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**0 **

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

edited by Guest Aug 13, 2018

#4**0 **

(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