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 From pie to numbers (Posted on 2015-12-27)
Let us use the first 3 digits of PI, i.e. 3,1,4 to create numbers from 1 to 50, using only basic math. operators.
Not all the digits must be used, but none can be repeated.

Examples (out of this range): 52= 13*4; 65=4^3+1; 84=14*3!

Only few numbers cannot be achieved.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 solutions for most numbers | Comment 1 of 14
From the examples, an inference is made that exponents and factorials are to be considered basic math operators, though these two type of operators are not considered by some experts to be basic. Concatenation of the numerals to represent numbers, pre-operation*, also appears to be permitted. (An assumption is made that post-operation* concatenation is not permitted).

Besides concatenation, and limited to arithmetic operations - addition, subtraction, multiplication, division - with the additional operations of exponentiation and factorials I was able to find solutions for all but six numbers (32, 39, 45, 46, 49, 50). Many of the numbers have more than one solution, yet I have only listed one for each:
1) 1
2) 3 - 1
3) 3
4) 4
5) 4 + 1
6) 3 + 4 - 1
7) 3 + 4
8) 3 + 4 + 1
9) 3! + 4 - 1
10) 3! + 4
11) 3! + 4 + 1
12) 3 × 4
13) 13
14) 14
15) (4 + 1) × 3
16) 4 × (3 + 1)
17) 13 + 4
18) 4! - 3!
19) 4! - 3! + 1
20) 4! - 3 - 1
21) 4! - 3
22) 4! - 3 + 1
23) 4! - 1
24) 4!
25) 4! + 1
26) 4! + 3 - 1
27) 4! + 3
28) 4! + 3 + 1
29) 4! + 3! - 1
30) 4! + 3!
31) 4! + 3! + 1
32) ?
33) 34 - 1
34) 34
35) 34 + 1
36) (4 - 1)! × 3!
37) 4! + 13
38) 41 - 3
39) ?
40) (4 + 1)! ÷ 3
41) 41
42) 43 - 1
43) 43
44) 43 + 1
45) ?
46) ?
47) 41 + 3!
48) 4! + (3 + 1)!
49) ?
50) ?

*Operation is defined here as one of the binary mathematical operations: a + b, a - b, a × b, a ÷ b, a^b; or the unary operation: a!. As an operation, concatenation may be considered a string operation, though strings of numerals are yet often considered numbers -- as understood in this problem.

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Solutions for the six numbers where the "basic" operations were not found, yet -- though limited to the implied "given" mathematical operator symbols -- using the "non-basic" operations subfactorial { !a } and multifactorials { a!!, a!!!! } in addition to the "basic" operations:
32 = (!3)^(4 + 1)
39 = 4! + (3! - 1)!!
45 = !(4 + 1) + !(!3)
46 = !(4 + 1) + !3
49 = ((3!)!!!! × 4) + 1
50 = !(4 + 1) + 3!

Edited on December 28, 2015, 9:00 am
 Posted by Dej Mar on 2015-12-27 13:07:04

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