A - Find the LCM of two Numbers LCM and HCF of two integers LCM of two numbers Example 1: Find the LCM of 66 and 28 . Find the prime factorization of the two numbers. 2 66 3 33 11 11 1 2 28 2 14 7 7 1 28 = 22 x 7 66 = 2 x 3 x 11 Any multiple of 66 will also have 2‚ 3 and 11 as its factors. Likewise any multiple of 28 well have 2 and 7 as its factors. The common multiple will have all the prime factors of the two numbers as its factors. Where a prime factor is found in both the two numbers‚ the
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Groups) • Applications of HCF and LCM to Problem Solving 2 Insights into Lesson Study • Introduction: Focus of Lesson • Student Learning : What we learned about students’ understanding based on data collected • Teaching Strategies: What we noticed about our own teaching • Strengths & Weaknesses of adopting the Lesson Study process 3 Introduction Applications of HCF and LCM to Problem Solving Reason for study: Realised we needed to go further than just the skills of HCF and LCM‚ because students tend
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LCM‚ HCF‚ GCD: Basic concept‚ calculation‚ applications explained Introduction Concept of LCM‚ HCF important for number theory and remainder based problems (generally asked in SSC CGL‚ CAT.) LCM is important for time and speed‚ time and work problems. LCM is also important for circular racetracks‚ bells‚ blinking lights‚ etc. HCF is important for largest size of tiles‚ largest size of tape to measure a land etc. But before getting into LCM‚ HCF‚ let’s understand What is Prime number
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its prime factors 2. Find the LCM and HCF of 12‚ 15 and 21 by the prime factorization method. 3. Find the LCM and HCF of 6 and 20 by the prime factorization method. 4. State whether13/3125 will have a terminating decimal expansion or a non-terminating repeating decimal. 5. State whether 17/8 will have a terminating decimal expansion or a non-terminating repeating decimal. 6. Find the LCM and HCF of 26 and 91 and verify that LCM × HCF = product of the two numbers. 7
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some integer. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? Sol. Hints: Find the HCF of 616 and 32 3. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [Hint : Let x be any positive integer then it is of the form 3q‚ 3q + 1 or 3q + 2. Now square each
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recognise this as the usual long division process. Although this result is quite easy to state and understand‚ it has many applications related to the divisibility properties of integers. We touch upon a few of them‚ and use it mainly to compute the HCF of two positive integers. The Fundamental Theorem of Arithmetic‚ on the other hand‚ has to do something with multiplication of positive integers. You already know that every composite number can be expressed as a product of primes in a unique way —
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find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 Solution. (i) In 135 and 225‚ 225 is larger integer. Using Euclid’s division algorithm‚ [Where 135 is divisor‚ 90 is remainder] 225 = 135 × 1 + 90 Since‚ remainder 90 ≠ 0 ‚ by applying Eudid’s division algorithm to 135 and 90 ∴ 135 = 90 × 1 + 45 Again since‚ remainder 45 ≠ 0 ‚ by applying Eudid’s division algorithm to 90 and 45 ∴ 90 = 45 × 2 + 0 Now‚ the remainder is zero so‚ our procedure stops. Hence‚ HCF of 135
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------------------------------------------------- Prime number A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example 5 is prime‚ as only 1 and 5 divide it‚ whereas 6 is composite‚ since it has the divisors 2 and 3 in addition to 1 and 6. The fundamental theorem of arithmetic establishes the central role of primes in number theory: any
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the HCF of 210 and 55. (1 Mark) (Ans) 5 Explanation: 5 ‚ Given integers are 210 and 55 such that 210 > 55. Applying Euclid’s division leema to 210 and 55‚ we get 210 = 55 x 3 + 45 ……….(1) 55 = 45 x 1 +10 ………(2) 45 = 10 x 4 + 5 ………..(3) 10 = 5 x 2 + 0 ………..(4) we consider the new divisor 10 and the new remainder 5 and apply division leema to get 10 = 5 x 2 + 0 The remainder at this stage is zero. So‚ the divisor at this stage or the remainder at the previous stage i.e.5 is the HCF of 210
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SUB TOPICS RECOMMENDED FOR IMPROVEMENT * Understanding of lines * Understanding of angles. 1 DA STUDENT REPORT CLASS 5 ROLL NO 25 Multiples and Factors; Lines and Angles DATE 27 August 2012 Misconceptions Concept: Application of HCF in Real Life 10 Q Reema has 3 pieces of ribbon which are 18 m‚ 36 m and 45 m long respectively. She wants to cut them into pieces of equal length. She wants to do this in such a way that no ribbon is wasted. What is the length of the LONGEST piece
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