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This problem trains for: SAT-I, AMC-8, GMAT.

The fraction:


is irreducible (it is in its lowest terms, it is in its simplest form). The decimal representation of this fraction consists of a non-repeating part and of a repeating part. Which of the following statements are false?

If the fraction is irreducible this means that the greatest common divisor of a and p is 1:


This is the definition of two coprime numbers. Therefore, I is true.

For the decimal representation of an irreducible fraction to have a non-repeating part, the denominator must have factors that divide 10 exactly. Therefore, there must be either a factor of 2, of 5 or both in the prime factorization of the denominator:


where f is some factor, different from 1, 2 or 5. Therefore, II is true.

Since p must have a factor of 2 or a factor of 5, the only ways for it to be prime is to be exactly equal to either 2 or 5. However, since there exists a repeating part in the decimal representation, p must have a factor that is not equal to 1, 2 or 5. Therefore, p cannot be prime and III is false.

If a is even, then p cannot also be even, since the fraction is irreducible. Therefore, since the decimal representation has a non-repeating part, this cannot be generated by a factor of 2 in the denominator. It follows that the denominator must have at least one factor of 5. IV is true.

If a is odd,then p may be odd or even. If it is even, then the decimal representation has a non-repeating part whether or not the denominator is also divisible by 5 or not. If it is odd, then it must have a factor of 5 since there is no factor of 2 that can generate the non-repeating part. Therefore, just from knowing that a is odd, it does not necessarily follow that p must be even. V is false.

Note that the properties of decimals are strictly dependent on the base of numeration, i.e. they are true only when counting in base 10 (the same goes for all divisibility rules).