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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 07 Oct 2011, 22:13
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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.
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Re: DS  [#permalink]

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New post 08 Oct 2011, 00:49
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kkalyan wrote:
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Help End Exam Review Section

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

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


Used brute force:

1/1!=1
1/2!=0.5
1/3!=0.1XX
1/4!=0.04XX
1/5!=0.008XX
1/6!=0.001XX

1)
x=4
Sufficient.

2)
x=4
x=5
Not Sufficient.

Ans: "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 Oct 2011, 20:26
1
Hi,
i am a bit confused, Can some body please elaborate why the answer is not D) ? Because from first statement, the x! is 4 leading to hundredths place on decimal, then similarly from 2nd statement 5! leads to thousandth place, therefore x+1 = 5, hence x= 4 ?
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Re: If x is a positive integer, what is the value of x ? The  [#permalink]

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New post 03 Apr 2014, 12:30
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1
Encountered this in Manhattan Quiz:

Posting OE for others. 700 level q.

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 (110) but at least as large as 0.01 (1100). Therefore, for 1x! 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 1x! indicates "1 divided by the factorial of x," not "the factorial of 1x," which would only have meaning if 1x 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 (1100) but at least as large as 0.001 (11000). Therefore, for 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.


The correct answer is 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 Apr 2015, 07:35
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.



1/100>1/x! >1/1000
x must be 4
if x=5, it is not ok
if x=3, it is not ok

2, x can be 6 or 6,
not sufficient

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

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New post 15 May 2015, 02:17
fluke wrote:
kkalyan wrote:
Previous Next
Help End Exam Review Section

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

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


Used brute force:

1/1!=1
1/2!=0.5
1/3!=0.1XX
1/4!=0.04XX
1/5!=0.008XX
1/6!=0.001XX

1)
x=4
Sufficient.

2)
x=4
x=5
Not Sufficient.

Ans: "A"


can not say a word for this excellence. very good.
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Re: If x is a positive integer, what is the value of x ?  [#permalink]

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