If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and : DS Archive - Page 2
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# If X is a repeating decimal = 0.TBCDBCD, where T, B, C, and

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Re: Challenge: Can you crack THIS? [#permalink]

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07 Jun 2009, 00:37
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]

Nobody is obliged to show you anything. I am closing this thread for now. If anyone has questions about it, please PM me and I will reopen.
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Re: Challenge: Can you crack THIS? [#permalink]

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07 Jun 2009, 07:40
1
KUDOS
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.
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Re: Challenge: Can you crack THIS?   [#permalink] 07 Jun 2009, 07:40

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