Даны четыре шифровки: 16153, 20121, 13225, 71205. Только одна из них расшифровывается единственным способом. Найдите и расшифруйте её.

Пошаговое решение:

Используем предоставленную таблицу для расшифровки:

Рассмотрим каждую шифровку:

1. 16153: 16=P, 15=O, 3=C. Получается слово POC. Но возможна и другая расшифровка: 1=A, 6=F, 15=O, 3=C. Получается слово AFOC. Так как есть несколько вариантов, эта шифровка не подходит.

2. 20121: 20=T, 12=L, 1=A, 21=U. Получается слово TLAU. Возможна другая расшифровка: 2=B, 0 - нет такой буквы, 12=L, 1=A, 21=U. Это тоже не единственный вариант.

3. 13225: 13=M, 22=V, 25=Y. Получается слово MVY. Рассмотрим другой вариант: 1=A, 3=C, 22=V, 25=Y. Получается слово ACVY. Этот вариант тоже не единственный.

4. 71205: 7=G, 12=L, 0 - нет такой буквы, 5=E. Не подходит.

Перепроверим шифровки, учитывая, что цифры могут означать как одну, так и две цифры.

1. 16153: 16=P, 15=O, 3=C => POC. Другой вариант: 1=A, 6=F, 15=O, 3=C => AFOC. Есть несколько вариантов.

2. 20121: 20=T, 12=L, 1=A, 21=U => TLAU. Другой вариант: 2=B, 0 - нет. Этот вариант не подходит.

3. 13225: 13=M, 22=V, 25=Y => MVY. Другой вариант: 1=A, 3=C, 22=V, 25=Y => ACVY. Есть несколько вариантов.

4. 71205: 7=G, 12=L, 0 - нет. Есть другой вариант: 7=G, 1=A, 20=T, 5=E => GATE. Есть другой вариант: 7=G, 12=L, 05 - нет. Этот вариант не единственный.

Давайте внимательнее посмотрим на условие: "Только одна из них расшифровывается единственным способом".

Пересмотрим варианты:

1. 16153: 16(P) 15(O) 3(C) = POC. Или 1(A) 6(F) 15(O) 3(C) = AFOC. Не единственный.

2. 20121: 20(T) 12(L) 1(A) 21(U) = TLAU. Или 2(B) 0(нет) 12(L) 1(A) 21(U). Не подходит.

3. 13225: 13(M) 22(V) 25(Y) = MVY. Или 1(A) 3(C) 22(V) 25(Y) = ACVY. Не единственный.

4. 71205: 7(G) 12(L) 05 - нет. Или 7(G) 1(A) 20(T) 5(E) = GATE. Или 7(G) 1(A) 2(B) 05 - нет.

Судя по всему, я неправильно интерпретирую цифры. Посмотрим на таблицу еще раз. Все цифры от 1 до 26 соответствуют буквам. Цифра 0 не используется.

1. 16153: 16(P) 15(O) 3(C) -> POC. 1(A) 6(F) 15(O) 3(C) -> AFOC. Не единственный.

2. 20121: 20(T) 12(L) 1(A) 21(U) -> TLAU. 2(B) 0(нет) 12(L) 1(A) 21(U). Не единственный.

3. 13225: 13(M) 22(V) 25(Y) -> MVY. 1(A) 3(C) 22(V) 25(Y) -> ACVY. Не единственный.

4. 71205: 7(G) 1(A) 20(T) 5(E) -> GATE. 7(G) 12(L) 05 - нет. 7(G) 1(A) 2(B) 05 - нет. 7(G) 12(L) 5(E) -> GLE. Не единственный.

Перечитываем условие: "Даны четыре шифровки: 16153, 20121, 13225, 71205. Только одна из них расшифровывается единственным способом."

Проблема может быть в том, что я не учитываю возможность разделения двузначных чисел.

1. 16153: 16(P) 15(O) 3(C) = POC. Также: 1(A) 6(F) 15(O) 3(C) = AFOC. Не единственный.

2. 20121: 20(T) 12(L) 1(A) 21(U) = TLAU. Или 2(B) 0(нет). Не подходит.

3. 13225: 13(M) 22(V) 25(Y) = MVY. Или 1(A) 3(C) 22(V) 25(Y) = ACVY. Не единственный.

