#1

    Ekadhikena Purvena

    "By one more than the previous number."

    💡 Quick Tip

    35² → 3×4 = 12, append 25 → 1225. Works for any number ending in 5!

    What It's Used For

    Calculating squares of numbers ending in 5, multiplying numbers whose first digits add up to 10 and last digits are 5.

    How It Works

    When squaring a number ending in 5, multiply the digit(s) before 5 by the next consecutive number, then append 25.

    This sutra is based on the algebraic identity: (10a + 5)² = 100a(a+1) + 25. The beauty is that you only need to multiply the tens digit by "one more than itself" and write 25 at the end. This extends to any number ending in 5 — for example 105² = 10×11 followed by 025 = 11025. The sutra also applies when multiplying two numbers whose tens digits add to 10 and units digits are both 5, e.g. 35 × 75 = 3×8|5×5 = 2625. In the Vedic system this is considered one of the most elegant and useful shortcuts for mental squaring.

    History & Origin

    This sutra originates from the ancient Indian text "Vedic Mathematics" reconstructed by Bharati Krishna Tirtha (1884–1960), the Shankaracharya of Puri, who claimed to have rediscovered these 16 sutras from the Atharva Veda's appendix (Parishishta). The concept of "one more than the previous" reflects a deep Indian mathematical tradition dating back to the Sulba Sutras (800–500 BCE), where geometric constructions required rapid mental squaring. Indian mathematicians like Aryabhata (476 CE) and Brahmagupta (598 CE) used similar shortcut principles in their astronomical calculations, where squaring numbers was essential for computing planetary positions.

    Scientific & Mathematical Basis

    The algebraic proof is rigorous: (10a+5)² = 100a² + 100a + 25 = 100a(a+1) + 25. This identity holds universally in any base-10 system. From a cognitive science perspective, this sutra reduces a multiplication problem to a simpler one — computing a(a+1) requires less working memory than full squaring. Research in mathematical cognition shows that "chunking" complex operations into simpler sub-operations (as this sutra does) significantly improves speed and accuracy. The pattern also connects to the theory of finite differences in numerical analysis, where consecutive products form predictable sequences.

    Real-World Use Cases

    • Mental math competitions and speed calculation contests — squaring numbers ending in 5 appears frequently
    • Quick area calculations in construction: computing (x5)² for plot dimensions like 25m, 35m, 45m
    • Financial calculations involving percentage computations where squaring 5-ending numbers arises
    • Computer science: optimizing modular arithmetic in cryptographic algorithms
    • Astronomical calculations for angular measurements (degrees often end in 5)

    Step-by-Step Method

    1. 1Identify the digit(s) before 5
    2. 2Multiply that number by the next consecutive number (one more)
    3. 3Append 25 to the result

    📝 Worked Example

    Example 1: 35²
    1. Step 1:Take the number before 5 → 3
    2. Step 2:Multiply by next number → 3 × 4 = 12
    3. Step 3:Append 25 → 1225
    Answer: 1225

    🌟 Beginner Level Examples

    Example 1: 15²
    1. Step 1:Digit before 5 → 1
    2. Step 2:1 × 2 = 2
    3. Step 3:Append 25 → 225
    Answer: 225
    Example 2: 25²
    1. Step 1:Digit before 5 → 2
    2. Step 2:2 × 3 = 6
    3. Step 3:Append 25 → 625
    Answer: 625
    Example 3: 45²
    1. Step 1:Digit before 5 → 4
    2. Step 2:4 × 5 = 20
    3. Step 3:Append 25 → 2025
    Answer: 2025

    🟡 Intermediate Level Examples

    Example 1: 65²
    1. Step 1:Digit before 5 → 6
    2. Step 2:6 × 7 = 42
    3. Step 3:Append 25 → 4225
    Answer: 4225
    Example 2: 85²
    1. Step 1:Digit before 5 → 8
    2. Step 2:8 × 9 = 72
    3. Step 3:Append 25 → 7225
    Answer: 7225
    Example 3: 115²
    1. Step 1:Digits before 5 → 11
    2. Step 2:11 × 12 = 132
    3. Step 3:Append 25 → 13225
    Answer: 13225

    🔴 Advanced Level Examples

    Example 1: 195²
    1. Step 1:Digits before 5 → 19
    2. Step 2:19 × 20 = 380
    3. Step 3:Append 25 → 38025
    Answer: 38025
    Example 2: 305²
    1. Step 1:Digits before 5 → 30
    2. Step 2:30 × 31 = 930
    3. Step 3:Append 25 → 93025
    Answer: 93025
    Example 3: 995²
    1. Step 1:Digits before 5 → 99
    2. Step 2:99 × 100 = 9900
    3. Step 3:Append 25 → 990025
    Answer: 990025

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