Problems & Puzzles: Puzzles

 

Puzzle 864. 10958, the only hole...

Brazilian mathematician Inder J. Taneja has found (Jan 2014) a way to render every number from 1 to 11,111 by starting with either of these ordered strings.

1234567869 (increasing 1->9) and 987654321 (decreasing 9->1

applying any of the following:

a) arithmetical operations permitted: addition, subtraction, multiplication, division, and exponentiation.
b) string operation permitted: concatenation.
c) auxiliary symbols permitted: brackets "(" and ")".


Examples:

10957 = (1 + 2)^(3+4) × 5 - 67 + 89.

and

11038 = (9 × 8 × 7 + 6^5) × 4/3 - 2 × 1.

Unexpectedly, there’s just one "hole" in this range: 10958. There doesn’t seem to be a way to render 10958 from the increasing sequence.

Here is his Taneja's paper entitled "Crazy Sequential Representation: Numbers from 0 to 11111 in terms of Increasing and Decreasing Orders of 1 to 9"

Q1. Find the representation of 10958 for the increasing sequence using only the permitted operations and auxiliary symbols listed above

Q2. Find the representation of
10958 for the increasing sequence using only the permitted operations and auxiliary symbols listed above AND the minimal additional arithmetical operators or string operations or auxiliary symbols as the suggested by Inder J. Taneja: factorial, decimal, square root, etc...


Contributions came from Emmanuel Vantieghem and Inder J. Taneja

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Emmanuel wrote:

The first question will be the most difficult. The second question is much  easier : 10958 = (1-2+3)*(456+7!-8-9). But, I continue my search.

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Inder's answer to a question that I send to him by email, the following:

- I wonder if the hole currently (Jan 2017) remains using only the five permitted operations (+, -, *, /, ^) and brackets in your paper. [CR]

- Without use of factorial or square-root, it is impossible to find. Examples using factorials:

10958=1+2+3!!+(-4+5!+6-7)*89
 
10958=1*2*(3!!-4!*(5+6)+7!-8-9)
Below is different way of writing 10958. (But) It uses all the 10 numbers from 1 to 10:
 
10958=1*2^3+(4^5+6+7*8+9)*10 [IJT]

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Chris Smith wrote on Jan 25, 2017:

My friend Fraser McKey and I have been playing about with 10958 this week after seeing Inder J. Taneja papers and he's uncovered this near miss: 1.2 + ((3/4) ^ (5 - (6 x 7 x (8/9))) ~ 10958.052439016

A few solutions using factorial including:

 -(1 + 2 - 3 + 4 - ((5! + 6) x (78 + 9))) = 10958

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On Jan 29, 2017 José de Jesús Camacho wrote:

10958=ceiling(1/2 + ((3/4) ^ (5 - (6 *7 * (8/9))))

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On April 23, 2018 Edward Kang sent a solution to Q1 -the first absolutely valid solution, as far as I know, using only three of the permitted arithmetic operator symbols "+", "x" and "||", the last, for the concatenation operation, and the auxiliary grouping symbols, "(" & ")".

 10958 = 1 x 2||3 + ((4*5*6)||7 + 8) x 9 = 23 + (120||7+8) x 9 = 23 + (1207 + 8) x 9 = 23 + 1215 x 9 =23 + 10935 = 10958

As far as I can see the key of Kang's clever solution is to use the concatenation symbol not only to concatenate digits but also to concatenate arithmetic expressions. Well, nobody told him not to do this!...

By my side I wrote to Inder J. Taneja about the result reported by Mr. Kang to know his opinion. Taneja responded:

This is the same procedure explain in video - numberphile my Matt Parker. The way it is worked is correct. "A Solution - https://www.youtube.com/watch?v=pasyRUj7UwM"

In the video by Matt Parker as a matter of fact it is described the same solution that was reported by Mr Kang to me by email the 23 of April. The video published by Mr. Parker in YT is dated on April 18.

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