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x is positive integer

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by j_shreyans » Mon Sep 15, 2014 3:38 am
If x is a positive integer, what is the value of x?

(1) The first nonzero digit in the decimal expansion of 1/x! is in the hundredths place.

(2) The first nonzero digit in the decimal expansion of 1/(x+1)! is in the thousandths place

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Source: — Data Sufficiency |
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by GMATGuruNY » Mon Sep 15, 2014 4:14 am
j_shreyans wrote:If x is a positive integer, what is the value of x?

(1) The first nonzero digit in the decimal expansion of 1/x! is in the hundredths place.

(2) The first nonzero digit in the decimal expansion of 1/(x+1)! is in the thousandths place
1/3! = 1/6 ≈ 0.16.
1/4! = 1/24 ≈ 4/100 = 0.04.
1/5! = 1/120 = 8/960 ≈ 8/1000 = 0.008.
1/6! = 1/720 ≈ 1/7 * 1/100 ≈ 0.14 * 10­ˉ² = 0.0014.

Statement 1:
Thus, x=4, with the result that 1/x! = 1/4! = 0.04.
SUFFICIENT.

Statement 2:
It's possible that x=4, with the result that 1/(x+1)! = 1/5! ≈ 0.008.
It's possible that x=5, with the result that 1/(x+1)! = 1/6! ≈ 0.0014.
Since x can be different values, INSUFFICIENT.

The correct answer is A.
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by Brent@GMATPrepNow » Mon Sep 15, 2014 6:42 am
j_shreyans wrote:If x is a positive integer, what is the value of x?

(1) The first nonzero digit in the decimal expansion of 1/x! is in the hundredths place.
(2) The first nonzero digit in the decimal expansion of 1/(x+1)! is in the thousandths place
Mitch's solution is great.
We can also solve the question without finding an approximate value of each term. Here's how:
First make the following observations:
1/1 = 1
1/10 = 0.1
1/100 = 0.01
1/1000 = 0.001

So, we get:
1/1 = 1
1/2! = 1/2 = 0.5
1/3! = 1/6 = 0.something [if 1/6 is BETWEEN 1/10 and 1/1, then 1/6 is between 0.1 and 1, which means 1/6 = 0.something]
1/10 = 0.1
1/4! = 1/24 [if 1/24 is BETWEEN 1/100 and 1/10, then 1/24 is between 0.01 and 0.1, which means 1/24 = 0.0something]
1/100 = 0.01
1/5! = 1/120 = 0.00something.
1/6! = 1/720 = 0.00something.
1/1000 = 0.001

Okay, onto the question.....

Target question: What is the value of x?

Statement 1: The first nonzero digit in the decimal expansion of 1/x! is in the hundredths place.
As we can see above, ONLY 1 value of x (x = 4) satisfies this condition.
So, x MUST EQUAL 4
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The first nonzero digit in the decimal expansion of 1/(x+1)! is in the thousandths place
As we can see above, 1/5! and 1/6! both satisfy this condition.
This means that EITHER (x+1) = 5 OR (x+1) = 6
In other words, EITHER x = 4 OR x = 5
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent
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by gmatcracker0123 » Mon Sep 15, 2014 11:07 am
Great understanding builder types solution Brent! Thanks!
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