c4countdown

I have been looking into learning the 4 large method.

it says on one guide 75 and 50 turn into 2 and 3.

and if you can solve the solution using 2 and 3 you can solve it using the 50 and 75.

so say 100, 75, 50, 25, 8, 4 turns into 100, 3, 2, 25, 8, 4 and the solution is 902

i can get it by doing 8+4-3=9X100=900 then add 2. but i dont know how to do it using the 75 and 50 apart from I know to subtract 50 at the end just before dividing by 25.

any help on this would be much appreciated.

Cheers

I think it should really say "75 and 50 turn into 3 and 2 under certain circumstances"

I'm not a 4 large specialist (6 small please!) but I don't think your sum can be turned into the solution.
I think it refers more to the following circumstance:

100, 75, 50, 25, 8, 3 turns into 100, 3, 2, 25, 8, 3 and the solution is 902

Here 100*3*3+2=902 becomes
100*3*75=22500
22500+50=22550
22550/25=902

In this case the 75 and the 50 can both use the 25 to become 3 and 2. In your sum they can't.

Someone keener on 4 large can probably give you a clearer explanation of the guide's statement.

I just remembered these formulae as it was far less confusing to me personally and something I could do and replicate the solution quickly:


25 * 50 * 75 / 100 = 937.5 (50/100 can be replaced by 2, 25/100 by 4)

25 * 50 * (75+x) / 100 = 937.5 + 12.5x (x +/- odd)
25 * (50+x) * 75 / 100 = 937.5 + 18.75x (x +/- 2, 6, 10, 14 etc.)
(25+x) * 50 * 75 / 100 = 937.5 + 37.5x (x +/- odd)

(25*75 + x) / (100/50) = 937.5 + 0.5x (x +/- odd)
(50*75 + x) / (100/25) = 937.5 + 0.25x (x +/- 2, 6, 10, 14 etc.)


Those are the simplest ones. There's more things you can do, assuming you were fast enough and could work out in-between answers.

Jerry Humphries web page has all the basic tricks http://jerryh.pwp.blueyonder.co.uk/coun ... umbers.htm.


Here is a slight variation on the method Craig was discussing.

25 goes into 937.5 37.5 times so if you have a 1, say, you can make 900 easily.

((25-1)*50*75)/100 - now with your other number you can order this sum to divide it by 2, 4 or three quarters such as ((25-1)*50+8*75)/25=906.

Or you can add it to one of the numbers before you multiply to get whatever factor that number divides into 900, eg, 75 goes into 900 12 times therefore ((25-1)*(75+8)*50)/100 = 996.

900 doesn't have to be your new root number you can mess about with subtractions from any of 25,50,75 and they will subtract 37.5, 18.75, and 12.5 from the 937.5 root number respectively. You have to bear in mind what factors of 25, 50 and 75 your new root number will take.
eg: ((75*50*(25+2))/100 = 975 so say we have to get 988 using 1,2
We can do ((75+1)*50*(25+2))/100= 988 because the new root number is 975 and 75 goes in at a factor 13 hence any change to the 75 will effect the result by 13 each time

thanks guys