Numerical Alphabet — Concept Explanation

In certain quantitative reasoning or aptitude questions, letters such as \(A, B, C, D, X, Y, Z\) are used to represent digits (0–9). These are called Numerical Alphabet or Letter-Digit Substitution problems. The rules of arithmetic—especially addition, carry, and place value—remain the same as for ordinary numbers.

1. Fundamental Rules

  • Each letter represents exactly one single digit (0–9).
  • Different letters represent different digits unless stated otherwise.
  • Leading digits (first digit of a number) cannot be zero.
  • Arithmetic is performed column-wise from the units place toward the left, obeying the carry-over rules.

Carry Rules in Addition:

  • Sum of two single-digit numbers ⇒ carry can be \(0\) or \(1\).
  • Sum of three single-digit numbers ⇒ carry can be \(0, 1,\) or \(2\).
  • When the sum in a column ≥ 10, we write the unit digit and carry the tens digit to the next column.
Example of Carry Logic:
\( 8 + 7 = 15 \Rightarrow \) write \(5\), carry \(1\). \( 9 + 9 + 1(\text{carry}) = 19 \Rightarrow \) write \(9\), carry \(1\) again.

2. Strategy for Solving Letter–Digit Addition Problems

  1. Start from the rightmost column (units place): Write the possible sum and deduce the carry.
  2. Move left: Add the digits in that column including the carry from the previous column.
  3. Translate conditions: Apply clues like “different digits”, “nonzero digits”, or “sum produces a carry”.
  4. Systematically eliminate contradictions: Check whether your assignments satisfy all columns simultaneously.
  5. Verify the entire sum: Substitute the digits back and confirm the full addition is correct.

This process is entirely logical—no trial and error is necessary if you track column carries carefully.

3. Worked Examples (Step-by-Step)

Example 1: \(XYZ + ZYX = 848\)

Here, \(X, Y, Z\) are distinct digits.

  • Units Column: \(Z + X\) gives a unit digit of \(8\). Thus, \(Z + X = 8 \text{ or } 18\). Carry can be \(0\) or \(1\).
  • Tens Column: \(Y + Y + \text{carry}\) gives the tens digit \(4\). So \(2Y + c_1 \equiv 4 \ (\text{mod }10)\). If \(c_1=0\), then \(2Y=4 \Rightarrow Y=2\). If \(c_1=1\), then \(2Y=3 \Rightarrow\) not possible (Y must be integer). Hence, \(Y=2,\;c_1=0\).
  • Hundreds Column: \(X + Z + c_2 = 8\). Since we already saw \(X+Z=8\), the hundreds place confirms \(c_2=0\).

\(X+Z=8,\;Y=2\Rightarrow X+Y+Z=10\). Final Answer: \(X+Y+Z=10\).

Example 2: \(3A + B7 + 4B + 27 = 178\)

We interpret this as a column addition. \(A\) and \(B\) are digits.

  • Units: \(A + 7 + B + 7 = A + B + 14\). The unit digit is \(8\), so \(A + B + 14 \equiv 8 \pmod{10}\). Hence \(A + B \equiv 4 \pmod{10}\) and the carry is \(2\) (since 14 + something > 10).
  • Tens: \(3 + B + 4 + 2 (\text{carry}) = 9 + B + 2 = 11 + B\). The tens digit of the sum is \(7\), so \(11 + B \equiv 7 \pmod{10} \Rightarrow B=6\).
  • Now \(A + B \equiv 4 \pmod{10} \Rightarrow A + 6 \equiv 4 \pmod{10} \Rightarrow A=8\).

Final Answer: \(A = 8, B = 6\).

Example 3: \(BA + 6B = CAB\)

Letters \(A, B, C\) are distinct. Expanding by place value:

\[ (10B + A) + (60 + B) = 100C + 10A + B \] Simplify: \[ 11B + A + 60 = 100C + 10A + B \Rightarrow 10B + 60 = 100C + 9A \] We look for integers \(A, B, C \in [0,9]\) satisfying this.
  • Try \(C=1 \Rightarrow 10B + 60 = 100 + 9A \Rightarrow 10B = 40 + 9A\).
  • For \(A=4, B=7\), \(10B = 70, 40+9A=76 \Rightarrow\) close but not exact.
  • For \(A=5, B=8 \Rightarrow 10B=80, 40+9A=85\) → too high.
  • For \(A=6, B=9 \Rightarrow 10B=90, 40+9A=94\) → close.
Using column analysis (from original source), the only consistent solution gives \(B=4\). Hence, \(B=4\).

Example 4: \(RP + 7T = 15P\)

Units column: \(P + T\) must give units digit \(P\). That means \(T = 0\) and carry = 0. Tens column: \(R + 7 + 0 = 15\), so tens digit = 5, carry = 1. Therefore \(R + 7 = 15 \Rightarrow R = 8.\)

Final Answer: \(R = 8.\)

Example 5: \(RS + SR = TR4\)

Each letter represents a distinct nonzero digit.

  • Units: \(S + R \equiv 4 \pmod{10}\).
  • Tens: \(R + S + c_1 \equiv R \pmod{10}\) ⇒ the carry \(c_1 = 9 – S\).
  • Hundreds: \(c_2 = 1\) since sum is a three-digit number.
Checking digit combinations that make \(S+R\) end with 4 and consistent carry chain gives \(R=5, S=9, T=1 \Rightarrow 2R + T = 11.\) Final Answer: \(2R + T = 11.\)

Example 6: \(X3Y + 5YX = 1,33X\)

This pattern produces a carry chain forming a four-digit number beginning with 1. Solving systematically gives \(X = 8, Y = 0.\)

Final Answer: \(X = 8, Y = 0.\)

4. Key Takeaways & Common Insights

  • Column Logic: Always work from right to left (Units → Tens → Hundreds).
  • Carry Constraints: The most powerful restriction in every column equation.
  • Symmetry Patterns: Sums like \(XYZ + ZYX\) repeat \(X+Z\) twice—exploit this symmetry to find relationships.
  • Distinct vs. Nonzero: Distinct letters mean unique digits; nonzero rules out zero in leading positions.
  • Verification Step: After deducing all digits, substitute back to confirm that the addition works numerically.
  • Elimination Logic: Quickly discard cases violating carry or modular consistency.

These questions test arithmetic precision and logical consistency more than calculation speed. Mastering column-wise carry analysis makes such puzzles straightforward.

5. Practice Recommendations

  • Practice 2-digit, 3-digit, and mixed addition patterns until column reasoning becomes automatic.
  • Try creating your own letter-digit puzzles and verify solutions programmatically.
  • After mastering addition, move on to subtraction and multiplication letter puzzles (similar principles apply but include borrows or multi-level carries).

Source reference: Based on “GMAT Algebra Topics 2” compiled by Manoj K. Singh (Numerical Alphabet section). Detailed logic and example walkthroughs expanded for conceptual clarity.