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If x is a positive integer, what is the value of x ?

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If x is a positive integer, what is the value of x ?  [#permalink]

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New post 18 Nov 2014, 08:38
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Tough and Tricky questions: Decimals.



If \(x\) is a positive integer, what is the value of \(x\)?


(1) The first nonzero digit in the decimal expansion of \(\frac{1}{x!}\) is in the hundredths place.

(2) The first nonzero digit in the decimal expansion of \(\frac{1}{(x+1)!}\) is in the thousandths place.

Kudos for a correct solution.

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Re: If x is a positive integer, what is the value of x ?  [#permalink]

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New post 19 Nov 2014, 07:51
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Bunuel wrote:

Tough and Tricky questions: Decimals.



If \(x\) is a positive integer, what is the value of \(x\)?

(1) The first nonzero digit in the decimal expansion of \(\frac{1}{x!}\) is in the hundredths place.

(2) The first nonzero digit in the decimal expansion of \(\frac{1}{(x+1)!}\) is in the thousandths place.

Kudos for a correct solution.


Official Solution:

If \(x\) is a positive integer, what is the value of \(x\)?

The question does not need rephrasing, although we should note that \(x\) is a positive integer.

Statement 1: SUFFICIENT. We should work from the inside out by first listing the first several values of \(x!\) (the factorial of \(x\), defined as the product of all the positive integers up to and including \(x\)).
\(1! = 1\)
\(2! = 2\)
\(3! = 6\)
\(4! = 24\)
\(5! = 120\)
\(6! = 720\)
\(7! = 5040\)

Now we consider decimal expansions whose first nonzero digit is in the hundredths place. Such decimals must be smaller than \(0.1\) (\(\frac{1}{10}\)) but at least as large as \(0.01\) (\(\frac{1}{100}\)). Therefore, for \(\frac{1}{x!}\) to lie in this range, \(x!\) must be larger than 10 but no larger than 100. The only factorial that falls between 10 and 100 is \(4! = 24\), so \(x = 4\).

(Note that factorials are akin to exponents in the order of operations, so \(\frac{1}{x!}\) indicates "1 divided by the factorial of \(x\)," not "the factorial of \(\frac{1}{x}\)," which would only have meaning if \(\frac{1}{x}\) were a positive integer.)

Statement 2: INSUFFICIENT. We consider decimal expansions whose first nonzero digit is in the thousandths place. Such decimals must be smaller than \(0.01\) (\(\frac{1}{100}\)) but at least as large as \(0.001\) (\(\frac{1}{1000}\)). Therefore, for \(\frac{1}{(x+1)!}\) to lie in this range, \((x+1)!\) must be larger than 100 but no larger than 1,000.

There are two factorials that fall between 100 and 1,000, namely \(5! = 120\) and \(6! = 720\). Thus, \(x+1\) could be either 5 or 6, and \(x\) could be either 4 or 5.

Answer: A.
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Re: If x is a positive integer, what is the value of x ?  [#permalink]

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New post 18 Nov 2014, 09:19
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Statement 1:

The first nonzero digit in the decimal expansion of \(\frac{{1}}{{x!}}\) is in the hundredths place.

Let us quickly check few factorials.

\(2! =2\) ---->\(\frac{1}{2}\) =.5

\(3!= 6\) ----> \(\frac{1}{6}\)= .16

\(4! = 24\) -----> \(\frac{1}{24}\) = .04 -- first nonzero digit is hundredths place
\(5!= 120\) -----> \(\frac{1}{120}\) = .008 ---- Hundredths place is zero.

With increasing value of denominator, place of nonzero digit in decimal expansion will shift to right.

So only \(4!\)has first non zero digit in hundredths place.

Therefore, \(x= 4\)

Statement1 is sufficient.



Statement 2:
The first nonzero digit in the decimal expansion of \(\frac{{1}}{{(x+1)!}}\) is in the thousandths place.

Let us try some number and check.

\(x= 4\) ----->\(\frac{1}{(4+1)!}\) \(=\) \(\frac{1}{5!}\) \(=\) \(0.008\) after decimal expansion first non zero digit is at thousandths place.

\(x= 5\) -----> \(\frac{1}{(5+1)!}\) = \(\frac{1}{6!}\) \(=\) \(0.0013\) after decimal expansion first non zero digit is at thousandths place.

Statement 2 not sufficient.

Answer: A

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Re: If x is a positive integer, what is the value of x ?  [#permalink]

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New post 27 Sep 2017, 00:17
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Re: If x is a positive integer, what is the value of x ? &nbs [#permalink] 27 Sep 2017, 00:17
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