Problem 093

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TripleM
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Re: Problem 093

Post by TripleM »

Yes, see the fourth example where * has been used more than once.
TiptTop
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Re: Problem 093

Post by TiptTop »

TripleM wrote:Yes, see the fourth example where * has been used more than once.
Seems like I did not understand the problem clearly.

It says ".... it is possible to obtain thirty-one different target numbers .... and each of the numbers 1 to 28 can be obtained before encountering the first non-expressible number"

From the rules stated in the problem statement, one can get 31 different numbers with largest number 36. Then what is 28? Please clarify. Thanks.
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jaap
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Re: Problem 093

Post by jaap »

TiptTop wrote:
TripleM wrote:Yes, see the fourth example where * has been used more than once.
Seems like I did not understand the problem clearly.

It says ".... it is possible to obtain thirty-one different target numbers .... and each of the numbers 1 to 28 can be obtained before encountering the first non-expressible number"

From the rules stated in the problem statement, one can get 31 different numbers with largest number 36. Then what is 28? Please clarify. Thanks.
All the numbers 1, 2, ..., 28 are expressible, 29 is the first number that is non-expressible using the set {1,2,3,4}.

The question asks you to find a different set of digits; chosen to make that first non-expressible number is as high as possible.
v6ph1
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Re: Problem 093

Post by v6ph1 »

As stated in the problem description:
You can express all numbers from 1 to 28 - but not 29.
30 = 2*3*(1+4), 32 = 2*(1+3)*4 and 36 are the other three.
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TiptTop
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Re: Problem 093

Post by TiptTop »

Thanks guys.
joachim-ma
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Joined: Fri Mar 11, 2016 1:50 pm

Re: Problem 093

Post by joachim-ma »

Hi, it's my first post in this forum :)
I don't get it, to build the 27 with the set in the example {1,2,3,4}
Could anybody help me?
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sjhillier
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Re: Problem 093

Post by sjhillier »

Think of the divisors of $27$, eg $27=3 \times 9$ . Now how can you reach $3$ and $9$ with the given inputs? I'm sure you'll work it out!
joachim-ma
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Re: Problem 093

Post by joachim-ma »

sjhillier wrote:Think of the divisors of $27$, eg $27=3 \times 9$ . Now how can you reach $3$ and $9$ with the given inputs? I'm sure you'll work it out!
Thank you, I was blind :-D
itteerde
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Re: Problem 093

Post by itteerde »

can someone plz check which case/s are wrong? Although I do not have any code fine tuned to the example it gets the exmple right but not the answer:

{{9},{0,1,2,3}}
{{10},{0,1,2,4}}
{{12},{0,1,2,5}}
{{14},{0,1,2,6}}
{{15},{0,2,3,7}}
{{28},{1,2,3,4}} //example
{{34},{1,2,3,8}}
{{37},{1,2,4,7}}
{{40},{1,2,4,9}}
{{43},{1,2,5,6}}

if that is what you get you are missing terms such as (9-(6-2))*3=15 and your shortcut does not work at all - back to proper tree of operators...
DJohn
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Re: Problem 093

Post by DJohn »

itteerde wrote: Fri May 05, 2017 7:13 pm can someone plz check which case/s are wrong?
This forum isn't supposed to be for hints or checking partial solutions, although it seems that rule can be a bit flexible sometimes. You have correctly understood the problem.

Is it possible to make 8 from {1,3,7,9}?
parasakin
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Joined: Tue Aug 01, 2017 1:47 am

Re: Problem 093

Post by parasakin »

Am I missing something????

1 = 6 + 8 / 9 - 5
2 = 6 + 9 - 5 - 8
3 = 8 + 6 / 9 - 5
4 = 8 + 9 / 6 - 5
5 = 9 + 8 / 6 - 5
6 = 8 + 9 - 5 - 6
7 = 8 * 9 / 6 - 5
8 = 8 + 5 / 6 * 9
9 = 9 + 5 / 6 * 8
10 = 5 + 6 + 8 - 9
11 = 5 + 6 + 8 / 9
12 = 5 + 6 + 9 - 8
13 = 5 + 8 + 6 / 9
14 = 5 + 8 + 9 / 6
15 = 5 + 9 + 8 / 6
16 = 5 + 8 + 9 - 6
17 = 8 + 9 + 5 / 6
18 = 6 + 8 + 9 - 5
19 = (5 + 8) * 9 / 6
20 = 6 + 8 * 9 / 5
21 = (8 - 5) * 9 - 6
22 = 5 * 9 / (8 - 6)
23 = (8 - 6) * 9 + 5
24 = 5 * 6 / 9 * 8
25 = 5 * 8 - 6 - 9
26 = 5 * 6 * 8 / 9
27 = 5 * 6 / 8 * 9
28 = 5 + 6 + 8 + 9
29 = 8 + 5 * 6 - 9
30 = 5 * 6 + 8 / 9
31 = 9 + 5 * 6 - 8
32 = (9 - 5) * 6 + 8
33 = 5 * 6 * 9 / 8
34 = 6 * 8 - 5 - 9
35 = (6 + 9 - 8) * 5
36 = (5 + 9 - 8) * 6
37 = 6 + 5 * 8 - 9
38 = (9 - 5) * 8 + 6
39 = 5 * 8 - 9 / 6
40 = 5 * 8 + 6 / 9
41 = 5 * 8 + 9 / 6
42 = 8 * 9 - 5 * 6
43 = 6 + 5 * 9 - 8
44 = 5 + 6 * 8 - 9
45 = 5 * 9 + 6 / 8
46 = 5 * 9 + 8 / 6
47 = 8 + 9 + 5 * 6
48 = 6 * 8 + 5 / 9
49 = 6 * 8 + 9 / 5
50 = (9 + 8 / 6) * 5
51 = 5 + 6 * 9 - 8
52 = 9 + 6 * 8 - 5
53 = 6 * 9 - 8 / 5
54 = 6 * 9 + 5 / 8
55 = 6 + 9 + 5 * 8
56 = 5 * 9 / 6 * 8
57 = 8 + 6 * 9 - 5
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jaap
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Re: Problem 093

Post by jaap »

parasakin wrote: Tue Aug 01, 2017 1:50 am Am I missing something????

1 = 6 + 8 / 9 - 5
6 + 8 / 9 - 5 is not a whole number (=1.88888....). Many computer languages will round an integer division such as 8/9 down to zero, but that is not a valid answer to the question. Take a good look at the examples given in the problem.
TexasRebel
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Re: Problem 093

Post by TexasRebel »

This is my last problem in the first 100...

I'm having a very difficult time not just guessing the 126 possible answers. :lol:
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RishadanPort
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Re: Problem 093

Post by RishadanPort »

What happens when if a number cannot be evenly divided. -- Does it just treat as int division.

AKA
1 / 2 = 0
or
1 / 2 = 0.5f

?
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RishadanPort
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Re: Problem 093

Post by RishadanPort »

Nevermind. The problem explains it.
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