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Meaning: "By one more than the previous one". This sutra simplifies squaring numbers ending in 5 and certain multiplications where the last digits sum to 10 and preceding digits are the same.

Why is it super useful?

Squaring Numbers Ending in 5: If you need to find the square of any number that ends with the digit 5 (like 15, 25, 35, 105, etc.), this sutra makes it super easy and fast! No long multiplication needed. The process is:

1. The last part of the answer is always 25 (since 5x5=25).
2. For the first part, take the digit(s) before the 5, let's call this 'N'. Multiply N by (N+1). Combine this result with 25 to get your answer.

Special Multiplications: It can also be used for some other specific types of multiplication, especially when numbers are close to multiples of 10, or when one number can be conveniently related to 'one more than' a part of the other. For instance, multiplying two numbers where the sum of their last digits is 10 and the preceding digits are the same (e.g., 43 x 47). Here, the last part is 3x7=21. The first part is 4 x (4+1) = 4x5 = 20. So, 43x47 = 2021.

Are there any rules or restrictions?

For Squaring: The main restriction (and the magic part!) is that the number must end in 5. It won't work directly for squaring numbers like 23 or 47 if you're trying to use the "ending in 5" shortcut.

For General Multiplication (sum of last digits is 10): The preceding digits of both numbers must be identical. For example, it works for 62 x 68 but not for 62 x 78.

Conceptual Understanding: This sutra is a specific application of algebraic identities. For numbers ending in 5, like (10N+5)², it expands to 100N² + 100N + 25, which is 100N(N+1) + 25. This directly shows why we multiply N by (N+1) for the first part and append 25.

Meaning: "All from 9 and the last from 10". This sutra is a cornerstone for subtraction from powers of 10 (like 100, 1000) and for multiplying numbers near these bases.

Subtraction from Powers of 10:

To subtract a number from a power of 10 (e.g., 1000 - 457):

1. Take the number being subtracted (minuend, here 457). Starting from its leftmost digit, subtract each digit from 9.
   (9 - 4 = 5), (9 - 5 = 4).
2. Subtract the very last digit of the minuend from 10.
   (10 - 7 = 3).
3. Combine these results: 543. So, 1000 - 457 = 543.

Note: The number of digits in the result should match the number of zeros in the power of 10. If the minuend has fewer digits than zeros in the base, pad it with leading zeros (e.g., 1000 - 57 becomes 1000 - 057).

Multiplication Near a Base:

For numbers close to a base (e.g., 98 x 97, base 100):

1. Find the 'deficiencies' from the base:
   98 is 2 less than 100 (deficiency = -2 or simply 2).
   97 is 3 less than 100 (deficiency = -3 or simply 3).
2. The answer has two parts:
   First part (LHS): Subtract one deficiency from the other number (or cross-subtract). E.g., 98 - 3 = 95 (or 97 - 2 = 95).
   Second part (RHS): Multiply the deficiencies: (-2) x (-3) = 6. Since the base 100 has two zeros, the RHS should have two digits. So, write 06.
3. Combine: 9506. So, 98 x 97 = 9506.

If numbers are above the base (e.g., 103 x 104), the 'deficiencies' are 'surpluses' (+3, +4). The process is similar:
LHS: 103 + 4 = 107 (or 104 + 3 = 107).
RHS: 3 x 4 = 12.
Answer: 10712.

Why it works (Conceptual):

For subtraction, 1000 - N is like 999 - N + 1. Subtracting each digit of N from 9 (except the last) and the last from 10 achieves this. For multiplication, (Base - a)(Base - b) = Base² - Base(a+b) + ab. The Nikhilam method cleverly arranges this.

Meaning: "Vertically and crosswise". This general method for multiplication applies to any two numbers, breaking down the process into simpler vertical and crosswise steps.

How it works (e.g., for 2-digit by 2-digit numbers like AB x CD):

1. Step 1 (Rightmost Vertical): Multiply the rightmost digits vertically: B x D. This gives the rightmost digit(s) of the answer.
2. Step 2 (Crosswise): Multiply diagonally and add the products: (A x D) + (B x C). This gives the middle digit(s) of the answer. If the sum is a two-digit number, carry over the tens digit to the next step.
3. Step 3 (Leftmost Vertical): Multiply the leftmost digits vertically: A x C. Add any carry-over from Step 2 to this product. This gives the leftmost digit(s) of the answer.
4. Combine: Arrange the results from these steps to form the final product.

