Final answer:
To find how many values of n²/2010 yield repeating decimals, we must factor 2010 and then count integers n² that do not share the same prime factors with the same or greater multiplicity. All integers n that result in a n² with different prime factors from 2010 will yield a repeating decimal when divided by 2010.
Step-by-step explanation:
The question asks us to determine how many integer values of n, where 1 < n < 2010, result in n²/2010 yielding repeating decimals. A decimal is repeating if and only if the denominator (after the fraction is in simplest form) has prime factors other than 2 and 5, the prime factors of 10. To address this, we need to factor 2010. We find that 2010 is equal to 2 × 3 x 5 x 67. Since we are investigating n²/2010, we are interested in the squares of integers, specifically those where n² does not have the same prime factors as 2010 with the same or greater multiplicity.
To have a non-repeating decimal, n must be a multiple of 2, 3, 5, and 67 to fully cancel out the prime factors in the denominator. If n includes any other prime factors, or does not include enough of these, the fraction cannot be simplified to remove all prime factors other than 2 and 5, and so we would have a repeating decimal. Therefore, all squares that do not have 2, 3, 5, and 67 as their prime factors to at least the same multiplicity as they appear in 2010 will yield a repeating decimal when divided by 2010. Counting those is a matter of determining the number of unique squares, which are not the product of the factors of 2010, between 1 and 2010.