Think of a natural number, multiply it by 13, draw the last digit of the obtained number, multiply this number by 7, draw the last digit of the result and get 21. What numbers were you thinking of?

Let's break down the problem step by step:

1. You start by thinking of a natural number, let's call it \( n \).
2. Multiply \( n \) by 13: 
   \[
   13n
   \]
3. Draw the last digit of \( 13n \), which we'll denote as \( d_1 \). So:
   \[
   d_1 = \text{last digit of } 13n
   \]
4. Multiply \( d_1 \) by 7:
   \[
   7d_1
   \]
5. Draw the last digit of \( 7d_1 \), which gives us \( d_2 \). So:
   \[
   d_2 = \text{last digit of } 7d_1
   \]
6. Finally, we are told that this last digit \( d_2 \) is 1. In other words:
   \[
   d_2 = 1
   \]

We need to determine what value of \( n \) satisfies this condition.

### Step 1: Determine Possible Values of \( d_1 \)
The last digit of \( 13n \) is based on the multiplication of 13 with \( n \), so let's check the last digits of multiples of 13:
- \( 13 \times 0 = 0 \) (last digit 0)
- \( 13 \times 1 = 13 \) (last digit 3)
- \( 13 \times 2 = 26 \) (last digit 6)
- \( 13 \times 3 = 39 \) (last digit 9)
- \( 13 \times 4 = 52 \) (last digit 2)
- \( 13 \times 5 = 65 \) (last digit 5)
- \( 13 \times 6 = 78 \) (last digit 8)
- \( 13 \times 7 = 91 \) (last digit 1)
- \( 13 \times 8 = 104 \) (last digit 4)
- \( 13 \times 9 = 117 \) (last digit 7)

So, \( d_1 \) could be 0, 3, 6, 9, 2, 5, 8, 1, 4, or 7.

### Step 2: Check \( 7d_1 \) and the Last Digit of the Result
Now, we multiply each possible \( d_1 \) by 7 and check the last digit:
- \( 7 \times 0 = 0 \) (last digit 0)
- \( 7 \times 3 = 21 \) (last digit 1)
- \( 7 \times 6 = 42 \) (last digit 2)
- \( 7 \times 9 = 63 \) (last digit 3)
- \( 7 \times 2 = 14 \) (last digit 4)
- \( 7 \times 5 = 35 \) (last digit 5)
- \( 7 \times 8 = 56 \) (last digit 6)
- \( 7 \times 1 = 7 \) (last digit 7)
- \( 7 \times 4 = 28 \) (last digit 8)
- \( 7 \times 7 = 49 \) (last digit 9)

From this, we see that only when \( d_1 = 3 \), the last digit of \( 7d_1 \) is 1, which is the condition given in the problem.

### Step 3: Find \( n \)
Since \( d_1 = 3 \), we know that the last digit of \( 13n \) is 3. From the list of possible last digits of multiples of 13, this happens when \( n = 1 \).

### Conclusion:
The number you were thinking of is \( n = 1 \).

Would you like more details, or do you have any other questions?

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Here are five follow-up questions to think about:
1. What would happen if you changed the multiplier from 13 to another number?
2. Can this process work with numbers larger than 1?
3. What would be the last digit if you used 11 instead of 7 in the second multiplication?
4. Can you generalize this problem for any multiplier?
5. What if the final digit were something other than 1—how would that affect the process?

**Tip:** Pay close attention to patterns in number properties, such as last digits, which can simplify complex multiplication problems.

Solve the problem of finding a natural number where multiplying it by 13 and using the last digit, followed by multiplying by 7, results in the last digit being 1. Learn modular arithmetic and last digit patterns in multiplication, suitable for Grades 8-10.

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