Given that HCF (306, 1314) = 18. Find LCM (306, 1314).
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Find the LCM and HCF of the following pairs of integer and verify that LCM×HCF=Product of the two numbers.
Solution : i) 26 and 91 26=2×13×1(expressing as product of it’s prime factors) 91=7×13×1(expressing as product of it’s prime factors) So, LCM(26,91)=2×7×13×1=182 HCF(26,91)=13×1=13 Verification: LCM×HCF=13×182=2366 Product of 26 and 91 =2366 Therefore,LCM×HCF=Product of the two numbers . i) 510 and...
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Find the LCM and HCF of the following integers by applying the prime factorization method. i) 12,15 and 21 ii) 17,23 and 29 iii) 8,9 and 25
Solution : i) 12,15 and 21 12=2×2×3 15=5×3 21=7×3 From the above ,HCF(12,15,21)=3and LCM(12,15,21)=420 ii)17,23,and 29 17=17×1 23=23×1 29=29×1 From the above ,HCF(17,23,29)=1and LCM(17,23,29)=11339 iii)8,9 and 25 8=2×2×2 9=3×3 25=5×5 From the above ,HCF(8,9,25)=1and LCM(8,9,25)=1800
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Given that HCF(306,657)=9 .Find LCM(306,657)?
Solution : We know, HCF×LCM=Product of two numbers i.e 9×LCM=306×657 LCM=306×657/9=22338
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State Fundamental theorem of arithmetic. Is it possible for the HCF and LCM of two numbers to be 18 and 378 respectively. Justify your answer.
Solution: Every composite number can be expressed (factorised) as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur. Yes. Justification: HCF...
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The HCF of 65 and 117 is expressible in the form 65m - 117. Find the value of m. Also find the LCM of 65 and 117 using prime factorization method.
Solution: 117 = 65 x 1 + 52 65 = 52 x 1 + 13 65 = 13 x 5 + 0 => HCF(65, 117) = 13 Since, HCF(65, 117) = 65m - 117 => 13 = 65m - 117 => 65m...
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If two positive integers x and y are expressible in terms of primes as x = p^2q^3 and y = p^3q, what can you say about their LCM and HCF.
Solution: x = p2q3 and y = p3q HCF(x, y) = p2 x q LCM(x, y) = p3 x q3 LCM(x, y) = (p x q2) x HCF(x, y) => LCM is a multiple of HCF.
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If a is an odd number, b is not divisible by 3 and LCM of a and b is P, what is the LCM of 3a and 2b?
The correct answer is 6p
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135.225 hcf lcm
135 and 225 Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 225 = 135 × 1 + 90 Since remainder 90 ≠ 0, we apply the...
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628.306 + 6.1325 × 44.0268 = ? (a) 820 (b) 970 (c) 1050 (d) 1175 (e) 900
(e) ? = 628.306 + 6.1325 × 44.0268 ? = 628 + 6 × 44 = 628 + 264 = 892 =900
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Let * be a ‘binary’ operation on N given by a*b = LCM (a, b) for all a, b∈N. Find 5 * 7.
5 * 7 = LCM of 5 and 7 = 35
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