decimal equivalent that terminates?

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decimal equivalent that terminates?

by ch0719 » Mon May 26, 2008 4:19 pm
I came across a question that asks about decimal equivalent, I couldn't figure it out what the question is asking for

Can anybody give a pointer on this? :)

Q: Which of the following fractions has a decimal equivalent that terminates?

49/224

22/189

37/196

25/513

17/175


the answer is 49/224, but i donno why..

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by chidcguy » Mon May 26, 2008 4:39 pm
If Q's like this are representative of the original GMAT, 2 min is not going to be enough.

Think about the opposite of termination. Recurring. Recurring decimals are the ones which do not end. some thing like .3333333 or .67777777777 and on

49/224= 7/32= .2175 and it ends there.

Glad the answer is A and not E.

Guru's

Any easy way of finding out if a fraction is going to yield a recurring decimal or terminating decimal. terminating decimal?? Is that a valid mathematical term? I have never heard of it, even though I could understand what the Q is trying to get at.

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by ch0719 » Mon May 26, 2008 5:17 pm
Thanks for the reply!

I was suspecting that was what the question is trying to ask, but I have never seen it written in such a way.

Btw its from princeton review

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by chidcguy » Tue Jun 03, 2008 3:33 pm
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

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by Ian Stewart » Tue Jun 03, 2008 4:36 pm
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.

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by chidcguy » Tue Jun 03, 2008 5:05 pm
Thanks Ian.

I should have looked deep enough

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by manishpal » Tue Sep 20, 2011 5:37 am
hey bud. Theres are rule that states if the denominator is ever in the form of 2^a or 5^b it will terminate.

For E. 128 = 2^7, so it terminates