Urdhva Tiryagbhyam — Vertically and Crosswise Multiplication (Vedic Math)

Urdhva Tiryagbhyam (“Vertically and crosswise”) is the most powerful Vedic multiplication sutra — it works for all numbers, not just special cases. It reduces multi-digit multiplication to a single-line mental calculation.

The Sutra

Urdhva Tiryagbhyam means “Vertically and crosswise.” The method generates partial products by combining digits in a cross-pattern, then sums them with carries.

2-Digit × 2-Digit Pattern

For multiplying AB × CD (two 2-digit numbers):

  A   B
  ×   ×
  C   D

The three partial products are:

1. Vertical: B × D (rightmost column) — gives the units digit (and carry)
2. Crosswise: A×D + B×C (middle cross) — gives the tens digit (and carry)
3. Vertical: A × C (leftmost column) — gives the remaining digits

    flowchart TD
    A["Two 2-digit numbers<br/>AB × CD"] --> B["Step 1: Vertical (B × D)<br/>Units digit with carry"]
    B --> C["Step 2: Crosswise (A×D + B×C)<br/>Tens digit with carry"]
    C --> D["Step 3: Vertical (A × C)<br/>Remaining digits"]
    D --> E["Combine all parts<br/>with carries"]
    E --> F["Final product ✓"]

    style A fill:#1a73e8,color:#fff,stroke:none
    style B fill:#34a853,color:#fff,stroke:none
    style C fill:#fbbc04,color:#333,stroke:none
    style D fill:#ea4335,color:#fff,stroke:none
    style E fill:#ab47bc,color:#fff,stroke:none
    style F fill:#1a73e8,color:#fff,stroke:none
  

Worked Examples

Example 1: 23 × 14 (2-digit × 2-digit)

    2   3
    ×   ×
    1   4

Step 1 — Vertical (right): 3 × 4 = 12

• Units digit: 2, carry 1

Step 2 — Crosswise: (2×4) + (3×1) = 8 + 3 = 11

• Plus carry 1: 11 + 1 = 12
• Tens digit: 2, carry 1

Step 3 — Vertical (left): 2 × 1 = 2

• Plus carry 1: 2 + 1 = 3

Answer: Combine: 3 | 2 | 2 = 322

Check: 23 × 14 = 322 ✓

Example 2: 87 × 93 (2-digit × 2-digit)

    8   7
    ×   ×
    9   3

Step 1: 7 × 3 = 21 → digit 1, carry 2

Step 2: (8×3) + (7×9) = 24 + 63 = 87 + carry 2 = 89 → digit 9, carry 8

Step 3: 8 × 9 = 72 + carry 8 = 80

Answer: 8091

Check: 87 × 93 = 8091 ✓

Example 3: 312 × 214 (3-digit × 3-digit)

For 3-digit numbers, the pattern extends naturally:

    3   1   2
    ×   ×   ×
    2   1   4

Step 1 — Vertical (right): 2 × 4 = 8 → digit 8, carry 0

Step 2 — Crosswise (last two): (1×4) + (2×1) = 4 + 2 = 6 → digit 6, carry 0

Step 3 — Crosswise (outer + inner): (3×4) + (2×2) + (1×1) = 12 + 4 + 1 = 17 → digit 7, carry 1

Wait — let me show this properly. For 3-digit, the pattern is:

    a   b   c
    ×   ×   ×
    d   e   f

The partial products (right to left):

1. c × f
2. b×f + c×e
3. a×f + b×e + c×d
4. a×e + b×d
5. a×d

Step 1: 2 × 4 = 8 → digit 8, carry 0

Step 2: (1×4) + (2×1) = 4 + 2 = 6 → digit 6, carry 0

Step 3: (3×4) + (1×1) + (2×2) = 12 + 1 + 4 = 17 → digit 7, carry 1

Step 4: (3×1) + (1×2) = 3 + 2 = 5 + carry 1 = 6 → digit 6, carry 0

Step 5: 3 × 2 = 6 → digit 6, carry 0

Answer: 66768

Check: 312 × 214 = 66768 ✓

Example 4: 1234 × 5678 (4-digit × 4-digit)

The pattern continues for any digit length:

    1   2   3   4
    ×   ×   ×   ×
    5   6   7   8

Partial products (right to left):

1. 4 × 8 = 32 → digit 2, carry 3
2. (3×8) + (4×7) = 24 + 28 = 52 + 3 = 55 → digit 5, carry 5
3. (2×8) + (3×7) + (4×6) = 16 + 21 + 24 = 61 + 5 = 66 → digit 6, carry 6
4. (1×8) + (2×7) + (3×6) + (4×5) = 8 + 14 + 18 + 20 = 60 + 6 = 66 → digit 6, carry 6
5. (1×7) + (2×6) + (3×5) = 7 + 12 + 15 = 34 + 6 = 40 → digit 0, carry 4
6. (1×6) + (2×5) = 6 + 10 = 16 + 4 = 20 → digit 0, carry 2
7. 1 × 5 = 5 + 2 = 7 → digit 7

Answer: 7006652

Check: 1234 × 5678 = 7,006,652 ✓

Example 5: Polynomial Multiplication (x² + 2x + 1)(x + 3)

The same crosswise pattern works for polynomials:

    1x² + 2x + 1
    ×     ×     ×
    0x² + 1x + 3

Step 1: 1 × 3 = 3 Step 2: (2×3) + (1×1) = 6 + 1 = 7 Step 3: (1×3) + (2×1) + (1×0) = 3 + 2 + 0 = 5 Step 4: (1×1) + (2×0) = 1 Step 5: 1 × 0 = 0

