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Укажите все номера годов текущего десятилетия (с 2...

Créé le : 5 novembre 2025

Répondu en utilisant GPT-5 Thinking par Chat01

Question

Укажите все номера годов текущего десятилетия (с 2021 по 2030), которые можно представить
суммой одно-, двух-, трёх- и четырёхзначного чисел так, чтобы каждая цифра была использована один раз

Python

python
import itertools

digits = '0123456789'

def num_from_digits(digs):
    return int('''.join(digs))

Réflexion

import itertools

digits = '0123456789'

def num_from_digits(digs): return int('''.join(digs))

Thought for 33s

Python

python
import itertools

digits = '0123456789'

def num_from_digits(digs):
    return int(''.join(digs))

targets = list(range(2021, 2031))
solutions = {t: [] for t in targets}

for d1 in digits:
    rem1 = set(digits) - {d1}
    n1 = int(d1)
    for d2pair in itertools.permutations(rem1, 2):
        if d2pair[0]=='0':
            continue
        rem2 = rem1 - set(d2pair)
        n2 = num_from_digits(d2pair)
        for d3trip in itertools.permutations(rem2, 3):
            if d3trip[0]=='0':
                continue
            rem3 = rem2 - set(d3trip)
            n3 = num_from_digits(d3trip)
            # remaining 4 digits: there are 4 chosen? Wait rem3

({2021: 0,
  2022: 0,
  2023: 0,
  2024: 0,
  2025: 2208,
  2026: 0,
  2027: 0,
  2028: 0,
  2029: 0,
  2030: 0},
 2208)

Python

python
# get one example
example = None
for sol in solutions[2025]:
    example = sol
    break
example

(0, 58, 274, 1693)

Python

python
solutions_nonzero_1digit = {t: [] for t in targets}
for d1 in '123456789':  # exclude 0
    rem1 = set(digits) - {d1}
    n1 = int(d1)
    for d2pair in itertools.permutations(rem1, 2):
        if d2pair[0]=='0':
            continue
        rem2 = rem1 - set(d2pair)
        n2 = num_from_digits(d2pair)
        for d3trip in itertools.permutations(rem2, 3):
            if d3trip[0]=='0':
                continue
            rem3 = rem2 - set(d3trip)
            n3 = num_from_digits(d3trip)
            for d4perm in itertools.permutations(rem3, 4):
                if d4perm[0]=='0':
                    continue
                n4 = num_from_digits(d4perm)
                s = n1 + n2 + n3 + n4
                if s in solutions_nonzero_1digit:
                    solutions_nonzero_1digit[s].append((n1,n2,n3,n4))

{k:len(v) for k,v in solutions_nonzero_1digit.items()}

{2021: 0,
 2022: 0,
 2023: 0,
 2024: 0,
 2025: 1920,
 2026: 0,
 2027: 0,
 2028: 0,
 2029: 0,
 2030: 0}

Python

python
# find example with one-digit != 0
ex2 = None
for sol in solutions_nonzero_1digit[2025]:
    ex2 = sol
    break
ex2

(2, 54, 876, 1093)

Réflexion

Confirming 2025 as the solution

The sum of 2 + 54 + 876 + 1093 equals 2025, using all digits 0–9 once, with no leading zeroes in the four-digit number. Using the mod 9 argument, only 2025 is divisible by 9 within the 2021–2030 range. I'll present this elegant solution concisely, showing that other years are not divisible by 9.

Presenting the solution clearly

For clarity, we can state that the sum of the four numbers is congruent to the sum of their digit sums, which is congruent to 0 mod 9. There's no need for external sources here—it's simply a math puzzle. I'll include just one decomposition, as that's sufficient based on the user's request.

Crafting the final response

I can structure the response as follows:

  • Answer: 2025.
  • Justification: The sum of all ten digits is 45 ≡ 0 mod 9. Any number is congruent to the sum of its digits modulo 9, so the sum must be divisible by 9. Looking at years from 2021 to

Thought for 1m 30s

Réponse

Только 2025.

Почему: по модулю 9 каждое число сравнимо с суммой своих цифр, а все десять цифр 0–9 в сумме дают 45 ≡ 0 (mod 9). Значит искомый год должен делиться на 9. В диапазоне 2021–2030 это выполняется только для 2025.

Конструкция (пример):

  • без нуля в одноразрядном числе: 2 + 54 + 876 + 1093 = 2025;
  • с нулём в одноразрядном числе: 0 + 58 + 274 + 1693 = 2025.

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