What is the most efficient way to solve this?

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by Ian Stewart » Sun Aug 24, 2008 8:42 pm
You probably know 1/3 = 0.333... already, and 2/3 = 0.666...., so you can rule them out.

Working out that 2/11 = 0.181818... is quite quick by long division. Notice:

If 2/11 = 0.181818....

then 1/11 is half of that: 0.090909...

and 1/99 is one ninth of that: 0.010101...

and 41/99 is 41 times that: 0.414141...

2/11 and 41/99 therefore both have two repeating digits- neither can be the correct answer, because then there would be two different correct answers to the question. So 23/37 must have the longest string of different digits.
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by beater » Mon Aug 25, 2008 11:27 am
Thanks Ian!

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by uptowngirl92 » Sun May 03, 2009 11:42 pm
hey guys can someone give me a more theoritical approach to this?I remember reading somewhere that dividing any number by 3 or 11(i.e if the denominator is 3 or 11)will always produce a recurring fraction.knowing that,we can eliminate options (a),(b),(d) straight away!However,my memory falls short there :? Can anybody tell me such observations which will help me eliminate (c)?Can it be said that since 99 is a multiple of 11 that too will provide a recurring decimal??

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by dtweah » Mon May 04, 2009 2:57 am
uptowngirl92 wrote:hey guys can someone give me a more theoritical approach to this?I remember reading somewhere that dividing any number by 3 or 11(i.e if the denominator is 3 or 11)will always produce a recurring fraction.knowing that,we can eliminate options (a),(b),(d) straight away!However,my memory falls short there :? Can anybody tell me such observations which will help me eliminate (c)?Can it be said that since 99 is a multiple of 11 that too will provide a recurring decimal??
The shortest way to that problem is to know about the repeating patterns of certain primes. I posted something about the number 7 and 11 some time earlier. In 10 seconds you can eliminate 1/3, 2/3 and 1/11 if you knew their properties: 1/3 and 2/3 we play with every day. 11 will always alternate with 0909. 7. will have 7 a period of 7. Division by 99 Always results in a pattern of two repeating numbers. That by 999 a pattern of 3 repeating numbers; by 9999 a pattern of four repeating numbers and so on. The only strange number in the answer choice is 23/37, division by two primes. I would suspect division by two primes will give the longest sequece of different numbers.

To see 99 in action, convert repeating decimals to fraction.
Convert .4141... to fraction. to two decimal places. Let N=.41
Then 100N= 41.41
Subtract N .41

You get 99N= 41
N=41/99
Convert. 2.0909 to decimal as in the above

100N= 209.09
subtract N= 2.09
99N=207.

Thus 207/99 is the fractions that gives 2.09090909.

You will alway divide by 99, 999 etc depending on the number of decimal places you want. If you want 4 decimal places then multiply N by 1000. Subtracting N gives 999. So you really did not have to divide 23/37 to know. You could solve this problem by eliminating the rest. Hope this helps. What I advice you do is go into excel and do all kinds of division with prime numbers and observ patterns it will help greatly. Below is an example:
41 99 0.414141414
23 37 0.621621622
39 41 0.951219512
7 9 0.777777778
11 12 0.916666667
17 14 1.214285714
37 23 1.608695652
43 53 0.811320755
57 97 0.587628866
23 99 0.232323232
57 99 0.575757576
24 999 0.024024024
47 9999 0.00470047
11 7 1.571428571
11 17 0.647058824

You can learn several things from above. Whenever you divide by two primes you see the longest string of different numbers. See the result by 99, 999 consistent with the treatment we did earlier. See division by 7. Hope this helps.

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by uptowngirl92 » Mon May 04, 2009 4:43 am
7. will have 7 a period of 7
11 7 1.571428571
could you please explain this division by 7 thing?

