# Sizning yechimingizdagi faktorlar sonini maksimal darajaga ko'taring 6 5 4 3 = 1

6 5 4 3 = 1 bilan bir xil qoidalarni qo'llang, lekin faktorlar sonini maksimal darajada oshirish. Siz 1 yoki 2 faktoriyani qabul qila olmaysiz.

Bu ko'rinishidan qiyin. Katta javobda oltita omil mavjud, ajoyib javobda yetti faktor bor, va ajoyib javoblar sakkizta yoki undan ko'p faktordan iborat.

Tartibga solish: Shoot, dasturda men kabi yigirmatadan kattaroq raqamlarni blokirovkalash kerak.

10
Ikki tomonli faktorlar ruxsat berilganmi va agar shunday bo'lsa, ularni qanday hisoblash mumkin?
qo'shib qo'ydi
@NeilG Ikkilangan faktorlar, albatta, royhatdan otgan operatsiyalardan farq qiladi. Lekin shaytonning advokati sifatida bitimga ruxsat berilsa va bu ikki raqamni birlashtiradigan raqamlar birlashuvini bildirmasligini bildirmaydi. turli operatsiyalarni beradi "!!".
qo'shib qo'ydi
Men shuni taxmin qilamanki, 2 faktoriyani ham qabul qilmasligimiz kerak
qo'shib qo'ydi
qo'shib qo'ydi
@shepler Men ularni boshqa savol berilsa ruxsat etilgan operatsiyalarning biri deb hisoblamadim. (Men ularga ruxsat beriladimi, deb o'ylardim, ular ro'yxatga o'zlarining kiritishi kerak edi.)
qo'shib qo'ydi
@ Sayidryry Buyuk savol. Aslida, ha, bu muhim! Ushbu qoidasiz cheksiz ko'p miqdordagi faktorlarga erisha olishingiz mumkin, ammo agar bu muammo haqida juda ko'p narsalarni qanday ochganimni aytib qo'ysam.
qo'shib qo'ydi
@xhienne Men sakkiz omillardan iborat echimni topdim. To'rt yoki undan ortiq faktorlar bilan 72 ta echim mavjud.
qo'shib qo'ydi
@BennettBernardoni rahmat, ha.
qo'shib qo'ydi
@ Saeidryl Siz (6 - 5) foydalana olmaysiz! na (6 - 4)!
qo'shib qo'ydi
Oh, afsus, bu tahrirlarni ko'rmadim. Vau, 8 faktorlar, qiziq!
qo'shib qo'ydi
Javob boshqa jumboqqa wolfram42ning javobidan farq qiladimi? (3 faktorlar)
qo'shib qo'ydi
@NeilG Factorial 2 qoidasi haqiqatdan ham foydalidirmi? Chunki 2 ta vakolatli raqamlar mavjud emas.
qo'shib qo'ydi
@xhienne Bu mening fikrim edi, lekin agar u 8 topsa, ehtimol yana bir savol uchun aniqdir.
qo'shib qo'ydi

## 6 javoblar

Javob:

