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If m is a 3-digit positive integer such that \(\mathop{\text{lcm}}[8m,10^{10}] = 4\cdot\mathop{\text{lcm}}[m,10^{10}]\), then what is the value of m?

 

The product of positive integers x, y and z equals 2004. What is the minimum possible value of the sum x+y+z?

 Jul 16, 2016
edited by higgsb  Jul 16, 2016

Best Answer 

 #8
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Thanks Alan! Now, I get this mumbo-jumbo!.

 Jul 16, 2016
 #1
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+5

The product of positive integers x, y and z equals 2004. What is the minimum possible value of the sum x+y+z?

 

The  divisors of 2004 are  1 | 2 | 3 | 4 | 6 | 12 | 167 | 334 | 501 | 668 | 1002 | 2004

 

The minimum sum will come from the factors of the middle two divisors, 12 and 167

 

167 is prime....it will be included in the sum.......the divisors of 12   are 1 | 2 | 3 | 4 | 6 | 12.......

 

The remaining two integers involved in minimizing the sum will come from the middle two divisors of 12, i.e, 3 and 4

 

So....the minumum sum is given when  the three integers x, y and z  are   3 , 4 and 167   ....the sum  =  174

 

 

 

cool cool cool

 Jul 16, 2016
 #2
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The PRIME factors of 2004=2^2 x 3 x 167=4+3+167=174

 Jul 16, 2016
 #3
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"If m is a 3-digit positive integer such that , then what is the value of m?"

 

512

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 Jul 16, 2016
 #4
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LCM[8,512, 10^10] =1,330,000,000,000 according to mighty Wolfram/Alpha!. Where does 4 fit in???

 Jul 16, 2016
 #5
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I think 8m is meant to be interpreted as 8*m, not 8000+m.

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Alan  Jul 16, 2016
 #6
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 lcm(8*512, 10^10) = 40 000 000 000

 

 lcm(512, 10^10) = 10 000 000 000

Alan  Jul 16, 2016
 #7
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Note that the prime factorisation of 10^10 is 2^10*5^10

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Alan  Jul 16, 2016
 #8
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+4
Best Answer

Thanks Alan! Now, I get this mumbo-jumbo!.

Guest Jul 16, 2016

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