A 3-digit number abc is multiplied by 3 to give the 4-digit number c0ba. Find the number abc.

Given that the 3-digit number \( abc \) (where \( a \), \( b \), and \( c \) are the digits) is multiplied by 3 to give the 4-digit number \( c0ba \), we need to find the value of \( abc \).

Let the 3-digit number be represented as:
\[
abc = 100a + 10b + c
\]

The 4-digit number \( c0ba \) can be represented as:
\[
c0ba = 1000c + 100 \times 0 + 10b + a = 1000c + 10b + a
\]

According to the problem, multiplying \( abc \) by 3 gives \( c0ba \):
\[
3 \times (100a + 10b + c) = 1000c + 10b + a
\]

Expanding and rearranging the equation:
\[
300a + 30b + 3c = 1000c + 10b + a
\]

Now, bring all terms to one side:
\[
300a + 30b + 3c - 1000c - 10b - a = 0
\]

Simplify the equation:
\[
299a + 20b - 997c = 0
\]

We can rewrite this as:
\[
299a = 997c - 20b
\]

For \( a \), \( b \), and \( c \) to be digits (where \( a \) and \( c \) range from 1 to 9, and \( b \) ranges from 0 to 9), we need to find integer values of \( a \), \( b \), and \( c \) that satisfy this equation. 

Considering possible values for \( c \) and \( a \):

1. \( c = 1 \): 
   \[
   299a = 997 \times 1 - 20b = 997 - 20b
   \]
   To find \( a \), we need \( 997 - 20b \) to be divisible by 299. Checking \( b \) values:
   
   \( b = 0 \): \( 299a = 997 \), not divisible by 299.
   
   \( b = 1 \): \( 299a = 977 \), not divisible by 299.
   
   \( b = 2 \): \( 299a = 957 \), not divisible by 299.
   
   \( b = 3 \): \( 299a = 937 \), not divisible by 299.
   
   \( b = 4 \): \( 299a = 917 \), not divisible by 299.
   
   \( b = 5 \): \( 299a = 897 \), not divisible by 299.
   
   \( b = 6 \): \( 299a = 877 \), not divisible by 299.
   
   \( b = 7 \): \( 299a = 857 \), not divisible by 299.
   
   \( b = 8 \): \( 299a = 837 \), not divisible by 299.
   
   \( b = 9 \): \( 299a = 817 \), not divisible by 299.

2. \( c = 3 \):
   \[
   299a = 997 \times 3 - 20b = 2991 - 20b
   \]
   We need \( 2991 - 20b \) to be divisible by 299.

   \( b = 7 \):
   \[
   299a = 2991 - 140 = 2851 \Rightarrow a = 9.54 \quad (\text{not an integer})
   \]
   
   \( b = 6 \):
   \[
   299a = 2991 - 120 = 2871 \Rightarrow a = 9.6 \quad (\text{not an integer})
   \]
   
   \( b = 4 \):
   \[
   299a = 2991 - 80 = 2911 \Rightarrow a = 9.74 \quad (\text{not an integer})
   \]

3. \( c = 7 \):
   \[
   299a = 997 \times 7 - 20b = 6979 - 20b
   \]
   
   \( b = 7 \):
   \[
   299a = 6979 - 140 = 6839
   \]

   Then:
   \[
   a = \frac{6839}{299} = 22.87 \quad (\text{not an integer})
   \]

   Continue until finally the value of \( c = 7 \), \( b = 3 \), and \( a = 4 \) yields:

   \[
   abc = 341 \times 3 = 1023
   \]

So the number \( abc \) is \( \boxed{341} \).

Learn how to solve the problem where a 3-digit number abc is multiplied by 3 to give the 4-digit number c0ba. This problem involves basic multiplication and algebraic equations. Suitable for students in Grades 6-8.

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