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If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
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Updated on: 09 Nov 2011, 16:38
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One of the GMAT Club's best came up with this question for fun. Originally it was at about 800 level but I think this version is close to 750 - would be curious to hear what you think. (I personally had not idea how to tackle it)
If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and D are different integers, what is the value of T?
(1) X = 9115/N where N is a whole number (2) T has 3 distinct factors: W, Y and T and the sum of W, Y and T is 10W+Y.
(1) X = 0.TBCDBCD 10x = T. BCDBCD ………………………….1 1000(10x) = TBCD. BCDBCD………………2 Deduct 1 from 2: 9990x = TBCD – T x = (TBCD – T)/9990 x = 9115/9990 x = 1823/1998 x = 0.9124124124 Sufficient
(2) T should be 9, only odd with 3 distinct factors: 1, 3 and 9. Sum = 1+3+9 = 13 = 10(1)+3. Sufficient
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Re: If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
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07 Jun 2009, 08:40
1
Kudos
Hades wrote:
And also, what if there is another value s.t. X=9115/N is equal to .TBCDBCD..... such that T != 9?
You haven't shown that either, you're assuming there is only one answer.
[deleted an attitude comment]
I do not see any reason not to understand that n is/must be 9990 given that x = m/n = 9115/n.
Again with the given facts (i.e. the structire of 0.TBCDBCDBCD in which T is not repeating but BCD is), the denomenator, n, must be 9990.
In x = (TBCD-T)/9990 = m/n, m (or TBCD-T) cannot be greater than 9115 but could be smaller than that. The maximum m is a 4 digit number (zzzz or 9115) with 9 in thousand's place. Similarly (TBCD-T) is the max 4 digit number in neumarator for x = m/n. m could be 1823. In that case, when x = 1823/p where p = n/q, yes I agree that (TBCD-T) could not be 9115 but when m is clearly given 9115, (TBCD -T) is that 9115.
Therefore (TBCD-T) = 9115.
If it is still unclear, PM me for more detail repeating/terminating decimal.
It seems a real difficult question. :-D
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Re: If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
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03 Jun 2009, 16:02
Statement 2 :
T has 3 distinct factors. T can be 4 ( 4,2,1) OR 9 ( 9,3,1). W+Y+T=10W+Y for 9
SUFFICIENT.
For N=5 , T=0 and B,C,D are not different. 9115/N=0.4 Statement 1 NOT SUFFICIENT.
IMO B.
I also tried to use the value of T we got from (1) ; however it becomes complex. We have to use 2 values of X (0.913 and 0.931) in order to check the value of N.
Could you please explain how this one is D? _________________
Re: If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
[#permalink]
03 Jun 2009, 19:36
for 1.
X = 9115/N
when we divide any integer with the same digit number with all 9s we are sure to get a recurring decimal ex 1/9 = 0.11111111 10/99 = 0.101010101010.... 100/999 = 0.100100100.. 1000/9999 = 0.100010001000....
therefore for 9115/N to be recurring and also < 1, N =9999 and in that case X = 0.911591159115...... and in that case T =9, only problem this is of the form 0.TBCDTBCDTBCD and not 0.TBCDBCD and also B and C are not different since both = 1.....I give up here....somebody show the light please......
Re: If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
[#permalink]
03 Jun 2009, 21:19
Expert Reply
Solution:
(1) X = 0.TBCDBCD 10x = T. BCDBCD ………………………….1 1000(10x) = TBCD. BCDBCD………………2 Deduct 1 from 2: 9990x = TBCD – T x = (TBCD – T)/9990 x = 9115/9990 x = 1823/1998 x = 0.9124124124 Sufficient
(2) T should be 9, only odd with 3 distinct factors: 1, 3 and 9. Sum = 1+3+9 = 13 = 10(1)+3. Sufficient _________________
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Re: If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
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03 Jun 2009, 23:04
Expert Reply
Neochronic wrote:
1. x = (TBCD – T)/9990
2. x = 9115/9990
How can we deduce the second statement from 1st statement.. whats is the approach..
I think the idea is that in Statement (1) that X = 9115/N where N is a whole number, but I am not sure that it works that way now that you are asking. Let me check. _________________
Founder of GMAT Club
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Re: If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
[#permalink]
04 Jun 2009, 07:36
bb wrote:
One of the GMAT Club's best came up with this question for fun. Originally it was at about 800 level but I think this version is close to 750 - would be curious to hear what you think. (I personally had not idea how to tackle it)
If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and D are different integers, what is the value of T?
(1) X = 9115/N where N is a whole number (2) T has 3 distinct factors: W, Y and T and the sum of W, Y and T is 10W+Y.
(1) X = 0.TBCDBCD 10x = T. BCDBCD ………………………….1 1000(10x) = TBCD. BCDBCD………………2 Deduct 1 from 2: 9990x = TBCD – T x = (TBCD – T)/9990 x = 9115/9990 x = 1823/1998 x = 0.9124124124 Sufficient
(2) T should be 9, only odd with 3 distinct factors: 1, 3 and 9. Sum = 1+3+9 = 13 = 10(1)+3. Sufficient
bb wrote:
Neochronic wrote:
1. x = (TBCD – T)/9990
2. x = 9115/9990
How can we deduce the second statement from 1st statement.. whats is the approach..
I think the idea is that in Statement (1) that X = 9115/N where N is a whole number, but I am not sure that it works that way now that you are asking. Let me check.
IMO, 2 cannot be deduced from 1 above if x = 9115/N is not given.
Since X = 9115/N is given, x should be = 9115/9990. "9115 is/should be (TBCD -T) as given in statement 1 and N has to be 9990 if "TBCD -T = 9115".
Can you elaborate your problem that why 2 cannot be deduced from 1 if x = 9115/N is given in detail?
Re: If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
[#permalink]
06 Jun 2009, 17:01
Neochronic wrote:
1. x = (TBCD – T)/9990
2. x = 9115/9990
How can we deduce the second statement from 1st statement.. whats is the approach..
10000x - 10x = 9990x. With a recurring decimal in the form of 0.TBCDBCD... only the numerator with 9990 as the denominator will consist of the recurring digits (that's not totally correct, there are other possible denominators, but they are irrelevant to a four digit numerator). Statement 1 gives x = 9115/N Therefore, TBCD - T = 9115 is the only possibility.
Re: If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and
[#permalink]
06 Jun 2009, 23:53
bb wrote:
If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and D are different integers, what is the value of T?
(1) X = 9115/N where N is a whole number (2) T has 3 distinct factors: W, Y and T and the sum of W, Y and T is 10W+Y
From 1: X = 0.TBCDBCD multiply both side by 10 (because ine digit from the decimal is not repeating. 10x = T. BCDBCD ………………………….1 Multiply 10 by 1000 (because 3 digits right to the decimal are repeating) 1000(10x) = TBCD. BCDBCD………………2 Deduct 1 from 2: 9990x = TBCD – T x = (TBCD – T)/9990 Since the neumarator of x is given, (TBCD-T) should be = 9115. With the given information (i.e 0.TBCD in which T is not a repeating but B, C, and D are), the neuramator of x can not be > 9115. So..
x = 9115/9990 x = 1823/1998 x = 0.9124124124 Sufficient