The highest intermediate total in a numbers game

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Gavin Chipper
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The highest intermediate total in a numbers game

Post by Gavin Chipper »

I think Paul "Tranmere" Thomson posted a similar thing the best part of a decade ago, but I was wondering:

1. What's the highest intermediate total ever used in a game by a contestant?
2. What's the highest total you would ever have to reach to get the solution spot on?
3. What's the highest total you can reach to get a solution spot on? I don't think it counts if you do something like *100/50 because that's the same as *(100/50), but you can go inefficient routes.
4. Other related questions that I haven't covered.

I devised a game yesterday when I was bored and according to WTP it only has one solution, which involves going very high, but you might think of it as a bit of a cheat.

996 - 100, 75, 50, 5, 3, 3

50*5 = 250
250*3 = 750
750-3 = 747
747*100 = 74700
74700/75 = 996

Obviously you have to multiply by 100 first because you can't do 747/75 or 100/75, but you might think that this is a bit sneaky.
Gavin Chipper
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Re: The highest intermediate total in a numbers game

Post by Gavin Chipper »

You can go even higher with this next one, but there are other routes:

938 - 100, 75, 50, 25, 5, 10

75*25 = 1875
1875*50 = 93750
5*10 = 50
93750+50 = 93800
93800/100 = 938
Gavin Chipper
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Re: The highest intermediate total in a numbers game

Post by Gavin Chipper »

999 - 100, 75, 50, 25, 3, 7

50+3 = 53
53*25 = 1325
1325+7 = 1332
1332*75 = 99900
99900/100 = 999


More than one solution though.
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JimBentley
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Re: The highest intermediate total in a numbers game

Post by JimBentley »

900 - 100, 75, 50, 25, 9, 8
(very inefficient route)

9 x 8 x 25 x 75 = 135,000
100 + 50 = 150
135,000 / 150 = 900
Gavin Chipper
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Re: The highest intermediate total in a numbers game

Post by Gavin Chipper »

JimBentley wrote:900 - 100, 75, 50, 25, 9, 8
(very inefficient route)

9 x 8 x 25 x 75 = 135,000
100 + 50 = 150
135,000 / 150 = 900
The only problem I'd have with that is that you can do the multiplication/division in a different order and avoid going so high. But I like your thinking.
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Rhys Benjamin
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Re: The highest intermediate total in a numbers game

Post by Rhys Benjamin »

100 6 3 75 25 50 --> 952

You can got up to 43,850(?)
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JimBentley
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Re: The highest intermediate total in a numbers game

Post by JimBentley »

Gavin Chipper wrote:
JimBentley wrote:900 - 100, 75, 50, 25, 9, 8
(very inefficient route)

9 x 8 x 25 x 75 = 135,000
100 + 50 = 150
135,000 / 150 = 900
The only problem I'd have with that is that you can do the multiplication/division in a different order and avoid going so high. But I like your thinking.
Heh yeah, knew you'd say that. Is that true for all intermediate stages over 100,000?
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