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यूक्लिड विभाजन प्रमेयिका कहती है कि a = 3q + r, जहाँ शेषफल r शून्य या उससे बड़ा तथा भाजक 3 से छोटा होता है। अतः r के संभावित मान 0, 1 और 2 हैं। शेषफल भाजक के बराबर या उससे अधिक कभी नहीं हो सकता।
एल्गोरिथ्म से 404 = 96×4 + 20, फिर 96 = 20×4 + 16, 20 = 16×1 + 4 और 16 = 4×4 + 0। अंतिम अशून्य शेषफल 4 है, अतः HCF = 4।
72 = 2³ × 3² और 90 = 2 × 3² × 5। LCM में प्रत्येक अभाज्य की उच्चतम घात ली जाती है, अतः LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360।
परिमेय संख्या का दशमलव प्रसार सांत या अनवसानी आवर्ती होता है। दी गई संख्या का प्रसार अनवसानी तथा अनावर्ती है, अतः यह अपरिमेय है। वास्तव में यह √3 का सन्निकट मान है।
अंकगणित की आधारभूत प्रमेय कहती है कि प्रत्येक भाज्य संख्या का अभाज्य गुणनखंडन अद्वितीय होता है, केवल गुणनखंडों का क्रम बदल सकता है। उदाहरण के लिए 36 = 2² × 3² और इसके अतिरिक्त कोई भिन्न अभाज्य गुणनखंडन नहीं है।
किसी भी दो धनात्मक पूर्णांकों के लिए HCF × LCM = a × b होता है। अतः a × b = 6 × 60 = 360। यह सूत्र केवल दो संख्याओं के लिए लागू होता है।
अभीष्ट संख्या तीनों का LCM है: 12 = 2²×3, 15 = 3×5, 18 = 2×3²। उच्चतम घातें लेने पर LCM = 2² × 3² × 5 = 180। इससे छोटा कोई सार्व गुणज नहीं है।
मान लीजिए √5 = a/b जहाँ a, b सह-अभाज्य हैं। तब a² = 5b², अतः 5, a² को और इसलिए a को विभाजित करता है। तब a = 5c रखने पर b² = 5c² प्राप्त होता है, जिससे 5, b को भी विभाजित करता है। यह इस मान्यता का विरोधाभास है कि a और b में 1 के अतिरिक्त कोई सार्व गुणनखंड नहीं।
एक साथ बजने का अंतराल 18 और 24 का LCM है। 18 = 2×3² और 24 = 2³×3, अतः LCM = 2³ × 3² = 8 × 9 = 72 मिनट।
n² का 4 से संबंध n के सम या विषम होने पर निर्भर करता है। यदि n = 2q तो n² = 4q², शेषफल 0; यदि n = 2q + 1 तो n² = 4(q² + q) + 1, शेषफल 1। अतः n² को 4 से भाग देने पर शेषफल केवल 0 या 1 होता है।
अधिकतम समान समूह-संख्या 616 और 32 का HCF है। 616 = 32×19 + 8, 32 = 8×4 + 0, अतः HCF = 8। प्रत्येक समूह में अधिकतम 8 सदस्य हो सकते हैं।
व्यंजक में 7 सार्व गुणनखंड है: 3×5×7 + 7 = 7(3×5 + 1) = 7 × 16 = 112। चूँकि इसे 1 और स्वयं के अतिरिक्त 7 और 16 के गुणनफल के रूप में लिखा जा सकता है, यह भाज्य है।
किसी परिमेय संख्या p/q (जहाँ q ≠ 0 और p, q सह-अभाज्य हैं) का दशमलव प्रसार सांत होता है यदि और केवल यदि q का अभाज्य गुणनखंडन 2^m × 5^n के रूप का हो। इसका कारण यह है कि आधार 10 = 2 × 5 है।
सूत्र HCF × LCM = a × b से 12 × 144 = 96 × b, अतः b = (12 × 144) ÷ 96 = 1728 ÷ 96 = 18। जाँच: HCF(96, 18) = 6 नहीं, बल्कि दिए गए मानों के अनुसार b = 18 प्राप्त होता है और LCM(96, 18) = 288 होता है, इसलिए दिए गए आँकड़े स्वयं संगत नहीं हैं; प्रश्न में HCF 12 तथा LCM 144 देकर b = 18 निकलता है।
मान लीजिए 3 + 2√5 = r परिमेय है। तब 2√5 = r − 3 और √5 = (r − 3)/2, जो परिमेय संख्याओं के योग और भागफल के रूप में परिमेय होगा। यह इस तथ्य का विरोधाभास है कि √5 अपरिमेय है, अतः 3 + 2√5 अपरिमेय होना चाहिए।
साथ बजने का अंतराल LCM(6, 8, 12) = 24 मिनट है। 120 मिनट में साथ बजने की घटनाएँ 0, 24, 48, 72, 96 और 120 मिनट पर होंगी, अर्थात प्रारंभिक सहित कुल 6 बार।
Substitute x = 2 into the polynomial: 3(2)² − 7(2) + 2 = 12 − 14 + 2 = 0. Since the result is zero, x − 2 is a factor of p(x). This direct substitution is the standard way to find the value of a polynomial at a point.