For larger numbers (e.g., 3-digit by 3-digit: ABC x DEF): The pattern extends. More crosswise products are involved.

• Rightmost: C x F
• Crosswise 1: (B x F) + (C x E)
• Crosswise 2 (Star product): (A x F) + (C x D) + (B x E)
• Crosswise 3: (A x E) + (B x D)
• Leftmost: A x D

Each step's result contributes to a part of the final answer, managing carries as you go from right to left.

Advantages:

• Universal: Works for all multiplication.
• Mental Math: With practice, it can be performed mentally.
• Efficient: Reduces the number of intermediate steps compared to traditional long multiplication.
• Foundation for algebraic multiplication: The same principle applies to multiplying algebraic expressions.

Meaning: "Transpose and apply" or "Transpose and adjust".

Primary Applications:

• Algebraic Division: This sutra provides an elegant method for division of polynomials, especially when the divisor is a simple linear binomial (like x - a or x + a) or can be easily related to one. It's also applicable for division of numbers when the divisor is slightly greater or slightly less than a power of 10.
  Example: Divide 123 by 11. Base is 10. Divisor 11 is 1 more than base. Transposed digit is -1.
  

    1 | 2  3 
      | -1 -1 
    ----------- 
    1   1 | 2

  (Quotient 11, Remainder 2).
  Bring down 1. Multiply 1 by -1 = -1. Add to 2 -> 1. Multiply 1 by -1 = -1. Add to 3 -> 2.
• Solving Linear Equations: It's used to solve linear equations by transposing terms. When a term moves from one side of the equation to the other, its sign changes. For example, in ax + b = c, transposing b gives ax = c - b. Then transposing 'a' (as a divisor) gives x = (c - b) / a.
• Solving Specific Types of Equations: The sutra is particularly powerful for equations of the form ax + b = cx + d. By transposing, we get ax - cx = d - b, so x(a-c) = d-b, leading to x = (d-b)/(a-c). More complex forms like (x+a)(x+b) = (x+c)(x+d) can also be solved by first expanding and then applying transposition, or by noticing specific relationships between a, b, c, d that simplify using Paravartya principles.

Core Idea: The sutra embodies the principle of "changing sides, changing signs" and using the "adjusted" or "transposed" values in subsequent operations. In division, the digits of the divisor (relative to a base) are effectively "transposed" (signs changed) and used in a simplified multiplication-addition process instead of conventional subtraction.

Meaning: "When the Samuccaya is the same, that Samuccaya is zero". 'Samuccaya' is a versatile term that can mean sum, product, combination, or a common factor/expression depending on the context.

Primary Application: This sutra is a powerful tool for solving certain types of algebraic equations where a commonality (the 'Samuccaya') can be identified. If this commonality behaves in specific ways across the equation, it implies that this common factor or expression itself must be equal to zero to satisfy the equation.

Key Scenarios where it's applied:

1. Common Factor in All Terms: If an expression 'X' is a common factor in every term of an equation (e.g., aX + bX = cX or aX + bX = 0), then X = 0 is one of the solutions.
2. Sum of Numerators and Denominators in Fractions:
   • If N1/D1 = N2/D2 and N1 + N2 = D1 + D2 (sum of numerators equals sum of denominators), then this sum is zero, leading to a solution.
   • In an equation like (ax+b)/(px+q) = (cx+d)/(rx+s), if (ax+b) + (cx+d) = (px+q) + (rx+s), then this sum equated to zero gives a solution.
3. If Numerators are Equal: In N/D1 = N/D2, if N is not zero, then D1 = D2. If the equation is N/D1 + N/D2 = 0, then D1 + D2 = 0 gives a solution (assuming N is not zero).
4. If Denominators are Equal: In N1/D = N2/D, then N1 = N2. If N1/D + N2/D = 0, then N1 + N2 = 0 gives a solution (assuming D is not zero).
5. Sum of Numerators is Same on Both Sides, and Sum of Denominators is Same on Both Sides: If N1/D1 + N2/D2 = N3/D3 + N4/D4, and N1+N2 = N3+N4, and D1+D2 = D3+D4, then this condition itself doesn't directly make something zero unless other specific forms are met. The classic application is simpler, e.g., if the sum of numerators of two fractions is equal to the sum of their denominators, then that sum is zero.

