Day 6 - Data Sufficiency Practice (OG, 13th Ed), Problem 148

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Hi guys! Can anyone help me with this?

"Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, .82, 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?
1) 90 < r < 100
2) s = 4

I understand why it's not 1 (aka, eliminate answer "A" and "D" as correct) but why is the answer to this "B"? According to the back of the book: "Division by the number 4 must terminate: the remainder when dividing 4 must be 0, 1, 2, or 3, so the quotient must end with .0, .25, .5, or .75, respectively." I think I'm missing something obvious, but...I don't understand that last line! Can someone please explain this to me in a simpler way? How do I know it will never repeat, and will definitely terminate, for any number that I divide by 4, because of the ".0, .25, .5 or .75" ? If I test a bunch of numbers, is that what will happen and I just need to "know" that this is a rule that applies for 4? Or is there a rule I can use to scale out to other questions like this, where I'm not dealing with "4" but with some other number?

Thanks so much!

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by aneesh.kg » Mon Oct 29, 2012 11:48 pm
ostrowskiamy wrote:Hi guys! Can anyone help me with this?

"Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 24, .82, 5.096 are three terminating decimals. If r and s are positive integers and the ratio r/s is expressed as a decimal, is r/s a terminating decimal?
1) 90 < r < 100
2) s = 4

I understand why it's not 1 (aka, eliminate answer "A" and "D" as correct) but why is the answer to this "B"? According to the back of the book: "Division by the number 4 must terminate: the remainder when dividing 4 must be 0, 1, 2, or 3, so the quotient must end with .0, .25, .5, or .75, respectively." I think I'm missing something obvious, but...I don't understand that last line! Can someone please explain this to me in a simpler way? How do I know it will never repeat, and will definitely terminate, for any number that I divide by 4, because of the ".0, .25, .5 or .75" ? If I test a bunch of numbers, is that what will happen and I just need to "know" that this is a rule that applies for 4? Or is there a rule I can use to scale out to other questions like this, where I'm not dealing with "4" but with some other number?

Thanks so much!
Let me try.
When you divide anything by 4, you're basically dividing that thing in four equal parts.

Let's take a rod of length 1 foot. When you divide it into four equal parts (same as performing 1/4), the length of each piece would be 0.25 feet.
If the rod were of length 2 feet, 3 feet and 4 feet, the lengths of each part would be 0.5 feet, 0.75 feet and 1 foot respectively.

Now let's take a BIG number.
Say 93.

What would 93/4 be?
Well, had the number been 92, I would divide it into 4 equal parts of 23 each. But now that I've a number that is 1 greater than 92, I would divide that 1 into equal parts of 0.25 and distribute them among each of 23s so that each of part is 23.25.

If you take 95, then it is 3 more than 92. This 3 can now be divided and distributed to make each piece of length 23.75.

This goes on for every integer.

Every integer, when being divided by 4 is either perfectly divisible by 4 or leaves a remainder of 1 or 2 or 3, which makes the resulting number ending with either no decimal, or a 0.25, or a 0.50 or a 0.75 respectively.

Did that help?
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by eaakbari » Thu Nov 08, 2012 5:24 am
Keep the below in mind as a general rule.

If a denominator has prime factors consisting of only 2 & 5, then the result will be a terminating decimal.

For e.g.

93/25 = 93 /(5*5) = 3.72 - terminating

93/20 = 93 / (2*2*5) = 4.65 - terminating

93/13 = 93 / (13) = some non-terminating decimal (try it)
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