n a 6-digit number, the digit in ones place is more than the digit in ten thousands place by 7. The digit in hundreds place is less than the digit in thousands place by 2. The digit in tens place is more than the digit in hundred thousands place by 4. The digit in hundred thousands place is less than the digit in hundreds place by 3. The digit in thousands place is less than the digit in ones place by 1. If the number at ones place is 8, find the number.

We are given a 6-digit number and several conditions about the relationships between the digits. Let's define the digits as follows:

- Let the 6-digit number be \( abcdef \), where:
  - \( a \) is the hundred thousands digit,
  - \( b \) is the ten thousands digit,
  - \( c \) is the thousands digit,
  - \( d \) is the hundreds digit,
  - \( e \) is the tens digit,
  - \( f \) is the ones digit.

The problem provides the following relationships:

1. **Ones place (f) and ten thousands place (b):**
   \[
   f = b + 7
   \]
   
2. **Hundreds place (d) and thousands place (c):**
   \[
   d = c - 2
   \]
   
3. **Tens place (e) and hundred thousands place (a):**
   \[
   e = a + 4
   \]

4. **Hundred thousands place (a) and hundreds place (d):**
   \[
   a = d - 3
   \]
   
5. **Thousands place (c) and ones place (f):**
   \[
   c = f - 1
   \]

6. **The ones place digit (f) is 8:**
   \[
   f = 8
   \]

---

### Step-by-Step Solution:

Using the information that \( f = 8 \), let's solve the equations step by step:

#### Step 1: Solve for \( b \)
From the first condition \( f = b + 7 \), we know:
\[
8 = b + 7 \implies b = 1
\]

#### Step 2: Solve for \( c \)
From the fifth condition \( c = f - 1 \), we know:
\[
c = 8 - 1 = 7
\]

#### Step 3: Solve for \( d \)
From the second condition \( d = c - 2 \), we know:
\[
d = 7 - 2 = 5
\]

#### Step 4: Solve for \( a \)
From the fourth condition \( a = d - 3 \), we know:
\[
a = 5 - 3 = 2
\]

#### Step 5: Solve for \( e \)
From the third condition \( e = a + 4 \), we know:
\[
e = 2 + 4 = 6
\]

---

### Final Answer:
The digits are \( a = 2 \), \( b = 1 \), \( c = 7 \), \( d = 5 \), \( e = 6 \), and \( f = 8 \).

Therefore, the 6-digit number is:
\[
\boxed{217568}
\]

Would you like more details or have any questions?

Here are 5 related questions to extend the problem:
1. Can you create a similar problem with a different digit relationship?
2. How would the solution change if the ones digit was different?
3. Can we form a 7-digit number using similar constraints?
4. How does the order of solving these equations affect the outcome?
5. Can this method be applied to finding relationships in larger numbers?

**Tip:** When solving word problems with multiple conditions, write down the relationships clearly and solve them step by step, substituting known values as soon as possible.

In a 6-digit number, the digit in ones place is more than the digit in ten thousands place by 7. The digit in hundreds place is less than the digit in thousands place by 2. The digit in tens place is more than the digit in hundred thousands place by 4. The digit in hundred thousands place is less than the digit in hundreds place by 3. The digit in thousands place is less than the digit in ones place by 1. If the number at ones place is 8, find the number.

Solve a 6-digit number puzzle where relationships between digits are provided, such as ones place being 7 more than the ten thousands place. This solution uses algebra and systems of equations. Suitable for students in Grades 6-8.

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