A classic "Shunyam Samuccaye" example: If (x+a) is a common factor in an equation, then x+a = 0 (i.e., x = -a) is a solution.

Meaning: "If one is in ratio, the other is zero" or "When there is proportionality, the other (variable part) is zero". This sutra deals with situations involving ratios or proportions.

Primary Application: This sutra is typically applied to solve a specific system of two simultaneous linear equations. The conditions are:

1. The ratio of the coefficients of 'x' in both equations is the same as the ratio of the coefficients of 'y'.
   (i.e., for equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂, if a₁/a₂ = b₁/b₂).
2. This common ratio (from condition 1) is different from the ratio of the constant terms.
   (i.e., a₁/a₂ = b₁/b₂ ≠ c₁/c₂).

If these conditions are met, the sutra implies that for a unique solution to exist under such "proportionality" of variable coefficients but "disproportionality" of constants, the variables themselves (x and y) must be zero. Geometrically, this represents two parallel lines that are distinct (not overlapping), which can only "meet" (have a common solution) at the origin if forced by constants being zero, or have no solution if constants are non-zero.

Simplified Interpretation: If the variable parts of two equations are proportional (one is a multiple of the other), but the constant parts are not proportionally related in the same way, then the only way to satisfy both equations simultaneously (if a unique solution is sought) is if the variables are zero. This assumes a system that would otherwise be inconsistent.

Example: Consider the system:
2x + 3y = 5
4x + 6y = 12

Ratio of x-coefficients: 2/4 = 1/2.
Ratio of y-coefficients: 3/6 = 1/2.
Ratio of constants: 5/12.

Here, 1/2 ≠ 5/12. The coefficients of x and y are in ratio, but the constants are not. This system has no solution (parallel distinct lines). The sutra "Anurupye Shunyamanyat" would imply x=0, y=0 if we were trying to force a solution under specific constraints related to a theoretical point of commonality when the system should otherwise be inconsistent. In its direct application, if such a system (2x+3y=k1, 4x+6y=k2 with k1/k2 not 1/2) must have a solution, that solution is x=0,y=0, implying k1 and k2 must then also be 0. The sutra is subtle and applies to specific forms or interpretations where such proportionality leads to a zero value for the variables.

Meaning:

"By addition and by subtraction".

Primary Application:

This sutra provides a very elegant and straightforward method for solving a specific type of system of two simultaneous linear equations. The key characteristic of such a system is that the coefficients of 'x' in the first equation and 'y' in the second equation are the same, AND the coefficients of 'y' in the first equation and 'x' in the second equation are the same. In other words, the coefficients of x and y are interchanged between the two equations.

Format of equations:
ax + by = c₁
bx + ay = c₂

The Method:

1. Addition (Sankalana): Add the two given equations. This will result in a new equation where the coefficients of x and y are (a+b).
   (ax + by) + (bx + ay) = c₁ + c₂
   (a+b)x + (a+b)y = c₁ + c₂
   (a+b)(x+y) = c₁ + c₂
   So, x+y = (c₁ + c₂) / (a+b). Let's call this Equation 3.
2. Subtraction (Vyavakalana): Subtract the second equation from the first (or vice-versa, ensuring consistency).
   (ax + by) - (bx + ay) = c₁ - c₂
   (a-b)x - (a-b)y = c₁ - c₂ (if a-b is taken common, or (a-b)x + (b-a)y )
   (a-b)(x-y) = c₁ - c₂
   So, x-y = (c₁ - c₂) / (a-b). Let's call this Equation 4.
3. Solve the New System: You now have a much simpler system of two linear equations (Equation 3 and Equation 4):
   x + y = P
   x - y = Q
   (where P = (c₁+c₂)/(a+b) and Q = (c₁-c₂)/(a-b))
   Adding these two new equations: 2x = P + Q => x = (P+Q)/2.
   Subtracting the second new equation from the first: 2y = P - Q => y = (P-Q)/2.

