## Problem 093

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ssp
Posts: 2
Joined: Sun Nov 11, 2007 7:49 am

### Problem 093

Hello. I've suppose it's easy problem, and i solve it, but my algorithm seems wrong:
it can't found equation for {1,2,3,4} set where answer is 22. Can anyone post those equation?
(Problem statement says it exists).
stijn263
Posts: 1505
Joined: Sat Sep 15, 2007 11:57 pm
Location: Netherlands

### Re: Problem 93 hint

2*(3*4-1) = 22
LarryBlake
Posts: 100
Joined: Sat Aug 29, 2009 8:49 pm

### Re: Problem 093

Well, I do get 28 for the example (digits 1, 2, 3, 4). But I also get the wrong overall answer.

Could someone please post another digit combination and its count, so that I can try to figure out where I'm going wrong? Thanks.
LarryBlake
Posts: 100
Joined: Sat Aug 29, 2009 8:49 pm

### Re: Problem 093

Does this count:
(6 / 9) * 3 = 2

I wouldn't find that, because I reject 6/9 as not an integer. (Yes, I know I could have rewritten this as (3 * 6) / 9. Maybe it doesn't matter.)
daniel.is.fischer
Posts: 2400
Joined: Sun Sep 02, 2007 11:15 pm
Location: Bremen, Germany

### Re: Problem 093

Yes, it does count. Intermediate results need not be integers, only the final result must be an integer.
Il faut respecter la montagne -- c'est pourquoi les gypa&egrave;tes sont l&agrave;.
LarryBlake
Posts: 100
Joined: Sat Aug 29, 2009 8:49 pm

### Re: Problem 093

Okay, thanks. Could you please also post another combination and its count?
LarryBlake
Posts: 100
Joined: Sat Aug 29, 2009 8:49 pm

### Re: Problem 093

Anyone?
zwuupeape
Posts: 189
Joined: Tue Jun 09, 2009 6:11 pm

### Re: Problem 093

Yeah: 2679 gives 21 consecutive numbers, and 2389 gives 33.
LarryBlake
Posts: 100
Joined: Sat Aug 29, 2009 8:49 pm

### Re: Problem 093

Thank you.
bobfin
Posts: 9
Joined: Thu Oct 22, 2009 7:06 am

### Problem 93

I have been struggling with problem 93 and think I have solution but my submitted solution is not correct. My method correctly finds that the digit set {1,2,3,4} yields a run of consecutive numbers 1 .. 28. Could anyone confirm the result for the set {4,5,7,8} is a sequence of 26 consecutive numbers. I hope this note is not a spoiler.
Fede
Posts: 1
Joined: Tue Mar 16, 2010 10:57 am

### Re: Problem 93

Hi,
my probabilistic algo gives me:
Expand
consecutives for [4, 5, 7, 8].

It may not be exactly that but if it's not it should be pretty close.
stijn263
Posts: 1505
Joined: Sat Sep 15, 2007 11:57 pm
Location: Netherlands

### Re: Problem 93

Problem 93 (View Problem)

I also get 17 for {4,5,7,8}. What solution does your program give for {4,5,7,8} --> 18 ?
bobfin
Posts: 9
Joined: Thu Oct 22, 2009 7:06 am

### Re: Problem 093

Thanks for the feedback. I have now located my problem (integer arithmetic) and have successfully completed the problem.
Tridecagon
Posts: 2
Joined: Sat Aug 07, 2010 4:26 pm
Location: Rome, Italy

### Re: Problem 093

Hi, could someone post the equation resulting in 15 for [2,6,7,9]? My algorithm doesn't find it and I can't figure out why.
Thanks!
harryh
Posts: 2091
Joined: Tue Aug 22, 2006 9:33 pm
Location: Thessaloniki, Greece

### Re: Problem 093

( 7 - ( 9 / 2 ) ) * 6
rayfil
Administrator
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Location: Quebec, Canada
Contact:

### Re: Problem 093

Would the following also be acceptable to you?

(9-6)*(7-2)
When you assume something, you risk being wrong half the time.
Tridecagon
Posts: 2
Joined: Sat Aug 07, 2010 4:26 pm
Location: Rome, Italy

### Re: Problem 093

Thanks for your help, I've found my mistake and solved the problem.
GenePeer
Posts: 112
Joined: Sat Apr 03, 2010 1:14 pm
Contact:

### Re: Problem 093

zwuupeape wrote:Yeah: 2679 gives 21 consecutive numbers, and 2389 gives 33.
My final solution was correct and I got the same value for 2679 but not for 2389 How do you evaluate 15 with [2,3,8,9]?
TripleM
Posts: 382
Joined: Fri Sep 12, 2008 3:31 am

### Re: Problem 093

3 * (9 - 8/2)
GenePeer
Posts: 112
Joined: Sat Apr 03, 2010 1:14 pm
Contact:

### Re: Problem 093

Thanks. I got lucky, my program was skipping a few combinations. I fixed the problem though.