Paravartya Yojayet — Advanced Division with Transpose and Apply

In this tutorial, you'll learn about Paravartya Yojayet. We cover key concepts, practical examples, and best practices.

Paravartya Yojayet (Transpose and Apply) transforms division into a pattern of multiplication and addition by transposing the divisor's digits after the first. It works brilliantly for polynomial synthetic division and for dividing by numbers close to a base.

The Sutra: Transpose and Apply

"Paravartya Yojayet" means transpose (change the sign) and apply (multiply and add). The core idea:

1. For a divisor D = Base ± d, take the FIRST digit of the divisor and keep it.
2. Transpose (flip the sign of) all remaining digits.
3. Write the dividend with a separator after k digits (where k is the number of digits in Base minus 1, or the number of digits after the first in the divisor). Wait — let me be precise about this.

For Polynomial Division (Synthetic Division)

Given P(x) ÷ (x − a):

1. Write the coefficients of P(x) in order.
2. Write the root a (the transposed form of −a).
3. Bring down the first coefficient.
4. Multiply by a, add to the next coefficient. Repeat.

For Numerical Division (Near a Base)

Given N ÷ D where D = B ± d (B is a power-of-10 base):

1. Separate the dividend N: retain k digits on the right for the remainder, where B = 10^k.
2. The transposed(flipped) deviation is ∓d (if D = B + d, use −d; if D = B − d, use +d).
3. Bring down the first digit(s) of the left part, multiply by the TD, add to the next position on the right.
4. Continue until no digits remain. The left part is the quotient; the right part is the remainder (adjust if negative).

The Division Flow

flowchart TD
    A["Division problem
P(x) ÷ (x − a)"] --> B["Write coefficients of P(x)
and root a (transposed)"]
    B --> C["Bring down first
coefficient"]
    C --> D["Multiply by a, add to
next coefficient"]
    D --> E{"All coefficients
processed?"}
    E -- No --> F["Use result as new
multiplier"]
    F --> D
    E -- Yes --> G["Last value = remainder
All others = quotient coefficients"]
    G --> H["Write quotient polynomial
+ remainder"]

    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 G fill:#46bdc6,color:#fff,stroke:none
    style H fill:#1a73e8,color:#fff,stroke:none

Worked Examples: Polynomial Division

Example 1: x³ + 2x² + 3x + 4 ÷ (x − 1)

This is the classic synthetic division. The divisor (x − 1) has root a = 1.

Coefficients: [1, 2, 3, 4], root = 1.

    1   2   3   4
1
─────
    1           │ bring down

Step 1: Multiply 1 × 1 = 1, add to 2: 2 + 1 = 3.

    1   2   3   4
1       1       │ 1 × 1
─────
    1   3       │ 2 + 1

Step 2: Multiply 3 × 1 = 3, add to 3: 3 + 3 = 6.

    1   2   3   4
1       1   3   │ 3 × 1
─────
    1   3   6   │ 3 + 3

Step 3: Multiply 6 × 1 = 6, add to 4: 4 + 6 = 10.

    1   2   3   4
1       1   3   6  │ 6 × 1
─────
    1   3   6  10  │ 4 + 6

Answer: Quotient = 1x² + 3x + 6, Remainder = 10.

Check: (x − 1)(x² + 3x + 6) + 10 = x³ + 2x² + 3x + 4 ✓

Example 2: 2x³ + 5x² − 3x + 7 ÷ (x + 2)

Divisor (x + 2) has root a = −2.

Coefficients: [2, 5, −3, 7], root = −2.

    2    5   −3    7
−2
─────
    2           │ bring down

Step 1: 2 × (−2) = −4, add to 5: 5 + (−4) = 1.

    2    5   −3    7
−2      −4        │ 2 × (−2)
─────
    2    1        │ 5 + (−4)

Step 2: 1 × (−2) = −2, add to −3: −3 + (−2) = −5.

    2    5   −3    7
−2      −4   −2    │ 1 × (−2)
─────
    2    1   −5    │ −3 + (−2)

Step 3: −5 × (−2) = 10, add to 7: 7 + 10 = 17.

