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If k is an integer and (.0025)(.025)(.00025)10^k is an integer, what is the least possible value of K?
(A) -12 (B) -6 (C) 0 (D) 6 (E) 12
Given: \(0.0025*0.025*0.00025*10^k=integer\)
There are 4 decimal places after zero in 0.0025, 3 decimal places after zero in 0.025 and 5 decimal places after zero in 0.00025 so in order the product to be an integer k must be at least 4+3+5=12 to convert all these fractions into the integers, in this case: \(0.0025*0.025*0.00025*10^{12}=25*25*25=integer\)
I.e. (k-12) = Integer I.e. Min (k-12)=0 I.e. (k)min = 12
Ans: Option E
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Re: If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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11 Dec 2015, 12:12
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since 25 to the power any integer will always yield last digit as 5 and not yield any 0, k is just summation of number of places we need to move the decimal point to get each individual number to be an integer i.e. 4+3+5=12
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If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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03 Mar 2017, 18:59
2
1
Bunuel wrote:
monirjewel wrote:
If k is an integer and (.0025)(.025)(.00025)10^k is an integer, what is the least possible value of K?
(A) -12 (B) -6 (C) 0 (D) 6 (E) 12
Given: \(0.0025*0.025*0.00025*10^k=integer\)
There are 4 decimal places after zero in 0.0025, 3 decimal places after zero in 0.025 and 5 decimal places after zero in 0.00025 so in order the product to be an integer k must be at least 4+3+5=12 to convert all these fractions into the integers, in this case: \(0.0025*0.025*0.00025*10^{12}=25*25*25=integer\)
Answer: E.
Hi. Can you please explain why it's not 10^-12? I understand how you get to 12, but have issues with how to know if it should be negative or positive.
Re: If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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04 Mar 2017, 02:15
1
mhill5446 wrote:
Bunuel wrote:
monirjewel wrote:
If k is an integer and (.0025)(.025)(.00025)10^k is an integer, what is the least possible value of K?
(A) -12 (B) -6 (C) 0 (D) 6 (E) 12
Given: \(0.0025*0.025*0.00025*10^k=integer\)
There are 4 decimal places after zero in 0.0025, 3 decimal places after zero in 0.025 and 5 decimal places after zero in 0.00025 so in order the product to be an integer k must be at least 4+3+5=12 to convert all these fractions into the integers, in this case: \(0.0025*0.025*0.00025*10^{12}=25*25*25=integer\)
Answer: E.
Hi. Can you please explain why it's not 10^-12? I understand how you get to 12, but have issues with how to know if it should be negative or positive.
10^(-12) = 1/10^12
It's not 10^(-12) because if you substitute this value there you won't get an integer.
You can check easier example: 0.12*10^(-2) = 0.12*1/10^2=0.0012 but 0.12*10^2 = 12.
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Re: If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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08 Dec 2017, 11:25
4
monirjewel wrote:
If k is an integer and (.0025)(.025)(.00025)10^k is an integer, what is the least possible value of K?
(A) -12 (B) -6 (C) 0 (D) 6 (E) 12
We are given the expression:
0.0025 x 0.025 x 0.00025 x 10^k = integer
To determine the least possible value of k, we want to use our rules of multiplication with decimals. When multiplying decimals, the final product has an equal number of decimal places to the decimal places of the numbers being multiplied. Let’s start by counting the number of decimal places.
0.0025 has 4 decimal places
0.025 has 3 decimal places
0.00025 has 5 decimal places
Thus, the result of 0.0025 x 0.025 x 0.00025 will have 12 decimal places.
In order for 0.0025 x 0.025 x 0.00025 x 10^k = integer, k would have to be at least 12, since 10^12 times any number with 12 decimal places would move the decimal point of that number 12 places to the right, making it an integer.
If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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28 Jul 2019, 14:29
ScottTargetTestPrep wrote:
monirjewel wrote:
If k is an integer and (.0025)(.025)(.00025)10^k is an integer, what is the least possible value of K?
(A) -12 (B) -6 (C) 0 (D) 6 (E) 12
We are given the expression:
0.0025 x 0.025 x 0.00025 x 10^k = integer
To determine the least possible value of k, we want to use our rules of multiplication with decimals. When multiplying decimals, the final product has an equal number of decimal places to the decimal places of the numbers being multiplied. Let’s start by counting the number of decimal places.
0.0025 has 4 decimal places
0.025 has 3 decimal places
0.00025 has 5 decimal places
Thus, the result of 0.0025 x 0.025 x 0.00025 will have 12 decimal places.
In order for 0.0025 x 0.025 x 0.00025 x 10^k = integer, k would have to be at least 12, since 10^12 times any number with 12 decimal places would move the decimal point of that number 12 places to the right, making it an integer.
Answer: E
I was able to solve the question, but just curious, Why can K not be Zero since if we raise any number to zero we get 1? and both 1 and 0 are integers.
If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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02 Aug 2019, 09:04
Shef08 wrote:
Hi,
Can someone explain why not C can be the correct answer? As 10^0 equals 1. So 0 is the least possible value of K
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Because if k=0, then \((0.0025)(0.025)(0.00025)*10^k\) is equal to \((0.0025)(0.025)(0.00025)*1\), which is not an integer.
See, \((0.0025)(0.025)(0.00025)\) is equal to \((25*10^-4)(25*10^-3)(25*10^-5)\) which is equal to \(25^3*10^-12\).
For it to be an integer by itself, then the numerator would have to be equal to or a multiple of the denominator and \(25^3\) is not equal to \(10^-12\).
Re: If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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16 Sep 2019, 07:59
monirjewel wrote:
If k is an integer and (.0025)(.025)(.00025)10^k is an integer, what is the least possible value of K?
(A) -12 (B) -6 (C) 0 (D) 6 (E) 12
If k is an integer and (.0025)(.025)(.00025)10^k is an integer, what is the least possible value of K? \(25 * 10^{-4} * 25* 10^{-3} * 25*10^{-5} * 10^k = 25*25*25 * 10^{-4-3-5+k}\) -4-3-5+k =0 k = 12
Re: If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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06 Nov 2019, 01:53
because the final number of the product is not 10, we can not reduce any time of 10. for example if we have 0.5*0.002 *k^x we do not need k^4, we need only K^3 because 5*2=10
Re: If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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13 Nov 2019, 08:51
can someone please help me understand the OG explanation. It says that "since the unit digit of (25)^3 is 5, it follows that if K=11, then the 10th digit of N would be 5, and thus N would not be an integer; and if K=12, then N would be (25)^3 x (10)^0, which is an integer". what do they mean by this?? won't (25)^3 also be an integer??
gmatclubot
Re: If k is an integer and (0.0025)(0.025)(0.00025) × 10^k is an integer,
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13 Nov 2019, 08:51
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76% (01:15) correct 
