Data Sufficiency

pepeprepa wrote:Ok for your principle:
"if denumerator can be expressed in prime factorisation of 2 and 5 it is always terminal dicimal"

But I don't understand what you mean by 6/3 and 14/7 I talk about reduced fractions...

this was a joke!

Was not sure
But my phrase is fool, 1/2 is a coutner-example.

Thanks Ian

Ian Stewart wrote:I'll try to state the rule as unambiguously as possible. If you need to know whether a fraction represents a terminating decimal:

1) Reduce the fraction completely;
2) Prime factorize the denominator;
3) Look at this prime factorization, and ignore the exponents. If there is a prime besides 2 or 5 in the denominator, the fraction represents a recurring (non-terminating) decimal. If the only primes in the denominator are 2, 5, or both, the fraction represents a terminating decimal.

So 9/125, 3/32, 7/80 and 3/30 are all terminating decimals (when you look at 3/30, you must reduce the fraction first, of course), while 7/121, 5/33, 7/60 and 19/99 all represent recurring decimals.

Your earlier explanation made sense, but this one is perfect. Thanks so much for your continued help Ian!

Ian Stewart wrote:I'll try to state the rule as unambiguously as possible. If you need to know whether a fraction represents a terminating decimal:

1) Reduce the fraction completely;
2) Prime factorize the denominator;
3) Look at this prime factorization, and ignore the exponents. If there is a prime besides 2 or 5 in the denominator, the fraction represents a recurring (non-terminating) decimal. If the only primes in the denominator are 2, 5, or both, the fraction represents a terminating decimal.

So 9/125, 3/32, 7/80 and 3/30 are all terminating decimals (when you look at 3/30, you must reduce the fraction first, of course), while 7/121, 5/33, 7/60 and 19/99 all represent recurring decimals.

Is there a shortcut to this problem?

If d= 1/ (2^3 * 5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

A. 1
B. 2
C. 3
D. 7
E. 10

[spoiler]OA: B[/spoiler]

The way I would do it is using the same method the previous poster mention.

Take the demonator and prime factorize it. If the prime factors of the denominator ONLY contain 2's OR 5's OR a comination of 2's and 5's, the fraction terminates!

As an example:

We know that 1/9 is approximately 0.1111 repeating. If you prime factorize the number '9', you get the prime factors: 3 and 3. Since it doesn't have only 2's or only 5's or a combination of 2's and 5's, we know it doesn't terminate.

Another example:

We know that 1/8 = 0.625. The prime factors of 8 are 2,2, and 2. Since the prime factors are ONLY 2, then we know the decimal terminates.