chidcguy wrote:Also found this on MGMAT forums
For fraction p/q to be a terminating decimal, the numerator must be an integer and the denominator must be an integer that can be expressed in the form of 2^x 5^y where x and y are nonnegative integers. (Any integer divided by a power of 2 or 5 will result in a terminating decimal.)
How ever 224 cannot be expressed in 2^x 5^y form. 224 comes down to 2^5 X 7^1
That's true, what you've quoted above, except it's incomplete. In any terminating decimal question like the above:
1. Make sure you reduce your fractions first.
2. Then prime factorize the denominators. If you see any prime besides 2 or 5, the decimal is recurring. If you only see 2s and/or 5s, it terminates.
So, in the text I quoted above, while it's true that 224 is divisible by 7, that 7 cancels with the 49 in the numerator- you need to reduce the fraction first.
You can see why step 2. above works- let's use the example answer choice A from the question:
49/224 = 7/32 = 7/2^5. Now multiply numerator and denominator by 5^5 so you have equal powers on 2 and 5 in the denominator:
(7/2^5)*(5^5/5^5) = 7*5^5/(2^5 * 5^5) = (7*5^5)/(10)^5
We have a fraction with a denominator of 100,000, and it's clear when you have a power of 10 in the denominator you get a terminating decimal. There's no need to multiply out the numerator, but if you do you can see exactly what decimal you get:
(7*5^5)/(10)^5 = 21,875/100,000 = 0.21875
Long story short, if your denominator is such that you could make a power of 10, you can get a terminating decimal. That is, the only primes you want to see in the denominator are 2 and/or 5 (***after you've reduced the fraction***). If there's any other prime down there, you have a recurring (infinite) decimal.
