MHB How Do You Calculate 'n' in This Permutation Problem?

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To find 'n' in the permutation problem, the equation 1/4! + 1/5! + 1/6! = n/7! is solved by multiplying both sides by 7!. This leads to the calculation of n as 7 * 6 * 5 + 7 * 6 + 7, resulting in n = 259. The discussion suggests that the number 259 might hold a special significance, possibly as an encrypted message. The problem is deemed solvable and not impossible.
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Find 'n' using permutations:

1/4! + 1/5! + 1/6! = n/7!
 
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Arunachaleshwar said:
1/4! + 1/5! + 1/6! = n/7!

Welcome on MHB Arunachaleshwar!...

... the problem You propose is not 'impossible'... multiplying the two sides for 7! You have...

$\displaystyle n= 7 \cdot 6 \cdot 5 + 7 \cdot 6 + 7 = 259$

... can it be that You are sending an encrypted message in which the number 259 has a special meaning? ...

Kind regards

$\chi$ $\sigma$
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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