o3-mini-twentyfive

gistfile1.txt

<USER>
### 4. Race to 2025!

```
   G N A J
+     N A
-----------
   2 0 2 5
```

In this coded addition, different letters represent different digits, and the same digit is always represented by the same letter. Additionally, none of the multi-digit numbers start with zero.

**What is the value of NA?**

<ASSISTANT>
We are given the addition problem:

\[
\begin{array}{r@{\;}r@{\;}r@{\;}r}
   & G & N & A & J\\[5mm]
+  &   & N & A & \\ \hline
   & 2 & 0 & 2 & 5\\
\end{array}
\]

Each letter represents a different digit (with no leading zero in any multi‐digit number), and we wish to determine the two-digit number \(NA\).

### Step 1. Set Up the Problem Column by Column

Since the numbers are aligned to the right, we label the columns from right (ones) to left (thousands):

1. **Ones column:**  
   \(J + A\) must yield a ones digit of \(5\) (plus a possible carry \(c_1\) into the tens column). So,  
   \[
   J + A = 5 + 10c_1 \quad \text{with } c_1 \in \{0,1\}.
   \]

2. **Tens column:**  
   The tens column has \(A\) (from \(GNAJ\)) and \(N\) (from \(NA\)) plus the carry \(c_1\) from the ones column. This sum gives the tens digit \(2\) and produces a carry \(c_2\) into the hundreds column:
   \[
   A + N + c_1 = 2 + 10c_2.
   \]

3. **Hundreds column:**  
   In the hundreds column, only the digit \(N\) from the number \(GNAJ\) contributes, along with the carry \(c_2\) from the tens column. The hundreds digit in the answer is \(0\):
   \[
   N + c_2 = 0 + 10c_3.
   \]
   Since \(N\) is a nonzero digit (it is the first digit of the two-digit number \(NA\)), the only possibility is:
   \[
   N + c_2 = 10.
   \]
   This forces:
   - \(c_2 = 1\) and hence \(N = 9\).

4. **Thousands column:**  
   In the thousands column, we have \(G\) (from \(GNAJ\)) plus the carry \(c_3\) from the hundreds column. The thousands digit in the answer is \(2\):
   \[
   G + c_3 = 2.
   \]
   We will determine \(c_3\) shortly.

### Step 2. Solve the Tens Column

We already found \(N = 9\) and \(c_2 = 1\). Now return to the tens column equation:
\[
A + 9 + c_1 = 2 + 10 \quad \Longrightarrow \quad A + 9 + c_1 = 12.
\]
Thus,
\[
A + c_1 = 3.
\]

Since \(A\) is a digit (and must be different from \(N=9\)) and \(c_1\) is either 0 or 1, we have two cases:

- **Case 1:** \(c_1 = 0\) then \(A = 3\).
- **Case 2:** \(c_1 = 1\) then \(A = 2\).

We will check these with the ones column.

### Step 3. Solve the Ones Column

The ones column equation is:
\[
J + A = 5 + 10c_1.
\]

- **Case 1 (\(c_1 = 0\), \(A = 3\)):**  
  Then,
  \[
  J + 3 = 5 \quad \Longrightarrow \quad J = 2.
  \]
  This gives a valid digit for \(J\).

- **Case 2 (\(c_1 = 1\), \(A = 2\)):**  
  Then,
  \[
  J + 2 = 15 \quad \Longrightarrow \quad J = 13,
  \]
  which is impossible because \(J\) must be a single digit.

Thus, only **Case 1** is acceptable. So, we have:
\[
c_1 = 0, \quad A = 3, \quad J = 2.
\]

### Step 4. Solve the Thousands Column

We had the thousands column:
\[
G + c_3 = 2.
\]

The carry \(c_3\) comes from the hundreds column. We already had:
\[
N + c_2 = 9 + 1 = 10,
\]
which means the hundreds column produced a carry of \(c_3 = 1\). Now substitute:
\[
G + 1 = 2 \quad \Longrightarrow \quad G = 1.
\]

### Step 5. Verify the Complete Addition

We now have:
- \(G = 1\)
- \(N = 9\)
- \(A = 3\)
- \(J = 2\)

Plug these into the original addition:
\[
\begin{array}{r@{\;}r@{\;}r@{\;}r}
   & 1 & 9 & 3 & 2\\[5mm]
+  &   & 9 & 3 & \\ \hline
   & 2 & 0 & 2 & 5\\
\end{array}
\]
Check:
- \(1932 + 93 = 2025\). This is correct.

### Step 6. Answer the Question

The two-digit number \(NA\) is formed by \(N\) and \(A\):
\[
NA = 93.
\]

Thus, the value of \(NA\) is:

\[
\boxed{93}
\]