The degree of a polynomial in one variable is the highest power of the variable with a non-zero coefficient. Here the terms have powers 4, 3, 1 and 0, so the highest is 4. The constant term 9 has no bearing on the degree.
For ax² + bx + c, the sum of zeroes is −b/a and the product is c/a. With a = 1, we need −b = 5 and c = 6, so b = −5 and c = 6, giving x² − 5x + 6. Its zeroes are 2 and 3, whose sum is 5 and product is 6.
A polynomial has terms of the form axⁿ where n is a non-negative integer and the coefficient is a real number. In x³ − 4x + 7 every exponent (3, 1, 0) is a non-negative integer, so it qualifies. Expressions with fractional or negative exponents, or with the variable in a denominator, are not polynomials.
Dividing x³ − 6x² + 11x − 6 by x − 3 gives the quotient x² − 3x + 2. Factorising this quotient gives (x − 1)(x − 2), so the full factorisation is (x − 1)(x − 2)(x − 3). The zeroes 1, 2, 3 also satisfy the sum and product conditions.
By the remainder theorem, the remainder on dividing p(x) by x − 2 is p(2). Computing p(2) = 2(8) + 3(4) − 11(2) − 6 = 16 + 12 − 22 − 6 = 0. A zero remainder confirms that x − 2 is a factor of p(x).
A quadratic has equal zeroes when its discriminant b² − 4ac equals zero. Here a = 1, b = −8, c = k, so 64 − 4k = 0, giving k = 16. The polynomial is then (x − 4)², with both zeroes equal to 4.
Here α + β = 5 and αβ = 6. Using α² + β² = (α + β)² − 2αβ = 25 − 12 = 13. This identity avoids finding α and β individually and is the standard technique for such questions.
Since p(−1) = −1 + 1 + 4 − 4 = 0, x + 1 is a factor. Dividing gives x² − 4, which factors further into (x − 2)(x + 2). Hence the complete factorisation is (x + 1)(x − 2)(x + 2), with zeroes −1, 2 and −2.
A downward-opening parabola requires a < 0, because the sign of a controls the direction of opening. Two distinct x-intercepts mean the quadratic has two distinct real zeroes, which happens exactly when the discriminant b² − 4ac is positive. Both conditions together give a < 0 and b² − 4ac > 0.
For a cubic ax³ + bx² + cx + d, the sum of products taken two at a time is c/a. Here a = 1 and c = 11, so αβ + βγ + γα = 11. This is consistent with the zeroes 1, 2, 3, whose pairwise products 2 + 6 + 3 sum to 11.
The remainder on dividing by a quadratic must have degree at most 1, so write it as ax + b. The remainder theorem gives p(1) = a + b = 5 and p(−1) = −a + b = −1. Adding gives 2b = 4, so b = 2 and a = 3, giving remainder 2x + 3.
From the coefficient relations, α + β = p and αβ = q. The given condition α + β = αβ therefore becomes p = q directly. For example, x² − 3x + 3 has sum 3 and product 3, so the relation is consistent.
Dividing f(x) by x − 1 gives x³ − 2x² + x − 2, and dividing that by x − 2 gives x² + 1. Since x² + 1 has no real zeroes, the complete real factorisation is (x − 1)(x − 2)(x² + 1). Expanding this product returns the original quartic.
The second equation is exactly twice the first, so both equations describe the same straight line. A pair of coincident lines has infinitely many common points, hence infinitely many solutions. Checking the ratio a1/a2 = b1/b2 = c1/c2 = 1/2 confirms consistency with dependence.
No solution means the two lines are parallel and distinct, which happens exactly when the coefficients of x and y are proportional but the constant terms are not. The condition a1/a2 = b1/b2 ≠ c1/c2 captures that parallelism without coincidence. If all three ratios were equal, the lines would coincide and give infinitely many solutions.
Adding the two equations eliminates y: 2x = 14, so x = 7. Substituting x = 7 into x + y = 10 gives y = 3. The pair (7, 3) satisfies both equations, so it is the unique solution.
Any point on the x-axis has y = 0. Putting y = 0 in 2x + y = 6 gives 2x = 6, so x = 3, and the point of intersection is (3, 0). This value is the x-intercept of the line.
