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How many positive perfect square integers are factors of the product \(\left(2^{10}\right)\left(3^{12}\right)\left(5^{15}\right)\)?

 

What is the smallest positive integer n such that, out of the n unit fractions  \(\frac{1}{k}\)where \(1 \le k \le n\), exactly half of the fractions give a terminating decimal?

mathtoo  Aug 12, 2018
 #1
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I think this is the number of perfect squares:

 

2^5  x  3^6  x  5^7.5 =4,075,233,888 Perfect squares??

Guest Aug 12, 2018
edited by Guest  Aug 13, 2018
edited by Guest  Aug 13, 2018
 #4
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(2^10)(3^12)(5^15) =16,607,531,250,000,000,000

Number of perfect squares that are FACTORS of the above product are calculated as follows:

 

The exponent of each term / 2 + 1, then multiplied together as follows:

2^10 =10 / 2 + 1 = 6,  3^12 =12/2 + 1 =7,   5^15 =15/2 + 1 =8

The total number of perfect squares that are FACTORS of the above product:

=6 x 7 x 8 =336  perfect squares.

Guest Aug 14, 2018
 #2
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If I understand the question, the smallest positive integer n = 10, where:

1/2=0.5, 1/4=0.25, 1/5=0.2, 1/8=0.125, 1/10=0.1 (Or 5 out of 10 give terminating decimal).

Guest Aug 12, 2018
 #3
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Are you sure? 

Guest Aug 13, 2018

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