Ekadhikena Purvena — Squaring Numbers Ending in 5 (Vedic Math)

Ekadhikena Purvena (“By one more than the previous one”) is a Vedic sutra that lets you square any number ending in 5 in under 2 seconds — mentally, without a calculator.

The Sutra

Ekadhikena Purvena translates to “By one more than the previous one.”

For a number ending in 5 (like 35, 85, 125, or 1005):

1. Take the digit(s) before the final 5 — call this the previous part.
2. Add 1 to it — this is ekadhikena (one more).
3. Multiply the previous part by this result.
4. Append 25 to the product.

That’s the square.

How the Method Works

    flowchart TD
    A["Number ending in 5<br/>e.g., 85"] --> B["Split: remove last 5<br/>Previous part = 8"]
    B --> C["Add 1: 8 + 1 = 9<br/>(Ekadhikena — one more)"]
    C --> D["Multiply: 8 × 9 = 72"]
    D --> E["Append 25: 7225"]
    E --> F["85² = 7225 ✓"]

    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: 35²

Step 1: Identify the previous part — the digits before the last 5.

• Number: 35 → Previous part = 3

Step 2: One more than the previous: 3 + 1 = 4

Step 3: Multiply: 3 × 4 = 12

Step 4: Append 25 → 1225

Answer: 35² = 1225

Example 2: 85²

Step 1: Previous part = 8

Step 2: One more: 8 + 1 = 9

Step 3: Multiply: 8 × 9 = 72

Step 4: Append 25 → 7225

Answer: 85² = 7225

Example 3: 125²

Step 1: Previous part = 12 (the digits before the 5)

Step 2: One more: 12 + 1 = 13

Step 3: Multiply: 12 × 13 = 156

Step 4: Append 25 → 15625

Answer: 125² = 15625

Example 4: 1005²

Step 1: Previous part = 100

Step 2: One more: 100 + 1 = 101

Step 3: Multiply: 100 × 101 = 10100

Step 4: Append 25 → 1010025

Answer: 1005² = 1,010,025

Example 5: 9995²

Step 1: Previous part = 999

Step 2: One more: 999 + 1 = 1000

Step 3: Multiply: 999 × 1000 = 999000

Step 4: Append 25 → 99900025

Answer: 9995² = 99,900,025

Example 6: 3.5² (Decimal Extension)

The sutra works for decimals too if the last digit of the whole part is 5:

3.5² = ?

Treat 35² = 1225, then place decimal: 12.25.

Alternatively: 3.5 × 3.5 = 3 × 4 + 0.25 = 12.25.

Why Does This Work? — Algebraic Proof

Let a number ending in 5 be written as (10n + 5):

[ (10n + 5)^2 = 100n^2 + 100n + 25 = 100n(n + 1) + 25 ]

(n(n+1)) is “previous part × (previous part + 1)”, and multiplying by 100 shifts it two places before appending 25. The sutra is algebraically exact.

Code Snippet: Python Implementation

Here’s a Python function that squares any number ending in 5 using Ekadhikena Purvena:

def ekadhikena_square(n):
    """
    Square a number ending in 5 using Ekadhikena Purvena.
    Works for integers and floats.
    """
    # Handle floats like 3.5
    if isinstance(n, float):
        str_n = str(n)
        # Count decimal places
        decimal_places = len(str_n.split('.')[1])
        integer_part = str_n.replace('.', '')
        # Square using sutra, then adjust decimal
        result = ekadhikena_square(int(integer_part))
        return result / (10 ** (2 * decimal_places))

    # Integer path
    str_n = str(n)
    if str_n[-1] != '5':
        raise ValueError("Number must end with 5")

    previous = int(str_n[:-1])  # digits before the 5
    one_more = previous + 1     # Ekadhikena
    product = previous * one_more
    return int(f"{product}25")