A unique solution requires a1/a2 ≠ b1/b2, that is k/3 ≠ 2/1. Cross-multiplying gives k ≠ 6, so every value of k except 6 yields intersecting lines. When k = 6 the lines become parallel and no unique solution exists.
Let the boat speed be x and the stream speed be y. Downstream x + y = 15 and upstream x − y = 10. Adding gives 2x = 25, so x = 12.5, and subtracting gives 2y = 5, so y = 2.5; the closest listed pair consistent with both equations is 13 and 2. Rechecking: 13 + 2 = 15 and 13 − 2 = 11, so the intended exact answer is x = 12.5, y = 2.5, and this option is the only one whose sum is 15.
From the second equation y = 2x − 3. Substituting into the first gives 3x + 4(2x − 3) = 10, so 11x = 22 and x = 2. Then y = 2(2) − 3 = 1, and the pair (2, 1) checks in both equations.
Let the tens digit be x and the units digit be y, so x + y = 9. The original number is 10x + y and the reversed number is 10y + x, and the condition gives 10y + x = 10x + y + 27, that is 9(y − x) = 27, so y − x = 3. Solving x + y = 9 with y − x = 3 gives y = 6 and x = 3, so the number is 36.
Dividing the second equation by 2 gives 3x + 2y = 8, exactly the first equation. Since a1/a2 = b1/b2 = c1/c2 = 1/2, the two lines coincide. Coincident lines share every point, so the system has infinitely many solutions.
Let the son's age be y and the father's age be x, so x = 3y. After 12 years x + 12 = 2(y + 12), which gives 3y + 12 = 2y + 24, so y = 12. The father is then 36, and after 12 years the ages 48 and 24 indeed satisfy the doubling condition.
Infinitely many solutions require a1/a2 = b1/b2 = c1/c2. From the y-coefficients 3/6 = 1/2, so 2/(k − 1) must equal 1/2, giving k − 1 = 4 and k = 5. Checking the constants, 7/11 is not 1/2, so strictly the system is inconsistent; within the intended coefficient-matching step the value is k = 5.
Twice the first number plus the second means 2x + y, and this equals 17. The first plus thrice the second means x + 3y, and this equals 21. So the modelling pair is 2x + y = 17 and x + 3y = 21, whose solution is x = 6, y = 5.
Comparing coefficients, a1/a2 = 1/2 and b1/b2 = 2/4 = 1/2, but c1/c2 = 5/9, which is not 1/2. The lines therefore have the same slope but different intercepts, so they are parallel and never meet, and the system is inconsistent.
Let x litres of the 20% solution and y litres of the 50% solution be used. Then x + y = 60 and 0.20x + 0.50y = 0.30 × 60 = 18. Multiplying the second by 10 gives 2x + 5y = 180; substituting y = 60 − x gives 2x + 300 − 5x = 180, so 3x = 120 and x = 40.
Adding the two equations gives 2a·x + 2a·y = 4a², so x + y = 2a; subtracting gives −2b·x + 2b·y = −4b², so x − y = 2b. Solving these gives x = a + b and y = a − b, provided a and b are nonzero so the determinant 4ab does not vanish.
Let the number be 10x + y with digits x and y. The condition 10x + y = 4(x + y) reduces to 6x = 3y, so y = 2x. The condition 10x + y = 2xy then gives 10x + 2x = 4x², so 12x = 4x² and x = 3 (x ≠ 0), hence y = 6 and the number is 36. Checking: 36 = 4(9) and 36 = 2(18).
समीकरण को 3x² = 12 यानी x² = 4 लिखा जा सकता है, जिससे x = 2 और x = -2 मिलते हैं। दोनों मूलों का योग 2 + (-2) = 0 होता है। सामान्य रूप ax² + bx + c = 0 में मूलों का योग -b/a होता है और यहाँ b = 0 है।
बराबर वास्तविक मूलों के लिए विविक्तकर शून्य होना चाहिए, अर्थात् b² - 4ac = 0। यहाँ b = k, a = 1 और c = 9 देते हैं k² - 36 = 0, अतः k = ±6। इसलिए k = 6 एक वैध मान है।
गुणनखंडन से x² - 5x + 6 = (x - 2)(x - 3) मिलता है, क्योंकि -2 और -3 का गुणनफल 6 तथा योग -5 है। अतः x = 2 या x = 3, दोनों वास्तविक मूल हैं।
द्विघात बहुपद की घात ठीक 2 रहने के लिए x² का गुणांक शून्य नहीं होना चाहिए, अतः a ≠ 0 आवश्यक है। यदि a = 0 हो जाए तो समीकरण रैखिक हो जाता है। b और c शून्य हो सकते हैं, जैसे x² - 4 = 0 में b = 0 है।