Advantages:

This method avoids complex multiplications or divisions often encountered in standard elimination or substitution methods for such specific systems, making calculations quicker and less prone to errors.

Meaning: "By the completion or non-completion (of the square or other forms)". This sutra often relates to the process of completing the square in quadratic equations or manipulating expressions to achieve a 'complete' or recognizable form.

Primary Applications:

• Solving Quadratic Equations: This is the most direct application, where the method of "completing the square" is used. For an equation like ax² + bx + c = 0, you manipulate it to form a perfect square trinomial on one side.
  Example: x² + 6x = 7. To "complete" x² + 6x into a square, take half the coefficient of x (6/2 = 3) and square it (3² = 9). Add this to both sides: x² + 6x + 9 = 7 + 9. This becomes (x+3)² = 16. Now it's easily solvable.
• Simplifying Expressions: It can be used to simplify algebraic expressions by recognizing or forcing parts of them into perfect squares or other "complete" forms. This can make factorization or further manipulation easier.
• Calculus (Integration): In integration, sometimes algebraic manipulation using completion of the square is necessary to transform an integrand into a standard form for which an integration formula is known. For example, integrals involving expressions like 1/(ax² + bx + c).
• Geometric Interpretations: The idea of "completing" can be visualized geometrically, for instance, by adding smaller squares or rectangles to an existing shape to form a larger, complete square.
• "Non-completion" aspect: This can refer to situations where an expression is close to a complete form but isn't quite. Recognizing this "non-completion" might lead to strategies like "difference of squares" factorization. For example, x² - 7 can be seen as "non-complete" for x² - a², but by thinking of 7 as (√7)², we can factor it as (x - √7)(x + √7).

Core Idea: The sutra emphasizes transforming expressions into more manageable or standard forms by adding or subtracting necessary components to achieve "completion," or by leveraging how close an expression is to a "complete" form.

Meaning: "Differences and Similarities" or more broadly, "By differentiation and integration" (Sequential calculus operations).

Context: This sutra is profound and directly points towards the principles of Calculus. 'Chalana' (चलन) refers to the study of change, rates of change, and differences, which is the domain of Differential Calculus. 'Kalana' (कलन) refers to collection, accumulation, or summation, which is the domain of Integral Calculus.

Applications & Interpretations:

1. Differential Calculus (Chalana):
   • Finding derivatives (instantaneous rates of change).
   • Analyzing slopes of curves, maxima and minima of functions.
   • Understanding how one quantity changes with respect to another.
2. Integral Calculus (Kalana):
   • Finding areas under curves, volumes of solids.
   • Summing up infinitesimally small quantities (accumulation).
   • The reverse process of differentiation.
3. Relationship between Differentiation and Integration: The sutra also implies the fundamental theorem of calculus, which links differentiation and integration as inverse operations.
4. Solving Differential Equations: These are equations involving functions and their derivatives. This sutra provides the conceptual basis for the techniques used to solve them.
5. Applications in Physics and Engineering: Many physical phenomena involving motion, flow, growth, decay, etc., are modeled using calculus, derived from these principles of "differences" and "accumulations."

Philosophical Aspect: Beyond mathematics, "Chalana-Kalana" can be interpreted as the interplay between analysis (breaking down into parts/differences) and synthesis (combining parts into a whole/similarities) in any field of study.

Vedic Context: While modern calculus was formalized later, the presence of such sutras suggests that the ancient Indian mathematicians had a deep understanding of the concepts of change and accumulation, which are the seeds of calculus.

Meaning: "By the deficiency" or "Whatever the extent of its deficiency". This is a specific case of the Nikhilam sutra, primarily applied to squaring numbers.

Primary Application: Squaring Numbers Near a Base

This sutra is exceptionally useful for squaring numbers that are close to a power of 10 (like 10, 100, 1000, etc.).

