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        riddles >> easy >> 6-Digit Number
        
(Message started by: THUDandBLUNDER on Jun 27^th, 2004, 4:13pm)

Title: 6-Digit Number
Post by THUDandBLUNDER on Jun 27^th, 2004, 4:13pm

Can you find a 6-digit number without writing a computer program such that:

(i) The first 3 digits minus last 3 digits = 665.
(ii) Within the number there is a 3 to the left of a 1.
(iii) There is a 7 to the right of a 9.
(iv) There is a 5 to the left of a 3.
(v) There is a 0.

Title: Re: 6-Digit Number
Post by Grimbal on Jun 27^th, 2004, 4:26pm

Yes.

Title: Re: 6-Digit Number
Post by THUDandBLUNDER on Jun 27^th, 2004, 4:30pm

That's that settled then.

::)

Title: Re: 6-Digit Number
Post by Noke Lieu on Jun 27^th, 2004, 6:27pm

Nice to see honour in action.

this got me to thinking, and I don't think it really warrants its own thread, so I'll tag it on here. (and not sure I've seen it)

A number, starting with two abritary digits, such that these two digits can be used to produce the third digit. Then the second and third used to produce the fourth.....
can only use each digit once. AND we're only staying decimal. Except maybe for an extension...

eg: 3257 (3+2=5; 5+2=7) or 8426

What is the biggest number that satisfies these requirements?

Title: Re: 6-Digit Number
Post by Leonid Broukhis on Jun 27^th, 2004, 8:49pm

Noke,
I got ::[hide]86235[/hide]::

Title: Re: 6-Digit Number
Post by Noke Lieu on Jun 28^th, 2004, 3:53am

9 pm, and still at work. Better go home before they switch the alarms on.

Leonid Broukhis, mine's bigger. ::)

Title: Re: 6-Digit Number
Post by Grimbal on Jun 28^th, 2004, 6:35am

I got 2 6-digit numbers.

Title: Re: 6-Digit Number
Post by Leonid Broukhis on Jun 28^th, 2004, 9:52am

What operations are allowed? Can division be inexact?

Title: Re: 6-Digit Number
Post by Grimbal on Jun 28^th, 2004, 10:09am

Ahaaa, but with integer operations I get a much longer one. :)

::[hide]972365410
9-7=2, 7/2=3, 2*3=6, 3^6=5, 6&5=4, 5-4=1, 4%1=0
(^ is xor)[/hide]::

Title: Re: 6-Digit Number
Post by Leonid Broukhis on Jun 28^th, 2004, 11:21am

So much for incomplete specifications. I limited myself to addition, subtraction, multiplication and division with no remainder.

Title: Re: 6-Digit Number
Post by Noke Lieu on Jun 28^th, 2004, 11:03pm

Yeah, I'm with you, Leonid Broukhis. Multiplying, dividing, subtracting and addition. No remainders.

Unsurprisingly, am stuck at 5 digits in decimal....

Title: Re: 6-Digit Number
Post by Eigenray on Jun 29^th, 2004, 6:05am

There are 2 of length 6, 16 of length 5, 40 of length 4, and 72 of length 3 . . .

Title: Re: 6-Digit Number
Post by Grimbal on Jun 29^th, 2004, 8:33am

Same number of solutions here: 2x6, 16x5, 40x4, 72x3, 90x2 and 10x1.
And surprise, surprise, one 6-digit number is just the reverse of the other! :o

Title: Re: 6-Digit Number
Post by Leonid Broukhis on Jun 29^th, 2004, 10:03am

I've made a mistake in my hand-drawn graph. Perl rules!