1 yoki 2 omillardan tashqari Infinite factorials, bu tenglama bilan $6! \ ldots * (5-4) \ div 3 !!!! \ ldots = 1 $$E'tibor bering, 3da yana bitta faktoriyaviy bor. Buning sababi 3! = 6 ni tashkil etadi va har ikkala jihatga ham qo'shilamiz. 43 qo'shib qo'ydi (Men buni eng yaxshi javob deb bilaman, lekin umid qilamanki, boshqa javobni men qabul qilganim sababli tanladim deb o'ylamasligingizga umid qilamanki.) qo'shib qo'ydi Uni yaxshi ko'ramanmi? qo'shib qo'ydi Vay, bu ajoyib qo'shib qo'ydi @NeilG Agar bu yashirincha ish topishmoq bo'lmasa, siz eng yaxshi javobni qabul qilishingiz kerak. Sizningcha, javobgarlarning biri sizdan ko'ra aqlli bo'lgan - bu mukofot! qo'shib qo'ydi Qanday qilib? Infinitely ko'p faktoring? Asosiy misol bo'la oladi: \ frac {6 * (5-4)} {3!} = 1 Ammo lotga ba'zi faktoringlarni qo'shishingiz mumkin: \ frac {6! * (5-4)} {3 !!} = 1 Yoki hatto: \ frac {6 !!!!!!! \ times (5-4)} {(3!) !!!!!!!} = 1 Haqiqatan ham, agar yana 3 faktoring orqasida bo'lsa, biz yaxshimiz! 11 qo'shib qo'ydi Xafa bo'lma, @Bennett Bernardoni! : p qo'shib qo'ydi Katta javob! Va 7k rep bo'yicha tanishing! : D qo'shib qo'ydi Men "20dan ortiq raqamlarda faktoring yo'q" qoida bilan harakat qilaman. Bu erda 7 bilan bir yechim bor. Ba'zi formatlash yordami qadrlanadi \ frac {6!} {(5! \ div4!)! \ div 3 !!$$ Boshqaruv: Kuzatuv bilan boshladim 6!/5! = 6, xuddi shunday 5!/4! = 5,3! = 6 so 6! = 3 !! Keyin u erdan qurilganman. Shunday qilib, 6! = (6!/(5! 4!)). Yo'qolgan faktoriyani aniqladi:$ \ left (\ frac {6!} {(5! \ div4!)! - 3 !! \ o'ng)! = 1 ta'rifi bo'yicha.

10
qo'shib qo'ydi
N, n - 2 k> 0} (n - n) ni anglatuvchi $i (n!)! 2 k)$?
qo'shib qo'ydi
Men shuni nazarda tutdimki, ikki tomonlama protsessual operatsiyalarga yo'l qo'yilmadi, chunki 8 dan ortiq usulni olish osonroq bo'ladi.
qo'shib qo'ydi
@NeilG Menimcha, bu safar ham bor
qo'shib qo'ydi
Aslida tez tahrir qilishni boshladim va men o'zimni aralashtirdim. Men bir daqiqada qayta tiklayman
qo'shib qo'ydi
Yaxshi, siz 8ga yaqinsiz
qo'shib qo'ydi
Menimcha, bir juft parentezni yo'qotishim mumkin, lekin ha ajoyib! Buni qanday qildingiz?
qo'shib qo'ydi

wolfram42 ning boshqa jumboqqa javobidan boshlab, eng ko'p ma'lumotli omillar soni:

cheksizdir, chunki teng sonli ! sonini qo'shishingiz mumkin   (6!)/(5 * 4! * 3!) Foydalanuvchining profili   Men matematik bo'lmaganim uchun, "cheksiz ortiqcha uch" deb ayta olaman; -)

Note: that's also true for any answer to the other puzzle that involved a division, like JonMark Perry's two solutions

6
qo'shib qo'ydi

Mana bu turdagi muammolarga umumiy dasturiy yechim. Shuningdek, u har qanday natija uchun chiqaradi, bu erda qanday echimlar mavjud, bu muammoni hal qilishda foydali bo'ladi:

from enum import IntEnum
from fractions import Fraction
from memoized import memoized
import math

NUMBERS = [6, 5, 4, 3]
RESULT = Fraction(1)

class Levels(IntEnum):
MULTIPLICATION = 1
NEGATION = 2
POWER = 3
FACTORIAL = 4
BARE = 5

class Result:

def __init__(self, value, way, level):
"""
* value is the value achieved.
* way is string representation of an expression.
* level is the outermost operator applied to the expression, which
determines how this expression should be braketed when it's part of a
larger expression.
"""
assert isinstance(value, Fraction)
assert isinstance(level, Levels)
self.value = value
self.way = way
self.level = level

def bracketed(self, for_level):
if for_level > self.level:
return f"({self.way})"
return self.way