Result: x³ + 5x² + 7x + 3

Code Snippet: Python Implementation

def urdhva_tiryagbhyam(a, b):
    """
    Multiply two integers using Urdhva Tiryagbhyam (vertically and crosswise).
    Works for any positive integers.
    """
    # Convert to digit lists (reversed for right-to-left processing)
    digits_a = [int(d) for d in str(a)][::-1]
    digits_b = [int(d) for d in str(b)][::-1]

    n = len(digits_a)
    m = len(digits_b)
    # Result can have up to n + m digits
    result_digits = [0] * (n + m)

    # Generate partial products using crosswise pattern
    for i in range(n):
        for j in range(m):
            result_digits[i + j] += digits_a[i] * digits_b[j]

    # Handle carries
    carry = 0
    for i in range(len(result_digits)):
        total = result_digits[i] + carry
        result_digits[i] = total % 10
        carry = total // 10

    # Remove leading zeros and convert back
    while len(result_digits) > 1 and result_digits[-1] == 0:
        result_digits.pop()

    return int(''.join(str(d) for d in result_digits[::-1]))

# Test
tests = [(23, 14), (87, 93), (312, 214), (1234, 5678), (999, 999)]
for a, b in tests:
    result = urdhva_tiryagbhyam(a, b)
    print(f"{a} × {b} = {result} (expected: {a*b})")

Expected output:

23 × 14 = 322 (expected: 322)
87 × 93 = 8091 (expected: 8091)
312 × 214 = 66768 (expected: 66768)
1234 × 5678 = 7006652 (expected: 7006652)
999 × 999 = 998001 (expected: 998001)

Code Snippet: JavaScript Implementation

function urdhvaTiryagbhyam(a, b) {
    const digitsA = String(a).split('').map(Number).reverse();
    const digitsB = String(b).split('').map(Number).reverse();
    const n = digitsA.length;
    const m = digitsB.length;
    const result = new Array(n + m).fill(0);

    // Crosswise multiplication
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < m; j++) {
            result[i + j] += digitsA[i] * digitsB[j];
        }
    }

    // Handle carries
    let carry = 0;
    for (let i = 0; i < result.length; i++) {
        const total = result[i] + carry;
        result[i] = total % 10;
        carry = Math.floor(total / 10);
    }

    // Remove leading zeros
    while (result.length > 1 && result[result.length - 1] === 0) {
        result.pop();
    }

    return parseInt(result.reverse().join(''));
}

// Test
[[23, 14], [87, 93], [312, 214], [1234, 5678], [999, 999]].forEach(([a, b]) => {
    console.log(`${a} × ${b} = ${urdhvaTiryagbhyam(a, b)} (expected: ${a * b})`);
});

Common Errors

1. Mixing up the cross pattern. For 3-digit, don’t jump straight to the full cross — build up step by step: right vertical → narrow cross → full cross → tapering cross → left vertical.
2. Forgetting carries between steps. Each partial product sum produces a carry to the next step on the left. Always add the carry before the next multiplication.
3. Off-by-one in digit alignment. The cross pattern must align digits correctly. For 2-digit × 2-digit, the middle cross pairs first×last and middle×middle correctly.
4. Using Urdhva when Nikhilam is faster. Urdhva works for everything, but Nikhilam is faster for numbers near powers of 10. Choose the right tool.
5. Treating 3-digit × 2-digit the same as 3-digit × 3-digit. For different digit lengths, pad the shorter number with leading zeros mentally.
6. Dropping the last carry. The final carry from the leftmost multiplication becomes part of the answer. Don’t drop it.
7. Not practicing mentally. Write the answer from right to left in one line — that’s the goal. Initially use paper, then transition to mental.

Practice Questions

1. 34 × 23 = ?
2. 96 × 87 = ?
3. 456 × 123 = ?
4. 2025 × 104 = ?
5. 9998 × 1234 = ?

Answers:

1. 34 × 23 = 782 (3×4=12→2/c1, (3×3+4×2)=17+1=18→8/c1, 3×2=6+1=7 → 782)
2. 96 × 87 = 8352 (6×7=42→2/c4, (9×7+6×8)=111+4=115→5/c11, 9×8=72+11=83 → 8352)
3. 456 × 123 = 56088
4. 2025 × 104 = 210600
5. 9998 × 1234 = 12337532

Mini Project: Vedic Multiplier Benchmark

Write a script that benchmarks Urdhva Tiryagbhyam against Python’s native multiplication for large numbers (1000+ digits) to see when Vedic approach wins:

import random
import time

def benchmark():
    for size in [10, 100, 500, 1000]:
        a = random.randint(10**(size-1), 10**size - 1)
        b = random.randint(10**(size-1), 10**size - 1)

        # Standard multiplication
        start = time.time()
        result_std = a * b
        time_std = time.time() - start

        # Vedic multiplication
        start = time.time()
        result_vedic = urdhva_tiryagbhyam(a, b)
        time_vedic = time.time() - start

        match = "✓" if result_std == result_vedic else "✗"
        print(f"Size {size}: Vedic={time_vedic:.6f}s Std={time_std:.6f}s {match}")

benchmark()

FAQ

Next Steps

Now learn Digital Roots and Casting Out Nines to verify all your multiplication results in seconds.

Related tutorials:

• Nikhilam — faster multiplication near a base
• Paravartya Yojayet — division using transpose and apply
• Python — explore more Vedic algorithms in code

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