riteshn Posted December 8, 2005 Share Posted December 8, 2005 1) If m = 2^5.3^7.5^10; n=2^7.3^8.7^12, then the HCF and LCM of m and n is: Answer given is: 2^5.3^7 :hmm: 2) Find the greatest number that divides 5140 and 3960 leaving remainder 7 in each case. A) 60 B) 59 C)56 D)69 E)None. I can get the GCM as 20 but its nowhere close to the given options. Answer is: 59 3) If the sum of the two numbers is 216 and their HCF is 27 then the numbers are: A) 27, 189 B)54,162 C)108, 108 D) 81, 189 E)None Ok. The obvious answers are A and B and by brute force method you can find out that GCM of B is 54 so the answer has to be A. But is there a elegant way to solve it? In some other question, the answers might not be that close.... Quote Link to comment Share on other sites More sharing options...
krishnakanthc Posted December 8, 2005 Share Posted December 8, 2005 1ans it would hv been good if options were there, anyways lcm involves taking smallest number out. i.e. 2. hcf must be (2^5)*(3^7 ) ( because thats the maximum you can take out of these two numbers) 2ans 59 , you did not take rem 7 into account since they asked hghest try 69 first then try 59 since 60 is too nice . you should get the ans in a min. 3ans hcf this is always true. that solves this question in a nanosec. what do you say ? Quote Link to comment Share on other sites More sharing options...
riteshn Posted December 8, 2005 Author Share Posted December 8, 2005 ****. I made a big blunder. For Q 1: you need to only find HCF. The options are: A) 2^5.3^7 B)2^7.3^8.7^12 C)40^5.3^7.5^10.7^13 D) 5^5.3^8.5^10.7^12. For Q 2: hcf Quote Link to comment Share on other sites More sharing options...
kronique Posted December 8, 2005 Share Posted December 8, 2005 Here u may find some basic intro to HCF & LCM:tup: Quote Link to comment Share on other sites More sharing options...
Dkumar Posted December 8, 2005 Share Posted December 8, 2005 1) If m = 2^5.3^7.5^10; n=2^7.3^8.7^12, then the HCF and LCM of m and n is: HCF= 2^5*3^7 2) Find the greatest number that divides 5140 and 3960 leaving remainder 7 in each case. A) 60 B) 59 C)56 D)69 E)None. It is the HCF of (5140-7)and(3960-7). By the long division method, ans is 59 (B) 3) If the sum of the two numbers is 216 and their HCF is 27 then the numbers are: A) 27, 189 B)54,162 C)108, 108 D) 81, 189 E)None a+b=216 27x+27y=216 x+y=8 x and y should the number such that their hcf is 1. It might be (1,7), (3,5), (5,3), (7,1) (1,7 is found in the answer); a= 27*1=27 and b=27*7=189. (A) Quote Link to comment Share on other sites More sharing options...
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