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Re: What is the 25th digit to the right of the decimal point in [#permalink]

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16 Aug 2012, 11:48

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C) 5 Same solution as above, but in case the repeating digits are more numerous then see: 1) Number of repeating digits (in this case 2 -> .5454..) 2) Divide the # of the digit you're supposed to view (25th digit) by # of repeats = (25/2)th digit 3) If no remainder, then its 4, if there's a Remainder = 1 then it has to be the next digit or 5, as in this case
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Re: What is the 25th digit to the right of the decimal point in [#permalink]

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16 Aug 2012, 12:35

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6/11 results in a Non-terminating decimal equal to .545454......54 Every even place (2nd,4th,6th,,,,,,,,,24th) digit is 4. Every Odd place (1,3,5.....25) digit is 5. Thus thus the odd place digit is 5 Hence Answer C.
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Re: What is the 25th digit to the right of the decimal point in [#permalink]

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29 Apr 2015, 06:29

Just want to be clear here. Are we expected to know non terminating fractions or expected to do the math of 6/11? I could do, but I felt it would be math intensive and wanted to avoid that. The rule that seemed relevant from the number theory link is that terminating fractions have a denominator that fits the 2^m*5^n formula. What's the best way to approach this, and what math is unnecessary? Please advise.

Re: What is the 25th digit to the right of the decimal point in [#permalink]

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29 Apr 2015, 09:37

I divided 6/11 and got to .545, after that I realized the pattern would repeat itself. As there's just two different digits the even decimals will be 5 and the odds will be 4. Thus 5 is the answer and C as the answer choice.
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The GMAT cannot realistically expect you to work through 25 digits one at a time, so this question MUST have a pattern to it.

There are certain pieces of information that you're expected to have memorized (formulas, math and grammar rules, etc.) and other types of information that would be beneficial to have memorized (fraction-to-decimal conversions, commonly tested 'patterns', etc.). It shouldn't be difficult to memorize the decimal equivalent of a set of fractions, BUT if you don't want to, then that's fine - the work will just take a little longer.

So the question in this case becomes "how quickly can you mathematically determine that 6/11 = .5454 repeating?" You were concerned that the work would be time-sensitive, so you didn't attempt it. But I'm curious about how quickly you COULD have done it. Grab a timer and find out. I bet it takes 10-15 seconds (at most) until you have the proof that you need.

What is the 25th digit to the right of the decimal point in the decimal form of 6/11?

(A) 3 (B) 4 (C) 5 (D) 6 (E) 7

To solve this question, we first have to use some long division. This long division allows us to get 6/11 in decimal form, which is 0.545454… where “54” is repeating.

We can see that the 1st, 3rd, 5th (and so on) digit to the right of the decimal point is a 5 and that the 2nd, 4th, 6th (and so on) digit to the right of the decimal point is a 4. In other words, each odd-positioned digit is a 5, and each even-positioned digit is a 4.

Because we are being asked about the 25th digit to the right of the decimal point and we see that 25 is odd, we know that the 25th digit is a 5.

Answer C.
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Re: What is the 25th digit to the right of the decimal point in [#permalink]

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23 Jun 2016, 03:42

Aximili85 wrote:

C) 5 Same solution as above, but in case the repeating digits are more numerous then see: 1) Number of repeating digits (in this case 2 -> .5454..) 2) Divide the # of the digit you're supposed to view (25th digit) by # of repeats = (25/2)th digit 3) If no remainder, then its 4, if there's a Remainder = 1 then it has to be the next digit or 5, as in this case

Re: What is the 25th digit to the right of the decimal point in [#permalink]

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30 Aug 2016, 15:25

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Another way of solving this one=> 6/11=> multiply 9 to both he numerator and denominator => 54/99 hence the decimal expansion would be 0.54545454545454 Clearly the 57th digit is 5=> Smash that C

Why did we write 6/11 as 54/99 => if denominator and numerator have same number of digits and if the denominator can be written as 10^n-1 ie 9,99,999,9999 etc then we need not carry any division process . The result would be given by the numerator

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