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 September 29th, 2013, 07:55 PM #1 Member   Joined: Sep 2011 Posts: 98 Thanks: 1 LCM & HCF doubts Hi, I have encountered some problem with the below questions. Thank you in advanced for the help. The numbers a, b and c written as a product of their prime factors, are a= 2^3 * 3 * 7 b= 2^3 * 2^2 * 5 c= 2* 3 * 5^ 2 *7 a). Find the largest integer that will divide a, b and c exactly, b). the smallest positive integer n for which n is a multiple of c,
 September 29th, 2013, 08:06 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: LCM & HCF doubts $2^3$ and $2^5 \cdot3\cdot5^2\cdot 7$
 September 29th, 2013, 08:23 PM #3 Member   Joined: Sep 2011 Posts: 98 Thanks: 1 Re: LCM & HCF doubts May i know how to get a) which is 2^3 and b) 2^5*3*5^2*7
 September 29th, 2013, 08:33 PM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: LCM & HCF doubts [color=#000000]Here is the trick that will make you remember it for every pair-triple etc for the first question watch for the common primes with the lowest power and for the second take each prime once put it in order and take its highest power. For example let $a=p_1^{\ell_1}p_2^{\ell_2}...p_n^{\ell_n},\;\;b=p_ 1^{\mu_1}p_2^{\mu_2}...p_n^{\mu_n}$ then $LCM=p_{1}^{\max(\ell_1,\mu_1)}p_2^{\max(\ell_2,\mu _2)}...p_n^{\max(\ell_n,\mu_n)}$ and $HCF=p_{1}^{\min(\ell_1,\mu_1)}p_2^{\min(\ell_2,\mu _2)}...p_n^{\min(\ell_n,\mu_n)}$ (for lcm look for the common $p_{i}$). [/color]

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