How it works (for a number N near a base B):

1. Find the Deficiency (or Surplus): Calculate the difference between the number (N) and the chosen base (B). Let this be 'd'.
   If N < B, then d = B - N (deficiency). The working value is -d.
   If N > B, then d = N - B (surplus). The working value is +d.
2. Calculate the First Part (LHS) of the Answer: Subtract the deficiency from the number (or add the surplus to the number). This is equivalent to Number ± deficiency/surplus from base.
   LHS = N - d (if N < B, so N - (B-N) which is wrong, it's N + (-d)) or more simply: N + (N-B). So, for 98 (base 100, deficiency 2), LHS = 98 - 2 = 96. For 103 (base 100, surplus 3), LHS = 103 + 3 = 106.
3. Calculate the Second Part (RHS) of the Answer: Square the deficiency (or surplus).
   RHS = d².
4. Adjust and Combine: The number of digits in the RHS should be equal to the number of zeros in the base. If d² has fewer digits, pad with leading zeros. If it has more digits, carry over the excess to the LHS.

Example 1 (Number below base): Square 98

• Base = 100. Number = 98.
• Deficiency (d) = 100 - 98 = 2.
• LHS = Number - Deficiency = 98 - 2 = 96.
• RHS = Deficiency² = 2² = 4. Since base 100 has two zeros, RHS should be 04.
• Answer: 9604.

Example 2 (Number above base): Square 103

• Base = 100. Number = 103.
• Surplus (d) = 103 - 100 = 3.
• LHS = Number + Surplus = 103 + 3 = 106.
• RHS = Surplus² = 3² = 9. Since base 100 has two zeros, RHS should be 09.
• Answer: 10609.

Why it works: For a number N = B-d, N² = (B-d)² = B² - 2Bd + d². This can be written as B(B-2d) + d². The sutra simplifies this to (N-d) | d². For N = B+d, N² = (B+d)² = B² + 2Bd + d² = B(B+2d) + d². The sutra simplifies this to (N+d) | d².

Meaning: "Part and Whole", "Specific and General", or "Individual and Total". This sutra relates to the idea of analyzing individual components (Vyashti) to understand or derive properties of the whole (Samashti), and vice-versa.

Applications & Interpretations:

1. Factorization: Identifying a common factor (Vyashti, the specific part) in an algebraic expression allows simplification of the whole expression (Samashti). For example, in 2x + 4y, '2' is the Vyashti, and factoring it out gives 2(x + 2y).
2. Summation of Series: Understanding the pattern of individual terms (Vyashti) in a series helps in finding the sum of the whole series (Samashti). For example, the sum of the first n natural numbers (1+2+...+n) is n(n+1)/2. Here, each number is a 'Vyashti', and the formula gives the 'Samashti'.
3. Calculus: Integration can be seen as summing up infinitesimal 'parts' (Vyashti) to find the 'whole' area or volume (Samashti). Differentiation involves looking at the specific rate of change (Vyashti) at a point within a function (Samashti).
4. Solving Equations: Sometimes, observing specific properties of parts of an equation (Vyashti) can lead to a solution for the entire equation (Samashti). For instance, if the sum of coefficients in a polynomial equation is zero, then x=1 is a root (a specific observation leading to a property of the whole).
5. Generalization from Specific Cases: In problem-solving, analyzing a few specific instances (Vyashti) can help in deducing a general formula or pattern (Samashti). This is a fundamental aspect of inductive reasoning.
6. Statistics: Individual data points (Vyashti) are analyzed to understand properties of the entire dataset or population (Samashti), like mean, median, or variance.

Core Idea: The sutra emphasizes the interconnectedness of the specific and the general. It's a methodological principle that can be applied across various mathematical (and even non-mathematical) domains to break down complex problems into simpler parts or to build up a general understanding from specific observations.

Meaning: "The remainders by the last digit".

Primary Application: Converting Fractions to Recurring Decimals

This sutra provides a method for converting fractions, especially those whose denominators end in 9 (or can be made to end in 9 by multiplying numerator and denominator by a suitable factor), into their exact recurring decimal representations.