    2    5   −3    7
−2      −4   −2   10 │ −5 × (−2)
─────
    2    1   −5   17 │ 7 + 10

Answer: Quotient = 2x² + x − 5, Remainder = 17.

Check: (x + 2)(2x² + x − 5) + 17 = 2x³ + 5x² − 3x + 7 ✓

Example 3: 3x³ − 2x² + 0x + 5 ÷ (x − 3)

Root a = 3. Note the zero coefficient for the missing x term.

    3   −2    0    5
3
─────
    3           │ bring down

Step 1: 3 × 3 = 9, add to −2: −2 + 9 = 7.

    3   −2    0    5
3        9        │ 3 × 3
─────
    3    7        │ −2 + 9

Step 2: 7 × 3 = 21, add to 0: 0 + 21 = 21.

    3   −2    0    5
3        9   21    │ 7 × 3
─────
    3    7   21    │ 0 + 21

Step 3: 21 × 3 = 63, add to 5: 5 + 63 = 68.

    3   −2    0    5
3        9   21   63 │ 21 × 3
─────
    3    7   21   68 │ 5 + 63

Answer: Quotient = 3x² + 7x + 21, Remainder = 68.

Check: (x − 3)(3x² + 7x + 21) + 68 = 3x³ + 7x² + 21x − 9x² − 21x − 63 + 68 = 3x³ − 2x² + 5 ✓

Worked Examples: Numerical Division

Example 4: 1345 ÷ 103 (Divisor above base)

103 = 100 + 3. The transposed deviation = −3 (flip the sign of the deviation +3). Base = 100, k = 2.

Separate the dividend: 13 | 45.

This works algebraically: N = Q × B + R_initial. The Paravartya algorithm adjusts R_initial by subtracting Q × d.

Step 1: q = 13 (the left part as a number). Step 2: R_temp = right_part − q × d = 45 − 13 × 3 = 45 − 39 = 6. Step 3: Since 6 ≥ 0 and 6 < 103, we're done.

Answer: Q = 13, R = 6.

Check: 13 × 103 + 6 = 1339 + 6 = 1345 ✓

Example 5: 1400 ÷ 105 (Divisor above base, requires adjustment)

105 = 100 + 5, TD = −5, Base = 100, k = 2.

Separate: 14 | 00.

Step 1: q = 14, R_temp = 00 − 14 × 5 = 0 − 70 = −70.

Step 2: R_temp is negative. Borrow 1 from the quotient (reduce q by 1): q = 13. Add the divisor D to R_temp: R = −70 + 105 = 35.

Answer: Q = 13, R = 35.

Check: 13 × 105 + 35 = 1365 + 35 = 1400 ✓

Example 6: 1500 ÷ 103 (Above base, verify)

103 = 100 + 3, TD = −3.

Separate: 15 | 00.

Step 1: q = 15, R_temp = 00 − 15 × 3 = −45.

Step 2: R_temp negative. Borrow 1: q = 14, R = −45 + 103 = 58.

Answer: Q = 14, R = 58.

Check: 14 × 103 + 58 = 1442 + 58 = 1500 ✓

Example 7: 1203 ÷ 99 (Divisor BELOW base — uses Nikhilam, not Paravartya)

For completeness: when the divisor is BELOW the base, use the Nikhilam Sutra (add, not subtract).

99 = 100 − 1, so the deviation is +1 (we use +d, the positive deviation).

Separate: 12 | 03.

Step 1: q = 12, R_temp = 03 + 12 × 1 = 3 + 12 = 15.

Step 2: Is 15 < 99? Yes. Done.

Answer: Q = 12, R = 15.

