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

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

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

Nothing in the question stem except that x is a positive integer.

(1) The first nonzero digit in the decimal expansion of $$\frac{1}{x!}$$ is in the hundredths place.
We can write it as $$\frac{1}{x!}=0.0xy$$, where x is non-zero digit.
In other words, $$0.001<\frac{1}{x!}< 0.1$$ or $$\frac{1}{0.01}>x!> \frac{1}{0.1}$$.
$$100>x!> 10$$.
3!=6
4!=24
5!=24*5=120
Thus, x=4.
Sufficient

(2) The first nonzero digit in the decimal expansion of $$\frac{1}{(x+1)!}$$ is in the thousandths place.
We can write it as $$\frac{1}{(x+1)!}=0.00xy$$, where x is non-zero digit.
In other words, $$0.001<\frac{1}{x!}< 0.01$$ or $$\frac{1}{0.01}>x!> \frac{1}{0.001}$$.
$$1000>x!> 100$$.
4!=24
(x+1)!=5!=24*5=120....x=4
(x+1)!=6!=120*6=720....x=5
Thus, x=4 or 5.
Insufficient

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