How it works (e.g., for 1/19):

1. The "Last Digit" (Charamena): The denominator is 19. The last digit is 9.
2. The Multiplier: The sutra implies working with a multiplier derived from the denominator. For denominators ending in 9, like D9 (e.g., 19, 29, 39), the "Ekadhika" of the part before 9 (i.e., D+1) is often used as a key multiplier or related to the process. For 19, the part before 9 is 1. Ekadhika of 1 is 1+1=2. This '2' will be our multiplier.
3. Process:
   a. Start with the numerator as the first (rightmost) digit of the recurring part of the decimal (or as the first remainder if setting up for division). For 1/19, start with 1.
   b. Multiply this digit by our multiplier (2): 1 x 2 = 2. This '2' is the next digit to the left in our decimal sequence.
   c. Take this new digit (2), multiply by 2: 2 x 2 = 4. '4' is the next digit.
   d. Take 4, multiply by 2: 4 x 2 = 8. '8' is the next digit.
   e. Take 8, multiply by 2: 8 x 2 = 16. Write down 6, carry over 1.
   f. Take 6, multiply by 2 = 12. Add carry-over 1: 12+1 = 13. Write down 3, carry 1.
   g. Take 3, multiply by 2 = 6. Add carry-over 1: 6+1 = 7. '7' is next.
   h. Continue this process: (7x2=14 -> 4 carry 1), (4x2+1=9 -> 9), (9x2=18 -> 8 carry 1), (8x2+1=17 -> 7 carry 1), (7x2+1=15 -> 5 carry 1), (5x2+1=11 -> 1 carry 1), (1x2+1=3 -> 3), (3x2=6 -> 6), (6x2=12 -> 2 carry 1), (2x2+1=5 -> 5), (5x2=10 -> 0 carry 1), (0x2+1=1 -> 1).
4. Result: The sequence of digits obtained, read from right to left as generated, then reversed for the decimal, is 052631578947368421. The process stops when the starting digit (1) reappears as the result of a step (after considering carries and the full cycle). Since we started with 1/19, the decimal is 0.052631578947368421 (the bar indicates recurring part).

Note: The "remainders by the last digit" part refers to how each step effectively processes a remainder using the last digit (or a related multiplier) to find the next digit of the decimal. The Ekadhika of the digit(s) preceding the 9 in the denominator plays a crucial role as the effective multiplier in this specific application.

Meaning: "The ultimate (last term) and twice the penultimate (second to last term)".

Primary Application: Solving Specific Algebraic Equations

This sutra is applied to solve certain types of algebraic equations, particularly those that can be structured or recognized to fit a specific pattern where the relationship between the "ultimate" (last term) and "twice the penultimate" (twice the second to last term) is key. It often involves setting an expression or a combination of terms to zero.

Common Scenarios:

1. Equations with Fractions: Consider an equation with two fractions on one side summing to zero, like:
   M / (ax+b) + N / (cx+d) = 0
   If M(cx+d) + N(ax+b) = 0, and if we expand this to (Mc+Na)x + (Md+Nb) = 0.
   Here, (Md+Nb) is the "ultimate" term (constant term), and (Mc+Na)x involves the "penultimate" concept (coefficient of x). If the sutra's condition, related to the sum of these being zero in a specific way, is met, it leads to a solution. A more direct interpretation is if Md+Nb = 0 and Mc+Na = 0 if M and N are related in a certain way, implying x is indeterminate or specific constraints.
2. Relationship between coefficients in products: If an equation of the form (x+a)(x+b) = (x+c)(x+d) is given, and if a+b = c+d (sum of independent terms in factors on LHS = sum on RHS), then this implies 2x + (a+b) = 0 if another condition (ab=cd implies 0=0) is met. However, a more direct application of Sopaantyadvayamantyam is seen when a specific structure of coefficients in a single polynomial equated to zero is present. For instance, if you have a polynomial P(x) = 0, and P(x) can be factored into Q(x) * R(x) = 0, then the sutra might apply to the structure of Q(x) or R(x) if they fit the "ultimate and twice penultimate" pattern leading to a root.
3. Specific condition for quadratic roots: For a quadratic equation Ax² + Bx + C = 0, if B = A+C (twice the coefficient of the middle term related to the sum of the first and last), this doesn't directly use the "twice the penultimate" wording. The sutra is more about structuring the equation or parts of it. If an expression like P(x) = (some factor related to 'ultimate') + 2 * (some factor related to 'penultimate') = 0, then this structure itself is the solution method.

