A Conversation for The Myth Of 42 [(5-3+0+5) * (6+0) = 42]

Help me.... please!

Post 1

cupati

I've tried+I've tried+I've tried+I've tried+I've tried+I've tried+I've tried+I've tried+I've tried+I've tried+I've tried+I've tried+I've tried......

1685894

Any hints?


Help me.... please!

Post 2

Argon0 (50 and feeling it - back for a bit)

You want a Hint? not a solution?

Well here goes...

First Hint: combine the second & third digit to make one number work out how to take 26 away from that...


Help me.... please!

Post 3

Potholer

Another hint.
Starting with the very simple formula of addition of all the individual digits, the sum is 41.
If you can find a way to fiddle with the formula to change the result by 1, you're there.


If a formula is of the form a+b+c+d+e, if you change one + to a -, you can lower the result by twice the number that follows, so you can easily get a result that is lower by 2a, 2b, 2c, 2d, or 2e.

To increase the result of a formula, the simplest 2 ways are by multiplying two adjacent digits, or concatenating them to make a 2-digit number.
If concatenating digits, if you change 'M+N' to 'MN', you increase the result by 9*M (adding M*10, and removing the original single M).
Effectively, that gives a way to increase a result in multiples of 9. For example, if you change '...+ 4+7 +...' to '...+ 47 +...', you increase the result by 36 (4*9), since 47 is 36 higher compared to 11 (4+7).


Help me.... please!

Post 4

cupati

1685894

5+8+9+4= 26

Oh, honestly. It just shows you can't trust the computerised ones.

1*68-5-8-9-4...


Help me.... please!

Post 5

Argon0 (50 and feeling it - back for a bit)

"computerised ones"?

Welcome, Mogget, to the knights of the DoQuaDecagonic Table.


Help me.... please!

Post 6

cupati

http://42.kaitak.org/

is the computerised one.


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