Countdown Game Show - Number Rounds (3 August 2022)
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- Опубліковано 17 жов 2024
- Colin Murray and Rachel Riley host as contestants race against the clock to pit their wits against vowels, consonants and numbers. Susie Dent in Dictionary Corner.
Background:
Countdown is Channel 4’s longest running show and the first show ever broadcast when it launched 40 years ago, in November 1982. The original version of Countdown is the French television programme « Des chiffres et des lettres » (literally "numbers and letters"), where the numbers round is referred to as « Le compte est bon » ("the total is right"). DCDL debuted in 1965 as « Le mot le plus long » ("the longest word") using letters only, and reached its present format in 1972. This year marks the shows 50th anniversary.
Spin-off:
"8 Out of 10 Cats does Countdown" is hosted by Jimmy Carr, with Rachel Riley and Susie Dent assuming their Countdown roles. The formation of the crossover with "8 Out if 10 Cats" began as a series of specials, the first of which was broadcast on 2 January 2012, when Channel 4 orchestrated a special "mash-up night", merging two shows to form a special edition of the pair, as part of its 30th-anniversary celebrations ten years ago.
Damn, several of these were difficult, I usually do pretty well with numbers but I struggled with some of these!
The way Rachel solved the 781 round was so impressive and she deserves every penny she earns
Yeah, there was no way I was going to get anywhere close to that! The "product of two primes" targets are the trickiest to solve especially with 6 small numbers.
Having said that, I do like it when people pick 6 little ones, as a nice challenge.
On some other episode Rachel "spilled a secret" to Colin that if you have a 3 digit number, then the sum of outside digits equals the middle digit, it is divisible by 11. So I have spotted that, but still didn't make the 71 in the 30 seconds, probably because I got my 11 as 5+6, instead of 7+4, and tried different combinations of 7, 8, 4 and 7
@@TheSpotify95 honestly when you have six smalls, ALWAYS factor it. I spent like 40 seconds going nowhere and then found it instantly after finally remembering it's 71 x 11
You're saying they don't run a computer program to solve the tough ones?
For the 3rd found I had:
1+9=10
10*4=40
75+40=115
115*5=575
For 575:
75 × 7 = 525
5 × (1 + 9) = 50
525 + 50 = 575
His 784 was correct. He was heading for 28x4x7 but Rachel mistook a 4 he made for the 4 that was already on the board. It wouldn't have changed the outcome of the game though.
I think he is correct that he used a 7 twice.
His solution was
7 - 6 = 1 (using 7,6)
5 - 1 = 4 (using 7,6,5)
4 x 7 = 28 (7,6,5,7)
28 x 4 = 112 (7,6,5,7,4 leaving the 8)
112 x 7(!) = 784
560:
6 * 100 = 600
4 * 10 = 40
600 - 40 = 460
6 * 75 = 450
450 + 100 + 10 = 560
100 / 4 = 25
75 - 25 = 50
50 + 6 = 56
56 * 10 = 560
575:
9 + 1 = 10
10 * 4 = 40
75 + 40 = 115
115 * 5 = 575
5 * 3 = 15
75 - 15 + 4 = 64
64 * 9 = 576
576 - 1 = 575
9 + 1 = 10
10 * 5 = 50
4 + 3 = 7
7 * 75 = 525
525 + 50 = 575
575:
9 + 1 = 10
10 x 4 = 40
75 + 40 = 115
115 x 5 = 575
483: (50-1)x(8+2)-7
560: (9-4)x100+6x10
575: (75-3)x(9-1)+4-5
483:
50 + 2 = 52
52 x 9 = 468
8 + 7 = 15
468 + 15 = 483
483:
8 + 2 = 10
50 * 10 = 500
9 + 7 + 1 = 17
500 - 17 = 483
9 + 1 = 10
50 * 10 = 500
8 + 7 + 2 = 17
500 - 17 = 483
50 + 2 = 52
52 * 9 = 468
8 + 7 = 15
468 + 15 = 483
50 + 8 + 2 + 9 = 69
69 * 7 = 483
50 - 1 = 49
49 * 10 = 490
490 - 7 = 483
560:
100 x 6 = 600
10 x 4 = 40
600 - 40 = 560
560
75 - 9 = 66
66 x 10 = 660
660 - 100 = 560
Another method for 560:
75 x 6 = 450
100 + 10 = 110
450 + 110 = 560
I had 75 - 9 - 6 - 4 = 56 x 10
560=6×100-4×10=(75-10-9)×(4+6)
75×(4+3)+(9+1)×5=575
575 9-1=8 8× 75
483 2×9=18+1=19+50=69×7=483
In the first numbers round, I saw both contestants' solutions, however, I also factored the target (23 x 7 x 3 = 483) for this solution:
9 x 2 = 18
50 + 18 + 1 = 69
69 x 7 = 483
And I had this one as well:
50 + 2 = 52
52 x 9 = 468
468 + 8 + 7 = 483
In the second numbers round, another easy solution similar to Quinn's is to approach the target from above:
75 - 9 = 66
66 x 10 = 660
660 - 100 = 560
And factoring the target (7 x 5 x 2^4 = 560) yields this solution:
100 + 9 + 6 = 115
115 - 75 = 40
10 + 4 = 14
40 x 14 = 560
In the third round, I got this from factoring the target (23 x 5^2 = 575):
9 + 1 = 10
10 x 4 = 40
75 + 40 = 115
115 x 5 = 575
And also this solution:
5 x 3 = 15
75 - 15 = 60
60 + 4 = 64
64 x 9 = 576
576 - 1 = 575
And in the final numbers round, factoring was absolutely indispensable, but Rachel had the only solution so it's not worth commenting here. However, I was pleased that I beat her to the solution because I got it in the 30 seconds. With solutions that are the product of two two-digit primes in a six-small selection, you rarely get more than one solution because you have to commit at least two of the given numbers to the creation of each prime.
(50+2)×9+8+7=483
For 575 I thought this was fairly obvious:
75(4+3)+5(9+1)
525+50
575
I did 5[4(9+1)+75].
483=50×(9+1)-8-7-2