4. 71205: 7(G) 12(L) 05(нет). Или 7(G) 1(A) 20(T) 5(E) = GATE. Или 7(G) 1(A) 2(B) 05(нет). Или 7(G) 12(L) 5(E) = GLE. Не единственный.

Что если в шифровке 20121, 20 это T, а 12 это L, 1 это A, 21 это U? TLAU. А что если 2=B, 0 - нет?

Давайте попробуем найти единственную расшифровку:

1. 16153: 16-P, 15-O, 3-C (POC). 1-A, 6-F, 15-O, 3-C (AFOC). Множество вариантов.

2. 20121: 20-T, 12-L, 1-A, 21-U (TLAU). 2-B, 0-нет. 20-T, 1-A, 21-U (TAU). 2-B, 0-нет. 20-T, 12-L, 21-U (TLU). 2-B, 0-нет. 20-T, 12-L, 1-A, 2-B, 1-A. Нет.

3. 13225: 13-M, 22-V, 25-Y (MVY). 1-A, 3-C, 22-V, 25-Y (ACVY). Множество вариантов.

4. 71205: 7-G, 12-L, 05-нет. 7-G, 1-A, 20-T, 5-E (GATE). 7-G, 1-A, 2-B, 05-нет. 7-G, 12-L, 5-E (GLE). Множество вариантов.

Кажется, что только одна шифровка может быть расшифрована однозначно. Проверим внимательно.

1. 16153: 16-P, 15-O, 3-C (POC). 1-A, 6-F, 15-O, 3-C (AFOC). Не единственный.

2. 20121: 20-T, 12-L, 1-A, 21-U (TLAU). 2-B, 0-нет. 20-T, 1-A, 21-U (TAU). 2-B, 0-нет. 20-T, 12-L, 21-U (TLU). 2-B, 0-нет.

3. 13225: 13-M, 22-V, 25-Y (MVY). 1-A, 3-C, 22-V, 25-Y (ACVY). Не единственный.

4. 71205: 7-G, 12-L, 05-нет. 7-G, 1-A, 20-T, 5-E (GATE). 7-G, 1-A, 2-B, 05-нет. 7-G, 12-L, 5-E (GLE).

Давайте пересмотрим 20121. Есть ли там еще варианты, кроме TLAU?

20 (T), 12 (L), 1 (A), 21 (U) -> TLAU.

2 (B), 0 (нет).

20 (T), 1 (A), 21 (U) -> TAU.

2 (B), 0 (нет), 12 (L), 1 (A), 21 (U).

20 (T), 12 (L), 21 (U) -> TLU.

2 (B), 0 (нет).

20 (T), 1 (A), 2 (B), 1 (A), 21 (U).

2 (B), 0 (нет).

Seems that 20121 is the one. Let's double check.

16153: 16(P) 15(O) 3(C) = POC. Also 1(A) 6(F) 15(O) 3(C) = AFOC. Not unique.

13225: 13(M) 22(V) 25(Y) = MVY. Also 1(A) 3(C) 22(V) 25(Y) = ACVY. Not unique.

71205: 7(G) 12(L) 5(E) = GLE. Also 7(G) 1(A) 20(T) 5(E) = GATE. Not unique.

20121: 20(T) 12(L) 1(A) 21(U) = TLAU. What if we try to split it differently? 2(B) ... but 0 is not in the alphabet. 20(T) ... 12(L) ... 1(A) ... 21(U). This seems to be the only way to form a valid word.

Let's assume the number 0 is not in the alphabet at all.

16153: POC. AFOC. (not unique)

20121: TLAU. TAU. TLU. (These are all unique ways of parsing if we consider allowed numbers)

Wait, the problem states "only one of them deciphers in a unique way". This means for that specific number, there is only one possible interpretation as a word from the alphabet. For other numbers, there might be multiple interpretations or none.

Let's re-evaluate each number's possible interpretations:

1. 16153:
- 16(P), 15(O), 3(C) = POC
- 1(A), 6(F), 15(O), 3(C) = AFOC
- 16(P), 1(A), 5(E), 3(C) = PAEC
- 1(A), 6(F), 1(A), 5(E), 3(C) = AFAEC
Multiple interpretations exist. Not unique.

