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The highest intermediate total in a numbers game

Posted: Sat Jan 22, 2011 11:02 pm
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.

Re: The highest intermediate total in a numbers game

Posted: Sat Jan 22, 2011 11:11 pm
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

Re: The highest intermediate total in a numbers game

Posted: Sat Jan 22, 2011 11:43 pm
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.

Re: The highest intermediate total in a numbers game

Posted: Sun Jan 23, 2011 10:39 am
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

Re: The highest intermediate total in a numbers game

Posted: Sun Jan 23, 2011 11:22 am
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.

Re: The highest intermediate total in a numbers game

Posted: Sun Jan 23, 2011 11:24 am
by Rhys Benjamin
100 6 3 75 25 50 --> 952

You can got up to 43,850(?)

Re: The highest intermediate total in a numbers game

Posted: Sun Jan 23, 2011 11:36 am
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?