# 15(13) РТ

avatar
user_8840555
python
2 years ago
189 B
5
Indexable
for a in range(1, 1000):
    for x in range(1, 10000):
        if ((x & 23 == 0)<= ((x & 13 != 0)<= (x & a != 0))) == False:
            break
    else:
        print(a)
# 15(13) РТ
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