Е8.79 при этом ни одна нечётная цифра не стоит рядом с цифрой D

Сколько существует 16-ричных четырёхзначных чисел , в записи которых ровно одна цифра D, при этом ни одна нечётная цифра не стоит рядом с цифрой D?

Ответ:

СтатГрад Вариант ИН2510501 14 апреля 2026 – задание №8

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Решение:

Решение —

from itertools import*
k = 0
alp = '0123456789ABCDEF'
for i in product(alp,repeat = 4):
    a = ''.join(i)
    if a[0] != '0' and a.count('D') == 1 :
        if a.count('1D') ==0 and  a.count('D1') ==0 :
            if a.count('3D') == 0 and a.count('D3') == 0:
                if a.count('5D') == 0 and a.count('D5') == 0:
                    if a.count('7D') == 0 and a.count('D7') == 0:
                        if a.count('9D') == 0 and a.count('D9') == 0:
                            if a.count('BD') == 0 and a.count('DB') == 0:
                                if a.count('FD') == 0 and a.count('DF') == 0:
                                    k += 1
print(k)

from itertools import*

k = 0

alp = '0123456789ABCDEF'

for i in product(alp,repeat = 4):

a = ''.join(i)

if a[0] != '0' and a.count('D') == 1 :

if a.count('1D') ==0 and a.count('D1') ==0 :

if a.count('3D') == 0 and a.count('D3') == 0:

if a.count('5D') == 0 and a.count('D5') == 0:

if a.count('7D') == 0 and a.count('D7') == 0:

if a.count('9D') == 0 and a.count('D9') == 0:

if a.count('BD') == 0 and a.count('DB') == 0:

if a.count('FD') == 0 and a.count('DF') == 0:

k += 1

print(k)

ИЛИ

from itertools import*
k = 0
alp = '0123456789ABCDEF'
for i in product(alp,repeat = 4):
    a = ''.join(i)
    if a[0] != '0' and a.count('D') == 1 :
        if all(a.count(f'{i}D') == 0 and a.count(f'D{i}') == 0 for i in ['1','3','5','7','9','B','F']) :
            k +=1
print(k)

from itertools import*

k = 0

alp = '0123456789ABCDEF'

for i in product(alp,repeat = 4):

a = ''.join(i)

if a[0] != '0' and a.count('D') == 1 :

if all(a.count(f'{i}D') == 0 and a.count(f'D{i}') == 0 for i in ['1','3','5','7','9','B','F']) :

k +=1

print(k)

ИЛИ

from itertools import *
k=0
for i in product('0123456789ABCDEF', repeat = 4):
    a = ''.join(i)
    if a.count('D') != 1 or a[0]=='0':
        continue
    nechet = '13579BDF'
    if a[0]=='D' and a[1] not in nechet:
        k+=1
    if a[1]=='D' and a[0] not in nechet and a[2] not in nechet:
        k+=1
    if a[2]=='D' and a[1] not in nechet and a[3] not in nechet:
        k+=1
    if a[3]=='D' and a[2] not in nechet:
        k+=1
print(k)

from itertools import *

k=0

for i in product('0123456789ABCDEF', repeat = 4):

a = ''.join(i)

if a.count('D') != 1 or a[0]=='0':

continue

nechet = '13579BDF'

if a[0]=='D' and a[1] not in nechet:

k+=1

if a[1]=='D' and a[0] not in nechet and a[2] not in nechet:

k+=1

if a[2]=='D' and a[1] not in nechet and a[3] not in nechet:

k+=1

if a[3]=='D' and a[2] not in nechet:

k+=1

print(k)

Ответ: 5216