# Test cases
tests = [15, 25, 35, 85, 125, 1005, 9995]
for t in tests:
    result = ekadhikena_square(t)
    print(f"{t}² = {result}")

print(f"3.5² = {ekadhikena_square(3.5)}")

Expected output:

15² = 225
25² = 625
35² = 1225
85² = 7225
125² = 15625
1005² = 1010025
9995² = 99900025
3.5² = 12.25

Code Snippet: JavaScript Implementation

function ekadhikenaSquare(n) {
    // Handle floats
    if (!Number.isInteger(n)) {
        const str = n.toString();
        const decimalPlaces = str.split('.')[1].length;
        const integerPart = parseInt(str.replace('.', ''));
        const result = ekadhikenaSquare(integerPart);
        return result / Math.pow(10, 2 * decimalPlaces);
    }

    const str = n.toString();
    if (str[str.length - 1] !== '5') {
        throw new Error("Number must end with 5");
    }

    const previous = parseInt(str.slice(0, -1));
    const oneMore = previous + 1;
    const product = previous * oneMore;
    return parseInt(product.toString() + '25');
}

// Test
[15, 25, 35, 85, 125, 1005, 9995, 3.5].forEach(n => {
    console.log(`${n}² = ${ekadhikenaSquare(n)}`);
});

Expected output:

15² = 225
25² = 625
35² = 1225
85² = 7225
125² = 15625
1005² = 1010025
9995² = 99900025
3.5² = 12.25

Common Errors

1. Including the 5 in the previous part. For 85, the previous part is 8, not 85. Only digits before the final 5.
2. Forgetting to append 25. The sutra always ends with 25 appended. Some beginners add 25 erroneously.
3. Using it on numbers not ending in 5. Ekadhikena Purvena only applies when the last digit is 5. For 36², use Urdhva Tiryagbhyam instead.
4. Mis-handling carries in multiplication. When the previous part × (previous+1) is large, carry properly. For 995², previous=99, 99×100=9900, append 25 → 990025.
5. Applying to negative numbers without sign handling. The sutra works for negatives too: (-35)² = 1225, but the sign becomes positive regardless.
6. Decimal errors. For 3.5², beginners err by writing 1225 or 122.5 instead of 12.25. Count decimal places carefully.
7. Zero-prefix confusion. 5² = 25. Previous part of 5 is 0 (not empty). 0 × 1 = 0, append 25 → 025 → 25. Correct.

Practice Questions

1. 65² = ?
2. 195² = ?
3. 5005² = ?
4. 99995² = ?
5. 4.5² = ?

Answers:

1. 65² = 4225 (6 × 7 = 42, append 25)
2. 195² = 38025 (19 × 20 = 380, append 25)
3. 5005² = 25,050,025 (500 × 501 = 250500, append 25)
4. 99995² = 9,999,000,025 (9999 × 10000 = 99990000, append 25)
5. 4.5² = 20.25 (45² = 2025, two decimal places → 20.25)

Mini Project: Interactive Squaring Calculator

Write a program that continuously accepts numbers from the user and squares any number ending in 5 using Ekadhikena Purvena, rejecting others with a helpful error:

def interactive_squarer():
    while True:
        inp = input("Enter a number ending in 5 (or 'q' to quit): ")
        if inp.lower() == 'q':
            break
        try:
            n = int(inp)
            print(f"{n}² = {ekadhikena_square(n)}")
        except ValueError:
            print("Invalid input. Enter an integer ending in 5.")

if __name__ == "__main__":
    interactive_squarer()

Try it: Enter 125 → outputs 15625. Enter 33 → outputs error message.

Extend this to generate a practice quiz with random numbers ending in 5, timing each answer.

FAQ

Next Steps

Now that you’ve mastered squaring numbers ending in 5, move on to Nikhilam — Multiplication Near a Base to multiply numbers close to 100, 1000, etc. in seconds.

Related tutorials:

• Urdhva Tiryagbhyam — general multiplication for any number
• Digital Roots — verify your squares with casting out nines
• Python — build more math utilities

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