def combine(left_rs, right_rs):
for l in left_rs:
for r in right_rs:
if r.level not in (Levels.NEGATION, Levels.ADDITION):
# We block a - (-b), and a + (-b) since these can be written as
# a + b, and a - b.  We also block a + (b + c), and a - (b + c)
# since these can be written as a + b + c, and a - b - c.
yield Result(l.value + r.value,
yield Result(l.value - r.value,
if r.level != Levels.MULTIPLICATION:
# We block a * (b * c) since this can be written as a * b * c.
# We block a * (b/c) since this can be written as a * b/c.
yield Result(l.value * r.value,
f"{l.bracketed(Levels.MULTIPLICATION)} * {r.bracketed(Levels.MULTIPLICATION + 1)}",
Levels.MULTIPLICATION)
if r.value != 0 and (r.level != Levels.MULTIPLICATION):
# We block a/(b/c) since this can be written as a/b * c.
# We block a/(b * c) since this can be written as a/b/c.
yield Result(l.value/r.value,
f"{l.bracketed(Levels.MULTIPLICATION)}/{r.bracketed(Levels.MULTIPLICATION + 1)}",
Levels.MULTIPLICATION)
new_value = None
power_okay = l.value >= 0 or r.value.denominator % 2 == 1
if l.value == 1:
new_value = Fraction(1)
elif l.value == -1:
if power_okay:
new_value = Fraction(-1
if r.value.numerator % 2 == 1
else 1)
elif l.value == 0:
if r.value > 0:
new_value = Fraction(0)
else:
# Only whole number powers are allowed for numbers other than
# -1, 0, 1.
if (power_okay
and r.value.denominator == 1
and -8 < r.value.numerator < 8):
new_value = l.value ** r.value
if new_value is not None:
# The left side is bracketed when it's a power because power is
# right associative.
yield Result(new_value,
f"{l.bracketed(Levels.POWER + 1)} ^ {r.bracketed(Levels.POWER)}",
Levels.POWER)
if (l.level == Levels.BARE
and r.level == Levels.BARE and len(r.way) == 1):
# We concatenate 123 as 1(23), but not as (12)3.
yield Result(l.value * 10 + r.value,
f"{l.way}{r.way}",
Levels.BARE)

def negate(rs):
for r in rs:
yield r
if r.level > Levels.MULTIPLICATION:
# We block -(a + b) since this can be written -a - b.
# We block -(a - b) since this can be written -a + b.
# We block -(a * b) since this can be written -a * b.
# We block -(a/b) since this can be written -a/b.
yield Result(-r.value,
f"-{r.bracketed(Levels.NEGATION)}",
Levels.NEGATION)

def factorialize(rs):
for r in rs:
yield r
x = 0
while (r.value.denominator == 1
and (r.value.numerator == 0 or (3 <= r.value.numerator <= 20))):
# We block fractional and negative factorials.
# We block 1 and 2 factorial.
# We block gigantic factorials.
r = Result(Fraction(math.factorial(r.value.numerator)),
f"{r.bracketed(Levels.FACTORIAL + 1)}!",
Levels.FACTORIAL)
yield r

# memoization makes the recursive solution into a dynamic programming solution.
@memoized(hashable=False)
def do(digits):
"""
Given a list of digits, produce a list of Result objects.
"""
l = len(digits)
if l == 1:
# Return a Result object for a single digit.
new_rs = [Result(Fraction(digits), str(digits), Levels.BARE)]
else:
# The list of results for a list of digits is the total results for every contiguous partition.
new_rs = []
for i in range(1, l):
new_rs.extend(list(combine(do(digits[:i]),
do(digits[i:]))))
return list(factorialize(negate(new_rs)))

rs = do(NUMBERS)
vs = {}
for r in rs:
vs.setdefault(r.value, []).append(r)
print(f"Solutions to {NUMBERS} = {RESULT}")
for r in sorted(vs[RESULT], key=lambda r: len(r.way)):
print(r.way)
print()
print(f"                         value solutions")
for v, l in sorted(vs.items(), key=lambda vl: (len(vl), len(str(vl)))):
print(f"{str(v):>30} {len(l):<12}")


Solutions to [6, 5, 4, 3] = 1
65 - 4 ^ 3
(6 - 5) ^ 43
65 + (-4) ^ 3
-(-6 + 5) ^ 43
-6 - 5 + 4 * 3
6 + 5 - 4 - 3!
6 - 5 * (4 - 3)
6 - 5/(4 - 3)
6 - 5 ^ (4 - 3)
6/(5 + 4 - 3)
6 - 5!/4/3!
(6 - 5) ^ 4 ^ 3
(6 - 5) ^ (-43)
...
(((-6 + 5 + 4)!)! - (3!)!)!
((6!/(5!/4!)!)! - (3!)!)!
((-(-6) ^ (5 - 4))! - (3!)!)!

