2 = 11
32 * 11 = 352
And that's how you do with using all the numbers 😂
I’m surprised that Susie mispronounced the conundrum word! I’ve always heard it pronounced the way it is in its original language 🤷♀️ (ie with the emphasis on the second syllable).
Round 1: 9 + 7 + 1 =17 x 6 = 102 + 5 = 107 x 3 = 213
Round 3: 9 x 4 = 36 + 2 + 3 = 41 x 8 = 328 + 1 = 329
7:54 what’s that noise?😂
352
(2*7*25)+(6/3)
587=25×(100-75)-50+9+3
352=(25+7)×(6+3+2)=7×50+2=2×(3×50+25+7-6)
To make 107
9*(6+5) + 1 + 7
In the first round, I saw Leo's solution, and had a solution similar to Florence's, except that she included a redundant step since you can do it this way:
7 + 5 = 12
12 x 9 = 108
108 - 1 = 107
107 x 3 = 321
I also had this solution:
7 - 1 = 6
9 x 6 x 6 = 324
324 - 3 = 321
In the second numbers round, it was so easy that I tried to use the 30 seconds to come up with a solution that used all the numbers and found this based on the factors (11 x 2^5 = 352):
6 x 3 = 18
50 - 18 = 32
7 x 2 = 14
25 - 14 = 11
32 x 11 = 352
In the third numbers round, I had a solution similar to Richard's, except I got to 330 this way:
8 + 4 = 12
12 x 9 = 108
108 + 2 = 110
110 x 3 = 330
330 - 1 = 329
Since this is the same starting step as my solution for the first problem, if Florence had spotted that 12 x 9 = 108 she might have solved this game as well.
And I had this solution from factoring (47 x 7 = 329):
9 + 2 = 11
11 x 4 = 44
44 + 3 = 47
8 - 1 = 7
47 x 7 = 329
In the final round, I saw Florence's solution-it's one of the main techniques for solving four-large puzzles with target numbers in the range of 625-but there are no other solutions.
352
50 x 7 = 350
350 + 2 = 352
For the first round, it is immediately noticed that 321 = 3 x 107, so one of the ways is to make 107. Apart from Florence's solution, there is another way to make 107:
7 + 5 = 12
9 x 12 = 108
108 - 1 = 107
107 x 3 = 321
Besides, there is one alternative way as well, apart from Leo's solution that I also got:
5 + 1 = 6
9 x 6 x 6 = 324
324 - 3 = 321
The second round is an easy round. I got the constestants' solution when Rachel declares the target.
Besides, there is one alternative way for the target:
50 x 3 = 150
150 + 25 + 7 - 6 = 176
176 x 2 = 352
The third round is similar to the first round, one way is to make 108 + 1 and then multiply by 3:
8 + 4 = 12
12 x 9 = 108
108 + 1 = 109
3 x 109 = 327
327 + 2 = 329
Besides, you may also obtain the target from 328:
3 + 2 = 5
4 x 9 = 36
36 + 5 = 41
41 x 8 = 328
328 + 1 = 329
3 + 2 = 5
5 x 9 = 45
45 - 4 = 41
41 x 8 = 328
328 + 1 = 329
I also got the way by Richard as well.
For the final round, I got the same as Florence's solution, but I cannot find any alternative for this one.
4×9=36-3=33 2+8=10×33=330-1=329
329
9 x 8 = 72
72 x 4 = 328
328 + 1 = 329
321
9 x 7 = 63
63 x 5 = 315
315 + 6 = 321
329=(4×9-3)×(8+2)-1
352 25-3=22 50+6=56÷7=8×2=16×22=352
321 7×5×9=315+6=321
Not relevant; but I do like Rachel’s dress today.
[4 × (1 + 8) × 9] + 2 + 3 = 329.
Anyone in here?
329: ((8+1)x9+2)x4-3
352
50 x 7 = 350
350 + 2 = 352