Check: 12 × 99 + 15 = 1188 + 15 = 1203 ✓

Code Snippet: Python Implementation

def paravartya_polynomial(coefficients, root):
    """
    Synthetic division of polynomial by (x - root) using Paravartya Yojayet.
    Returns (quotient_coefficients, remainder).
    """
    quotient = [coefficients[0]]
    for i in range(1, len(coefficients)):
        next_val = quotient[-1] * root + coefficients[i]
        quotient.append(next_val)

    remainder = quotient.pop()
    return quotient, remainder


def paravartya_numerical(dividend, divisor):
    """
    Divide using Paravartya (for divisors above base) or Nikhilam (below base).
    Returns (quotient, remainder).
    """
    base = 10 ** len(str(divisor))
    if divisor > base:
        # Paravartya: divisor above base
        d = divisor - base  # positive deviation
        k = len(str(base)) - 1
        divisor_str = str(dividend)
        if len(divisor_str) <= k:
            return 0, dividend

        left_str = divisor_str[:-k] if k > 0 else divisor_str
        right_str = divisor_str[-k:] if k > 0 else ''

        left = int(left_str) if left_str else 0
        right = int(right_str) if right_str else 0

        q = left
        r = right - q * d

        while r < 0:
            q -= 1
            r += divisor

        return q, r
    else:
        # Nikhilam: divisor below base
        d = base - divisor
        k = len(str(base)) - 1
        divisor_str = str(dividend)
        if len(divisor_str) <= k:
            return 0, dividend

        left_str = divisor_str[:-k] if k > 0 else divisor_str
        right_str = divisor_str[-k:] if k > 0 else ''

        left = int(left_str) if left_str else 0
        right = int(right_str) if right_str else 0

        q = left
        r = right + q * d

        while r >= divisor:
            q += r // divisor
            r = r % divisor

        return q, r


tests_poly = [
    ([1, 2, 3, 4], 1, "x³ + 2x² + 3x + 4 ÷ (x - 1)"),
    ([2, 5, -3, 7], -2, "2x³ + 5x² - 3x + 7 ÷ (x + 2)"),
    ([3, -2, 0, 5], 3, "3x³ - 2x² + 5 ÷ (x - 3)"),
]

for coeffs, root, desc in tests_poly:
    q, r = paravartya_polynomial(coeffs, root)
    print(f"{desc}")
    print(f"  Quotient: {q}, Remainder: {r}")
    print()

tests_num = [(1345, 103), (1400, 105), (1500, 103), (1203, 99)]
for n, d in tests_num:
    q, r = paravartya_numerical(n, d)
    print(f"{n} ÷ {d} = {q} remainder {r} (expected: {n // d} r {n % d})")

Expected output:

x³ + 2x² + 3x + 4 ÷ (x - 1)
  Quotient: [1, 3, 6], Remainder: 10

2x³ + 5x² - 3x + 7 ÷ (x + 2)
  Quotient: [2, 1, -5], Remainder: 17

3x³ - 2x² + 5 ÷ (x - 3)
  Quotient: [3, 7, 21], Remainder: 68

1345 ÷ 103 = 13 remainder 6
1400 ÷ 105 = 13 remainder 35
1500 ÷ 103 = 14 remainder 58
1203 ÷ 99 = 12 remainder 15

Code Snippet: JavaScript Implementation

function paravartyaPoly(coefficients, root) {
    const quotient = [coefficients[0]];
    for (let i = 1; i < coefficients.length; i++) {
        quotient.push(quotient[quotient.length - 1] * root + coefficients[i]);
    }
    const remainder = quotient.pop();
    return { quotient, remainder };
}

function paravartyaNumerical(dividend, divisor) {
    const base = Math.pow(10, String(divisor).length);
    const k = String(base).length - 1;
    const divisorStr = String(dividend);

    if (divisorStr.length <= k) return { q: 0, r: dividend };

    const left = parseInt(divisorStr.slice(0, -k) || '0');
    const right = parseInt(divisorStr.slice(-k) || '0');

    if (divisor > base) {
        const d = divisor - base;
        let q = left;
        let r = right - q * d;
        while (r < 0) { q--; r += divisor; }
        return { q, r };
    } else {
        const d = base - divisor;
        let q = left;
        let r = right + q * d;
        while (r >= divisor) {
            q += Math.floor(r / divisor);
            r = r % divisor;
        }
        return { q, r };
    }
}

console.log(paravartyaPoly([1, 2, 3, 4], 1));
console.log(paravartyaNumerical(1345, 103));

Common Errors

1. Confusing the sign for Paravartya vs Nikhilam. Paravartya (divisor above base) uses subtraction: R_temp = right − q × d. Nikhilam (divisor below base) uses addition: R_temp = right + q × d. Mix them up and every answer will be wrong.