Key Idea: The sutra is used when an algebraic expression can be set up or observed such that if the last term (Antyam) and twice the second to last term (Sopaantyadvayam) sum up to zero (or satisfy a similar specific relationship), then the value of the variable that makes this true is the solution. It's a pattern-recognition technique for specialized equation forms.

Meaning: "By one less than the previous one". This sutra is a powerful and elegant method for specific types of multiplication.

Primary Application: Multiplication by Nines

This sutra is extremely useful and efficient for multiplying any number by a series of nines (e.g., x 9, x 99, x 999, etc.).

The Method: The answer is obtained in two parts:

1. First Part (LHS - Left Hand Side): Subtract 1 from the multiplicand (the number being multiplied by nines). This is the "Ekanyunena" part – "one less than" the original number.
   LHS = Multiplicand - 1.
2. Second Part (RHS - Right Hand Side): Subtract the LHS (obtained in Step 1) from the multiplier (the number consisting of nines).
   Alternatively, and often easier, apply the "Nikhilam Navatashcaramam Dashatah" (All from 9, the last from 10) sutra to the digits of the LHS, ensuring the number of digits in the RHS matches the number of nines in the multiplier. If the LHS has fewer digits than the number of nines, pad it with leading zeros before applying Nikhilam or subtraction from the nines.

Case 1: Number of digits in multiplicand = Number of nines in multiplier (e.g., 46 x 99)

• Multiplicand = 46, Multiplier = 99.
• LHS = 46 - 1 = 45.
• RHS = 99 - 45 = 54. (Or, Nikhilam on 45: (9-4)(9-5) -> 54).
• Answer: 4554.

Case 2: Number of digits in multiplicand < Number of nines in multiplier (e.g., 7 x 99)

• Multiplicand = 7, Multiplier = 99.
• LHS = 7 - 1 = 6. (Treat as 06 for RHS calculation if needed for clarity).
• RHS = 99 - 06 = 93. (Or, Nikhilam on 06 for two digits: (9-0)(9-6) -> 93).
• Answer: 693. (The LHS is just 6, not 06 in the final answer unless it's part of a larger number).

Case 3: Number of digits in multiplicand > Number of nines in multiplier (e.g., 345 x 99) - This requires a slight modification or extension of the basic method, often involving splitting the multiplicand or using general multiplication if simpler.

One way for 345 x 99 (345 x (100-1)): 34500 - 345 = 34155.

Using Ekanyunena variation:
Split 345 based on number of 9s (two 9s): 3 | 45.
LHS part 1: (Digit(s) to left of split + 1) => 3+1 = 4.
345 - 4 = 341 (This forms the initial part of the LHS of the final answer).
For RHS: Nikhilam of the split part (45) from 100 => 100 - 45 = 55.
Answer: 34155. (This variation is more complex and shows limitations of direct application for this case).

Simpler general rule for Case 3 (like 345 x 99):
LHS: (345 - 1) = 344. This is NOT the full LHS yet.
Append as many zeros as there are nines: 34500.
Subtract original number: 34500 - 345 = 34155. (This is essentially (N x 100) - N). Ekanyunena is more direct when digits match or multiplicand is smaller.

Advantages:

• Extremely fast for multiplications by 9, 99, 999, etc., especially when digit counts are favorable.
• Reduces complex multiplication to simple subtraction.

Meaning: "The product of the sum (of coefficients in factors) is equal to the sum of the coefficients (in the product)". More simply, "Samuccaya (sum of coefficients) of the product of factors is equal to the Samuccaya of the product polynomial."

Primary Application: Checking Polynomial Multiplication and Factorization

This sutra serves as an excellent and quick verification tool in algebraic multiplication and factorization of polynomials. 'Samuccaya' in this context most often refers to the sum of the numerical coefficients of the terms in a polynomial. To get this sum, you can substitute the variable (e.g., x) with 1.

