h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

I want another number - I mean, come on - ending with 76? Well, here goes:

(10-7-2)*6/6*7*6=42

You haven't got a harder one for me?

Traveller in Time trying to find another
"(10-7-2)*6/6*7*6=42
Well apparently your mind is already flexed in calculations; try to find sme more

As far as I know the current recors is combining three digits and reducing it back to 42.

We also have brilliant Researchers making poetry with their calculations < A557246 > and < A597314 > "

Of course I could include the two sixes inside the brackets and substitute them with +6-6 or *sqrt(6*6)...
Maybe I should try to avoid multiplying by 7 and 6 at the end i.e. get to 42 without the first six equalling 1 or 0? Might be a challenge.

I meant of course sqrt(6/6)
Oh well. Would't have happened if I still went to school

It might be easy, but its also fun

Not sure about being a 'brilliant Researcher' though

Got one without using 7*6 at the end:

(1+0)*(7-2)+6*6+7-6=42

By the way - I think using inv tan is cheating because the division of the circle into 360 degrees is arbitrary and not universal.

Well yes, InvTan is kinda cheating, so is adding loads of leading zeroes, but try telling that to the italics .

Its all just creative bookkeeping - not everyone has 8 digits to play around with you know. When Argon0 first made the page, 5 or 6 digits was the norm.

May be interesting to know from which number on it's always possible to reach 42... How about a mathematical proof?

For 1072667, the simplest answer seems to be
10-7+26+6+7

If you're using simple operations, (+-*\, combining digits, power, factorial and sqrt) I think there won't be a number above which all other numbers work, since 10^n will always fail.
maybe it'd work with some other conditions (minimum sum of digits?), but it might just not be a soluble question.

Well, you forgot my last 6...
But factorial saves every '0'
So 10000000 will work:

10*(0!+0!+0!+0!)+0!+0!=42

And has anybody ever used log?

0! does feel like a bit of a cheat.

Even if it is defined to be 1 (for the purposes of certain calculations?) it's not really in the spirit of "Product of all the numbers between 1 and n, inclusive".

I'd say so too - but of course only because it doesn't concern me...

You can't use 7*6 anyway, because of course, 6*9=42, anything else is heresy

Yes, of course I know that. But what do you get if you multiply six by seven?

Well, 6*9 = 42 in base 13

If 6*7 = 42 in base 10, then in base 13 it would be:


1:1
2:2
3:3
...
9:9
10:a
11:b
12:c
13:10
14:11
...
22:19
23:1a
24:2b
25:2c
26:20
...
35:29
36:2a
37:2b
38:2c
39:30
40:31
41:32
42:33

so 6*7=33

Or not.

Oh, I understand base eight well enough - just forget about your thumbs. But base thirteen - I just wouldn't know which toes I'd have to use!