Question

All possible three digits even numbers which can be formed with the condition that if 5 is one of the digit, then 7 is the next digit is

Answer 1

365

Answer 2

The problem asks us to find all possible three-digit even numbers such that if 5 is one of the digits, then 7 is the next digit. Let the three-digit number be $d_1 d_2 d_3$.

Conditions:

1. It's a three-digit number: $d_1 \in \{1, 2, ..., 9\}$, $d_2, d_3 \in \{0, 1, ..., 9\}$.
2. It's an even number: $d_3 \in \{0, 2, 4, 6, 8\}$.
3. Conditional rule: If 5 is one of the digits, then 7 is the next digit. This means:
   • If $d_1 = 5$, then $d_2 = 7$.
   • If $d_2 = 5$, then $d_3 = 7$.
   • If $d_3 = 5$, this part of the condition doesn't apply as there is no "next digit". However, $d_3$ must be an even digit, so $d_3$ cannot be 5.

We can solve this by considering two main cases:

Case 1: The digit 5 is not used in the number.

If 5 is not used, the condition "if 5 is one of the digit, then 7 is the next digit" is vacuously true (the "if" part is false, so the implication holds). We need to form a three-digit even number using digits from $\{0, 1, 2, 3, 4, 6, 7, 8, 9\}$ (excluding 5).

• For $d_1$: It cannot be 0 or 5. So, $d_1 \in \{1, 2, 3, 4, 6, 7, 8, 9\}$ (8 choices).
• For $d_2$: It cannot be 5. So, $d_2 \in \{0, 1, 2, 3, 4, 6, 7, 8, 9\}$ (9 choices).
• For $d_3$: It must be an even digit and cannot be 5. So, $d_3 \in \{0, 2, 4, 6, 8\}$ (5 choices).

Number of such numbers = $8 \times 9 \times 5 = 360$.

Case 2: The digit 5 is used in the number.

If 5 is used, the condition (if 5 is one of the digit, then 7 is the next digit) must be strictly followed for every occurrence of 5.

• Subcase 2.1: $d_1 = 5$.

  According to the condition, if $d_1 = 5$, then $d_2$ must be 7. The number will be of the form $57d_3$. For the number to be even, $d_3$ must be an even digit: $d_3 \in \{0, 2, 4, 6, 8\}$. (Note: $d_3$ cannot be 5 as it must be even). Also, we check if any other 5s appear. Since $d_1=5$ and $d_2=7$, and $d_3$ is even, there are no other 5s in the number. The possible numbers are $570, 572, 574, 576, 578$. Number of such numbers = 5.

• Subcase 2.2: $d_2 = 5$.

  According to the condition, if $d_2 = 5$, then $d_3$ must be 7. The number will be of the form $d_1 57$. For the number to be even, $d_3$ must be an even digit. However, $d_3 = 7$ (which is an odd digit). This creates a contradiction. Therefore, no even numbers can be formed in this subcase. (Also, $d_1$ cannot be 5, because if $d_1=5$, then $d_2$ would have to be 7, not 5. So, no multiple 5s like 557 are considered here). Number of such numbers = 0.

• Subcase 2.3: $d_3 = 5$.

  This is not possible, as $d_3$ must be an even digit for the number to be even. Number of such numbers = 0.

• Subcase 2.4: Multiple 5s in the number.

  This situation is implicitly covered by the above subcases.

  • If $d_1=5$ and $d_2=5$: Not possible, because if $d_1=5$, then $d_2$ must be 7.
  • If $d_1=5$ and $d_3=5$: Not possible, because $d_3$ must be even.
  • If $d_2=5$ and $d_3=5$: Not possible, because $d_3$ must be even. So, no numbers with multiple 5s are possible under the given conditions.

Total Numbers:

The total number of possible three-digit even numbers is the sum of numbers from Case 1 and Subcase 2.1. Total = $360 + 5 = 365$.

The final answer is $\boxed{365}$.