2
qo'shib qo'ydi
Xo'sh, bu nimani anglatadi?
qo'shib qo'ydi
@haccks Men ba'zi izohlar qo'shdim.
qo'shib qo'ydi
@aschepler qo'shildi
qo'shib qo'ydi
Bu kod faqat boshimni yopib qo'ydi. Sharhlarda ba'zi izohlar qo'shilsa yaxshi bo'lardi.
qo'shib qo'ydi

# Qoidalarga rioya qilgan holda 9 ta faktor (quyidagi javob)

## Ba'zi fokuslar ishlatiladi

Birinchi hiyla:

$0! = 1$ ( aniqlash bo'yicha ). Biz $\ forall x \ in \ mathbb {R}: 0 \ cdot x = 0 \ Rightarrow (0 \ cdot x)! = 1$ haqiqatini ishlatamiz.

Ikkinchi hiyla:

Since we cannot use $1!$ or $2!$ and we cannot use $n!$ for $n>20$, we have to look the other way: think smaller. Lucky for us, we have $x!=\Gamma(x+1)$ where $\Gamma$ is the gamma function and this holds for $x\in\mathbb{R}$.

## Bizning javobimizni tuzing

Birinchi qadam (birinchi hiyla ishlatish):

Bizga nol kerak. Bu juda oson, bizda $6! - (3!)! = 6! -6! = 0$.

Ko'p ma'lumotlarga ehtiyoj bor, lekin biz faqatgina $4 va$ 5 dan foydalanishimiz mumkin, chunki biz birinchi qadamda boshqalarni qo'lladik. Ularni olish uchun biz natijani tashvishga solmaymiz, chunki biz baribir $0 ga ko'payamiz. Biz faqat$ 0 $yoki$ 1 $(biz ulardan kamida$ 0.01 $masofada qolamiz) va$ 20 dan oshmasligi uchun yaqin bo'lishga ehtiyot bo'lishimiz kerak.

birinchi faktorial $\ left (\ frac {-5} {4} \ o'ng)! \ taxminan -4,9$.   $\ chap (\ chap (\ frac {-5} {4} \ o'ng)! \ o'ng)! \ taxminan 0.50$.   "nofollow noreferrer"> uchinchi faktorial $\ chap (\ chap (\ chap (\ frac {-5} {4} \ o'ng)! \ o'ng)! \ o'ng)! \ taxminan 0.89$.   "nofollow noreferrer"> to'rtinchi faktorial $\ chap (\ chap (\ chap (\ chap (\ frac {-5} {4} \ o'ng)! \ o'ng)! \ o'ng)! taxminan 0.96$.   $\ chap (\ chap (\ chap (\ chap (\ chap (\ frac {-5} {4} \ o'ngda! \ o'ng)! \ o'ng)! \ o'ng)! \ o'ng)! \ taxminan 0.983$.   oltinchi faktorial $\ chap (\ chap (\ chap (\ chap (\ chap (\ chap (\ frac {-5} {4} \ o'ng)! \ o'ng)! \ o'ng)! \ o'ng)! \ o'ng)! \ o'ng)! \ taxminan 0,993$.  Shuning uchun biz bu erda 5 ta faktordan foydalanishimiz mumkin, chunki oltinchi qadam bilan $1$ ga yaqinlashamiz.

## Javob

$$x=6!-(3!)!=0$$ $$y=\left(\left(\left(\left(\left(\frac{-5}{4}\right)!\right)!\right)!\right)!\right)!\approx 0.983$$ $$\textit{final Javob}=(x\cdot y)!=0!=1$$

1
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