How it Works for Multiplication Verification:

1. Take two (or more) polynomials that are being multiplied.
2. For each polynomial factor, find the sum of its coefficients. (Let these sums be S₁, S₂, etc.)
3. Multiply these sums together: Product_of_Sums = S₁ x S₂ x ...
4. Now, look at the resulting product polynomial obtained after multiplication.
5. Find the sum of the coefficients of this product polynomial (Sum_of_Product_Coefficients).
6. Verification: If the multiplication is correct, then Product_of_Sums should be equal to Sum_of_Product_Coefficients.

How it Works for Factorization Verification:

1. You have a polynomial and its supposed factors.
2. Find the sum of coefficients of the original polynomial.
3. Find the sum of coefficients for each of the supposed factors.
4. Multiply the sums of coefficients of the factors.
5. Verification: If this product matches the sum of coefficients of the original polynomial, the factorization is likely correct (it's a necessary but not always sufficient condition, as different incorrect factorizations could coincidentally yield the same sum product).

Example: Check if (x+2)(x+3) = x² + 5x + 6.

• For (x+2): Coefficients are 1 (for x) and 2. Sum S₁ = 1+2 = 3.
• For (x+3): Coefficients are 1 (for x) and 3. Sum S₂ = 1+3 = 4.
• Product_of_Sums = S₁ x S₂ = 3 x 4 = 12.
• For the product polynomial x² + 5x + 6: Coefficients are 1, 5, and 6.
• Sum_of_Product_Coefficients = 1+5+6 = 12.

Since 12 = 12, the multiplication is verified by this sutra.

Note: This is related to the property that if P(x) = Q(x) * R(x), then P(1) = Q(1) * R(1), because P(1) is the sum of coefficients of P(x), and so on.

Meaning: "The Samuccaya of the factors is equal to the Samuccaya of the product". This is very closely related to Gunitasamuchyah. While Gunitasamuchyah often refers specifically to the sum of coefficients (by setting variable=1), Gunakasamuchyah can be interpreted more broadly, sometimes referring to the sum of the digits of the numbers (often using the "casting out nines" or "digit sum" method) as a check for numerical multiplication.

Primary Application: Checking Numerical Multiplication using Digit Sums

This is a classic and simple method to quickly check if a multiplication of numbers is likely correct.

How it Works (Digit Sum Method / Casting Out Nines):

1. Find the Digit Sum of the Multiplicand: Add all the digits of the first number. If the sum is a multi-digit number, add its digits again until you get a single digit (this is equivalent to finding the remainder when divided by 9, where a result of 9 is treated as 0 or 9 itself). Let this be DS₁.
2. Find the Digit Sum of the Multiplier: Do the same for the second number. Let this be DS₂.
3. Multiply the Digit Sums: Multiply DS₁ x DS₂. Find the digit sum of this product. Let this be DS_P.
4. Find the Digit Sum of the Product: Take the result of your original multiplication and find its digit sum. Let this be DS_R.
5. Verification: If DS_P = DS_R, then the multiplication is likely correct. If they are not equal, the multiplication is definitely incorrect.

"Casting Out Nines" aspect: When summing digits, any 9s or combinations of digits that sum to 9 can be ignored (cast out) as they don't affect the final single-digit sum (modulo 9).

Example: Check 123 x 45 = 5535.

• Digit Sum of Multiplicand (123): 1+2+3 = 6. (DS₁ = 6).
• Digit Sum of Multiplier (45): 4+5 = 9. (DS₂ = 9 or 0. Let's use 9 for now).
• Multiply Digit Sums: DS₁ x DS₂ = 6 x 9 = 54. Digit sum of 54 is 5+4 = 9. (DS_P = 9).
• Digit Sum of Product (5535): 5+5+3+5 = 18. Digit sum of 18 is 1+8 = 9. (DS_R = 9).

Since DS_P (9) = DS_R (9), the multiplication is likely correct.

Limitation: This method can only prove a multiplication incorrect; it cannot definitively prove it correct. For example, if digits are transposed in the answer (e.g., 5355 instead of 5535), the digit sum might still match, but the answer is wrong. However, it's a very good quick check.