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# If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and

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If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and [#permalink]  02 Jun 2009, 22:11
Expert's post
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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.

[Reveal] Spoiler: Answer
Official Answer: D
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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Last edited by wizardsasha on 09 Nov 2011, 15:38, edited 2 times in total.
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Senior Manager
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Re: Challenge: Can you crack THIS? [#permalink]  02 Jun 2009, 23:01
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Is it b,

1, insufficient.. I have no clue how to proceed with this..
2, sufficient..

the only single digit numbers with 3 distinct factors are 4 ( 4,2,1) and 9 ( 9,3,1)..

and 4 doesnt satisfy the other condition but 9 satisfies..

9+1+3 = 10*1 + 3
13 = 13.

Hence T = 9.

The answer must be B or D..
but still dont have how to proceed withe the option 1.
Founder
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Re: Challenge: Can you crack THIS? [#permalink]  03 Jun 2009, 09:17
Expert's post
You are on the right Track. The OA is D.
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Re: Challenge: Can you crack THIS? [#permalink]  03 Jun 2009, 15:02
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?
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Re: Challenge: Can you crack THIS? [#permalink]  03 Jun 2009, 16:27
I got B/D right away, and I would have gone with D.

(1) is too difficult to do with a calculator and far too time consuming, but I'd like to see the solution.
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Re: Challenge: Can you crack THIS? [#permalink]  03 Jun 2009, 18:36
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......
Founder
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Re: Challenge: Can you crack THIS? [#permalink]  03 Jun 2009, 20:19
Expert's post
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: Challenge: Can you crack THIS? [#permalink]  03 Jun 2009, 21:43
1. x = (TBCD – T)/9990

2. x = 9115/9990

How can we deduce the second statement from 1st statement..
whats is the approach..
Founder
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Re: Challenge: Can you crack THIS? [#permalink]  03 Jun 2009, 22:04
Expert's post
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.
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Re: Challenge: Can you crack THIS? [#permalink]  04 Jun 2009, 06:36
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.

[Reveal] Spoiler: Answer
Official Answer: D
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

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?
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Re: Challenge: Can you crack THIS? [#permalink]  04 Jun 2009, 20:30
TBCD - T............. I agree with neochronic, that doesn't make any sense
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Re: Challenge: Can you crack THIS? [#permalink]  06 Jun 2009, 15:52
Good question, I got D quite quickly.

For statement 1 use the standard method to convert a recurring decimal to a fraction:

x = 0.TBCDBCD

Using x form two different numbers whose difference will eliminate the recurring digits:
10000x = TBCD.TBCD...
10x = T.BCD...

Take the second equation from the first:

9990x = TBCD - T

x = (TBCD - T)/9990

So TBCD - T = 9115 - The only possibility is T=9.

Statement 2 points to answer very quickly - only 2 single digit numbers have 3 distinct factors, 4 and 9. Easy to eliminate 4.
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Re: Challenge: Can you crack THIS? [#permalink]  06 Jun 2009, 16:01
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.
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Re: Challenge: Can you crack THIS? [#permalink]  06 Jun 2009, 16:12
Quote:
10000x = TBCD.TBCD...
10x = T.BCD...

Take the second equation from the first:

9990x = TBCD - T

Still doesn't make any sense. You're making a pretty big jump in logic. If you were to literally subtract T.BCD from TBCD.TBCD...
we'd get

TBC(D-T).(T-B)(B-C)(C-D).........ignoring any carrying etc........................ how that converts to TBCD - T is beyond me
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Re: Challenge: Can you crack THIS? [#permalink]  06 Jun 2009, 16:16
Hades wrote:
Quote:
10000x = TBCD.TBCD...
10x = T.BCD...

Take the second equation from the first:

9990x = TBCD - T

Still doesn't make any sense. You're making a pretty big jump in logic. If you were to literally subtract T.BCD from TBCD.TBCD...
we'd get

TBC(D-T).(T-B)(B-C)(C-D).........ignoring any carrying etc........................ how that converts to TBCD - T is beyond me

It's TBCD.BCD not TBCD.TBCD. So using your method that'll give us:

TBCD - T. (T-T)(B-B)(C-C) giving us TBCD - T
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Re: Challenge: Can you crack THIS? [#permalink]  06 Jun 2009, 16:24
I disagree with the logic there... I guess it works

TBCD - T isn't proper mathematics
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Re: Challenge: Can you crack THIS? [#permalink]  06 Jun 2009, 22:53
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

(2) No dispute.

Sufficient:- D.
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Re: Challenge: Can you crack THIS? [#permalink]  06 Jun 2009, 22:57
mujimania wrote:
Good question, I got D quite quickly.

For statement 1 use the standard method to convert a recurring decimal to a fraction:

x = 0.TBCDBCD

Using x form two different numbers whose difference will eliminate the recurring digits:
10000x = TBCD.BCD...
10x = T.BCD...

Take the second equation from the first:

9990x = TBCD - T

x = (TBCD - T)/9990

So TBCD - T = 9115 - The only possibility is T=9.

Statement 2 points to answer very quickly - only 2 single digit numbers have 3 distinct factors, 4 and 9. Easy to eliminate 4.

Thats much clearer.
Good job.
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Re: Challenge: Can you crack THIS? [#permalink]  07 Jun 2009, 00:24
Can anyone show a proper Mathematical proof as to why (1) is sufficient?

TBCD - T is not based in any solid Mathematical framework... You can interpret TCBD as literally T*C*B*D or 10^3*T+10^2*C+10^1*B+10^0*D - 10^0*D...

But I still have no idea in hell how it magically morphs to 9115/X....... Sorry but nobody has provided any concrete Mathematics behind it.
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Re: Challenge: Can you crack THIS? [#permalink]  07 Jun 2009, 00:26
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]
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Re: Challenge: Can you crack THIS?   [#permalink] 07 Jun 2009, 00:26

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# If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and

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