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by dtweah » Mon May 04, 2009 5:38 am
uptowngirl92 wrote:
7. will have 7 a period of 7
11 7 1.571428571
could you please explain this division by 7 thing?
11/7

Divide 11 by 7. The quoitient after the decimal is .5714285714285.....
Make a sequence of the remainders as well. You will get 4 (7 into 11 goes in remainder 4. continue like this and you get) 513264513264.......


You can see that eachsequence repeats itself for the 7th term.

Supposed you are asked this: 11/7 is expressed to 200 digits after the decimal point. What is the 198th digit after the decimal point. How would you attack? Hope this helps

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by Ian Stewart » Mon May 04, 2009 6:21 am
uptowngirl92 wrote:hey guys can someone give me a more theoritical approach to this?I remember reading somewhere that dividing any number by 3 or 11(i.e if the denominator is 3 or 11)will always produce a recurring fraction.knowing that,we can eliminate options (a),(b),(d) straight away!However,my memory falls short there :? Can anybody tell me such observations which will help me eliminate (c)?Can it be said that since 99 is a multiple of 11 that too will provide a recurring decimal??
It's actually true that *every time* you divide one integer by another, one of two things must happen:

-you get a 'terminating decimal', i.e. a decimal that stops. 1/2 = 0.5, or 1/8 = 0.125 are two examples;

-you get a 'recurring decimal', i.e. a decimal that eventually 'loops' in a repeating pattern forever. 1/3 = 0.3333.... or 1/11 = 0.090909.... or 5/27 = 0.185185185.... are three examples.

So in the above question, 23/37 will also produce a recurring decimal (if you calculate it, you'll find it equals 0.621621621...). And if you do calculate it using long division, you should see why the decimal must eventually repeat. When we divide 23 by 37, to find the tenths digit of the decimal, we ask "how many times can we divide 230 by 37?". The answer is 6, with a remainder of 8. We then use that remainder to find the next digit. Since at each stage, we are just finding remainders when dividing by 37, eventually we have to get back to a remainder we've already used, and the digits will begin to repeat. I hope that's clear - it would be easier to demonstrate if I could write out a long division properly here - but it should be easier to see why the digits must repeat if you actually perform the steps in the long division.

____________

While it's not likely to be important to know the following for the GMAT, there are numbers which do not have any repeating pattern of digits in their decimal. Pi is one famous example, as is the square root of 2 (or any other root which isn't an integer). This is what makes memorizing the digits of Pi considerably more challenging than memorizing the digits of, say, the decimal equivalent of 1/3.

Because the decimal equivalents of Pi and root(2) do not repeat, it is impossible to write these numbers as fractions involving integers. That is, you cannot possibly find two integers a and b for which Pi = a/b, or for which root(2) = a/b. Decimals which cannot be written as fractions involving integers are known as 'irrational numbers' in mathematics, while numbers that can be written as fractions using only integers are called 'rational numbers'. The 'real numbers' include all of the rational and irrational numbers, and nothing else.
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by Ian Stewart » Mon May 04, 2009 6:45 am
dtweah wrote:
The only strange number in the answer choice is 23/37, division by two primes. I would suspect division by two primes will give the longest sequece of different numbers. ...
Whenever you divide by two primes you see the longest string of different numbers.
It's not especially relevant that the numbers in the fraction are prime. What can be important is that the numbers are relatively prime - that is, that they do not share any divisors, meaning that your fraction is completely reduced.

You might notice, for example, that 2/27 (with a prime numerator) has a pattern of three repeating digits, just as 4/27 (with a non-prime numerator) does. Indeed, whenever you divide an integer x by 27, as long as x is not a multiple of 3 you'll always have three repeating digits; it doesn't matter if x is prime. It only matters that the fraction x/27 is completely reduced. Further, 4/27, with a non-prime denominator and a non-prime numerator, has a longer pattern of repeating digits than 2/3, which has both a prime numerator and denominator; simply looking for prime numerators and denominators will not reliably let you determine which fraction has the longest pattern of repeating digits.
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