2. 20121:
- 20(T), 12(L), 1(A), 21(U) = TLAU
- 20(T), 1(A), 21(U) = TAU
- 20(T), 12(L), 21(U) = TLU
- 2(B) - not possible as next digit is 0.
- 20(T), 12(L), 1(A), 2(B), 1(A) = TLABA
It seems there are multiple ways to parse this. Let's assume the rule is to form valid English words using the numbers. If we assume that, then TLAU, TAU, TLU are valid number sequences. However, the question says "deciphers in a unique way". This implies that for ONE of the numbers, there's only one possible way to map the digits to letters to form a sequence of letters. For the other numbers, there might be NO valid way or MORE THAN ONE valid way.

Let's strictly look at number-to-letter mapping only. No word formation assumed unless it's implied by uniqueness.

1. 16153:
16-P, 15-O, 3-C => POC
1-A, 6-F, 15-O, 3-C => AFOC
16-P, 1-A, 5-E, 3-C => PAEC
1-A, 6-F, 1-A, 5-E, 3-C => AFAEC
Many ways to parse. Not unique.

2. 20121:
20-T, 12-L, 1-A, 21-U => TLAU
20-T, 1-A, 21-U => TAU
20-T, 12-L, 21-U => TLU
2-B (0 invalid)
This implies that if we can split it in different ways, it's not unique. The way we split it must be unique. Let's assume the goal is to find THE sequence of numbers that can ONLY be split in one way.

For 20121:

Possibility 1: 20, 12, 1, 21 -> T, L, A, U

Possibility 2: 20, 1, 21 -> T, A, U

Possibility 3: 20, 12, 21 -> T, L, U

This number has multiple interpretations. So it's not unique.

Let's reconsider 71205.

1. 71205:
7-G, 12-L, 5-E => GLE
7-G, 1-A, 20-T, 5-E => GATE
7-G, 1-A, 2-B (0 invalid)
7-G, 12-L, 05 (invalid)
Multiple interpretations. Not unique.

Let's re-examine 16153.

16-P, 15-O, 3-C = POC

1-A, 6-F, 15-O, 3-C = AFOC

16-P, 1-A, 5-E, 3-C = PAEC

1-A, 6-F, 1-A, 5-E, 3-C = AFAEC

This is also not unique.

Let's check 13225.

13-M, 22-V, 25-Y = MVY

1-A, 3-C, 22-V, 25-Y = ACVY

13-M, 2-B, 25-Y = MBY (0 invalid)

13-M, 22-V, 2-B, 5-E = MVBE

This is also not unique.

The problem must be simpler. Only one number has exactly one way of being broken down into valid numbers (1-26). Let's check that.

1. 16153:
16, 15, 3 (valid)
1, 6, 15, 3 (valid)
16, 1, 5, 3 (valid)
1, 6, 1, 5, 3 (valid)
Not unique segmentation.

2. 20121:
20, 12, 1, 21 (valid)
20, 1, 21 (valid)
20, 12, 21 (valid)
2, 0 (invalid)
This implies that 20121 can be interpreted as 20-12-1-21, 20-1-21, 20-12-21. Since there's more than one way to segment it, it's not unique.

3. 13225:
13, 22, 25 (valid)
1, 3, 22, 25 (valid)
13, 2, 25 (valid)
13, 22, 2, 5 (valid)
Not unique segmentation.

4. 71205:
7, 12, 5 (valid)
7, 1, 20, 5 (valid)
7, 12, 05 (invalid)
7, 1, 2, 05 (invalid)
Not unique segmentation.

Rethink. Which number, when broken into pieces, ONLY forms numbers between 1 and 26, and this partitioning is unique?

1. 16153:
16, 15, 3 (all valid)
1, 6, 15, 3 (all valid)
16, 1, 5, 3 (all valid)
1, 6, 1, 5, 3 (all valid)
This number has multiple valid segmentations.

2. 20121:
20, 12, 1, 21 (all valid)
20, 1, 21 (all valid)
20, 12, 21 (all valid)
2 (valid), 0 (invalid)
This number also has multiple valid segmentations.

3. 13225:
13, 22, 25 (all valid)
1, 3, 22, 25 (all valid)
13, 2, 25 (all valid)
13, 22, 2, 5 (all valid)
This number has multiple valid segmentations.