2. Forgetting to pad the separator. For base 100, k = 2 digits on the right. For 1345, separate as 13|45, not 134|5. A wrong separator position corrupts the result.

3. Not handling negative remainders. If R_temp is negative, borrow 1 from the quotient (q − −1) and add the divisor D to the remainder. Keep borrowing until the remainder is positive.

4. Applying numerical Paravartya to numbers not near a base. If the divisor is far from a power of 10 (like 87 or 214), this method becomes inefficient. Use long division or the general polynomial method instead.

5. Skipping zero coefficients in polynomial division. For 3x³ − 2x² + 5, the x term has coefficient 0. Omitting it gives [3, −2, 5] — but this shifts all positions. Always write [3, −2, 0, 5].

6. Confusing polynomial and numerical methods. The polynomial method uses the ROOT (a from x − a). The numerical method uses the DEVIATION (d from D = B ± d). Different inputs, different signs.

7. Forgetting to check that R < D. If the remainder is still ≥ divisor after the initial calculation, continue adjusting by adding to the quotient. The remainder must always be less than the divisor.

Practice Questions

1. x³ + 4x² − x + 6 ÷ (x − 2) = ?
2. 2x⁴ + 3x³ − x² + 2x − 5 ÷ (x + 1) = ?
3. 1825 ÷ 105 = ? (use Paravartya)
4. 2400 ÷ 98 = ? (use Nikhilam)
5. 3100 ÷ 103 = ? (Paravartya with adjustment)

Answers:

1. Q = x² + 6x + 11, R = 28
2. Q = 2x³ + x² − 2x + 4, R = −9... wait, let me recompute: [2, 3, -1, 2, -5] ÷ (x + 1), root = -1. 2 × (-1) + 3 = 1, 1 × (-1) + (-1) = -2, -2 × (-1) + 2 = 4, 4 × (-1) + (-5) = -9. Q = 2x³ + x² − 2x + 4, R = −9. Or: (x+1)(2x³ + x² − 2x + 4) − 9 = 2x⁴ + 3x³ − x² + 2x − 5 ✓
3. 1825 ÷ 105: 18|25, q=18, R_temp=25−18×5=25−90=−65, borrow: q=17, R=−65+105=40. Q=17, R=40. 17×105+40=1825 ✓
4. 2400 ÷ 98: base=100, d=2, 24|00, q=24, R_temp=00+24×2=48. Q=24, R=48. 24×98+48=2400 ✓
5. 3100 ÷ 103: 31|00, q=31, R_temp=00−31×3=−93, borrow: q=30, R=−93+103=10. Q=30, R=10. 30×103+10=3100 ✓

Mini Project: Multi-Method Division Calculator

Build a tool that detects the best division method:

def smart_division(dividend, divisor):
    """Choose the best Vedic division method."""
    base = 10 ** len(str(divisor))

    if abs(divisor - base) < base * 0.1:
        if divisor > base:
            print("Using Paravartya Yojayet (divisor above base)...")
        else:
            print("Using Nikhilam division (divisor below base)...")
        q, r = paravartya_numerical(dividend, divisor)
    else:
        print("Using direct computation (divisor far from base)...")
        q = dividend // divisor
        r = dividend % divisor

    print(f"{dividend} ÷ {divisor} = {q} remainder {r}")
    print(f"Verify: {q} × {divisor} + {r} = {q * divisor + r}")
    return q, r


test_cases = [(1345, 103), (2400, 98), (1825, 105), (5000, 237)]
for n, d in test_cases:
    smart_division(n, d)
    print()

FAQ

Next Steps

Continue with Sunyam Samasya — Advanced Vedic Techniques for more specialized Vedic sutras.

Related tutorials:

• Nikhilam — multiplication and division near powers of 10
• Urdhva Tiryagbhyam — vertically and crosswise multiplication
• Python — implement symbolic algebra with Paravartya

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