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If each of the following fractions were written as a [#permalink]

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20 Sep 2005, 20:59

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If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits ?
A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/37
F) 1/7

(I added one more choice to the original question.)

Is there an easy way to get result without calculating each choice ?

If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits ? A) 2/11 B) 1/3 C) 41/99 D) 2/3 E) 23/37 F) 1/7

(I added one more choice to the original question.)

Is there an easy way to get result without calculating each choice ?

It should be 1/7 which is having a sequence 0.142857

You can eliminate a couple of choices right off the bat before going to the step of calculating them out. For example, 1/3 is an obvious one to eliminate, because it is simply 0.3333333 and therefore would have the shorterst string of numbers. Same with 2/3.

Since 1/7 wasn't part of the original question, that leaves 3 choices to calculate. Then, since 2/11 is 0.181818, 41/99 is 0.414141 and 23/37 is 0.621621621, the answer is E, or 23/37.

With your added choice of F, F becomes the answer. But this wasn't part of the original question if I remember correctly, so no need to go there.

Alternately, you could note that dividing a number by 99 is the same as dividing it by 11x9. So you could simply get the decimal of 1/11 (0.090909) and then note that multiplying it by 2 would give you a repeating decimal of 2 digits, while dividing it by 9 and then multiplying it by an integer also yields a repeating decimal of 2 digits. Then you only need to calculate E to see that it has more than 2 digits in the decimal.

Also 37 = 111/3
So that will repeat fairly quickly.
In general things like 3,9,11 do.
It is the bigger primes like 7 and 13 that give longer series.
There is some interest maths underneath all this.
But to be honest working it out in each case is the safest.
You might get one question like this, or you might not get any.

thanks,
it seems that if m is a prime number, then n/m will have the same number of repeating digits as 1/m. That way the calcuation can be cut down a little bit

Re: If each of the following fractions were written as a [#permalink]

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11 Sep 2016, 05:50

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Re: If each of the following fractions were written as a [#permalink]

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15 Nov 2017, 20:52

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