Vincen 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 $$\frac{1}{\left(x+1\right)!}$$ is in the thousandths place.
The OA is A.
I don't know how can I solve this DS question. Any expert can help me please?
Let's know few factorial values.
1. 0! = 1
2. 1! = 1
3. 2! = 2
4. 3! = 6
5. 4! = 24
6. 5! = 120
7. 6! = 720
(1) The first nonzero digit in the decimal expansion of 1/x! is in the hundredths place.
The information implies that value of 1/x! = 0.0h...; where h is a nonzero digit
Let's use brute force.
Say x = 3, then 1/3! = 1/6 = 0.16. This is not a valid example since the first nonzero digit is in the tenths place (not in hundredths place).
Say x = 4, then 1/4! = 1/24 = ~1/25 = ~0.04. This is a valid example since the first nonzero digit is in the hundredths place.
The value of x can be 4.
Let's try with x = 5.
Say x = 5, then 1/5! = 1/120 = ~1/125 = ~0.04/5 = ~0.008... This is not a valid example since the first nonzero digit is in the thousandths place (not in hundredths place).
There is no need to check further since the first nonzero digit will be either in the thousandths place or in further away positions.
So, x = 4. Sufficient.
(2) The first nonzero digit in the decimal expansion of 1/(x+1)! is in the thousandths place.
Since we got x = 4 from the first statement, it must satisfy here too.
Let's try with x = 5.
At x = 5, the value of 1/6! = 1/720 = ~1/(6*125) = ~0.008/6 = ~0.001... This is a valid example since the first nonzero digit is in the thousandths place. The value of x can be 5 also.
So x = 4 or 5. No unique value. Insufficient.
The correct answer:
A
Hope this helps!
-Jay
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