4. 71205:
7, 12, 5 (all valid)
7, 1, 20, 5 (all valid)
7, 12, 05 (05 is invalid)
7, 1, 2, 05 (05 is invalid)
This is the one! Let's see why.

We are looking for a number that can be broken down into a sequence of numbers, where each number is between 1 and 26 (inclusive), and there is ONLY ONE such sequence of valid numbers possible.

Let's try to parse each number into valid numbers (1-26):

1. 16153:
- 16, 15, 3 (Valid: P, O, C)
- 1, 6, 15, 3 (Valid: A, F, O, C)
- 16, 1, 5, 3 (Valid: P, A, E, C)
- 1, 6, 1, 5, 3 (Valid: A, F, A, E, C)
Multiple valid segmentations. Not unique.

2. 20121:
- 20, 12, 1, 21 (Valid: T, L, A, U)
- 20, 1, 21 (Valid: T, A, U)
- 20, 12, 21 (Valid: T, L, U)
- 2, 0... (0 is not valid, so this path is invalid from the start for any segmentation starting with 2)
Multiple valid segmentations. Not unique.

3. 13225:
- 13, 22, 25 (Valid: M, V, Y)
- 1, 3, 22, 25 (Valid: A, C, V, Y)
- 13, 2, 25 (Valid: M, B, Y)
- 13, 22, 2, 5 (Valid: M, V, B, E)
Multiple valid segmentations. Not unique.

4. 71205:
- 7, 12, 5 (Valid: G, L, E)
- 7, 1, 20, 5 (Valid: G, A, T, E)
- 7, 12, 05 (05 is invalid)
- 7, 1, 2, 05 (05 is invalid)
- 7, 1, 20, 05 (05 is invalid)
Let's re-check the parsing. What if the number 0 means that the digit before it can't be part of a two-digit number ending in 0?

The key is "unique way". For 71205, let's analyze the possible valid number segmentations (1-26):
- We can take 7 (G). Remaining: 1205. From 1205, we can take 12 (L). Remaining: 05. 05 is invalid. We can take 1 (A). Remaining: 205. From 205, we can take 20 (T). Remaining: 5. 5 is valid (E). So, 7, 1, 20, 5 => GATE.
- We can take 7 (G). Remaining: 1205. From 1205, we can take 12 (L). Remaining: 05. 05 is invalid.
- Let's try another path for 71205. Could it be 71? No, >26. So it must start with 7.
- After 7 (G), we have 1205.
- Option A: Take 12 (L). Remaining: 05. 05 is invalid.
- Option B: Take 1 (A). Remaining: 205. From 205, we can take 20 (T). Remaining: 5. 5 is valid (E). So: 7, 1, 20, 5 => GATE.
- Are there other ways to segment 205? We can take 2 (B). Remaining: 05. 05 is invalid. We can take 20 (T). Remaining: 5. 5 is valid (E).
- Let's re-examine the first segmentation of 71205. If we take 7, then 12, we get 05. This IS invalid. So this path leads to no word.
- The ONLY valid segmentation for 71205 seems to be 7, 1, 20, 5.
Let's confirm this.
- Start with 71205.
- If we try to take 71, it's invalid (>26).
- So we must take 7 (G). Remaining: 1205.
- From 1205:
- Try to take 12 (L). Remaining: 05. 05 is invalid.
- Try to take 1 (A). Remaining: 205.
- From 205:
- Try to take 20 (T). Remaining: 5. 5 is valid (E). This gives GATE.
- Try to take 2 (B). Remaining: 05. 05 is invalid.
- Therefore, the ONLY possible valid segmentation for 71205 is 7, 1, 20, 5.
This leads to the word GATE.

Let's confirm why other numbers are not unique.

1. 16153: 16, 15, 3 (POC). Also 1, 6, 15, 3 (AFOC). Not unique.

2. 20121: 20, 12, 1, 21 (TLAU). Also 20, 1, 21 (TAU). Not unique.

3. 13225: 13, 22, 25 (MVY). Also 1, 3, 22, 25 (ACVY). Not unique.

So, 71205 is indeed the unique one.

Расшифровка:

7 -> G

1 -> A

20 -> T

5 -> E

Получается слово: GATE

Ответ: 71205, GATE