Since Prime factors of 210 = 2*3*5*7.hdk2104 wrote:The sum of four natural numbers a1, a2, a3 and a4 is 210. What is the minimum possible LCM of these 4 mentioned distinct natural number ??
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The key is try to do away with large factors i.e.5 & 7 and try to get 4 distinct nos.adding 210.
By Hit and Trial, we can get nos. as 108, 54, 36 & 12.
108 = 2^2.3^3;
54 = 2.3^3;
36 = 2^2.3^2;
12 = 2^2.3;
Minimum possible LCM with distinct nos. will be 108.
Lets try taking maximum 2s. In that case nos. would be 2K, 2^2K, 2^3K & 2^4K.
Now 2K + 2^2K + 2^3K + 2^4K = 210. => WHICH GIVES k= 7. But in this case Largest no. would be 2^4K = 2^4.7 = 126 or LCM would be 126, which is greater than 108.Not acceptable.
You may try this question with 'Not Necessarily Distinct no. too." What's the answer then?
