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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