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3 HCF and LCM problems


riteshn

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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....

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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 ?

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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)

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