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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
Answer = A. 0

\(0.0024 = 24 * 10^{-4}\)
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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
Bunuel wrote:
If .0024 is represented in its shortest possible decimal form, how many 0’s to the right of the decimal point does it contain?

A. 0
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.


I would go with A as well here.

0,0024 = 24 (10^-4) = 3(5^-3)(10^-1)
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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
PareshGmat wrote:
Answer = A. 0

\(0.0024 = 24 * 10^{-4}\)


I was lost on the phrase "shortest possible decimal form" what does that mean?
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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
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Bunuel wrote:
If .002^4 is represented in its shortest possible decimal form, how many 0’s to the right of the decimal point does it contain?

A. 0
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION

Correct Answer: (D)

Start by determining how many decimal places there are. Remember that the product of any number of decimals has as many decimal places as all the decimals being multiplied put together, so if .002 is multiplied by itself four times, the product will contain 12 decimal places. Next, get the result. 2^4 = 16, and since that 2 comes at the end of a decimal containing nothing other than zeros, it will come at the very end of the result, and EVERY other digit will be a 0. So there will be twelve digits, two of which are 1 and 6 and the rest of which are 0’s -- the decimal is .000000000016 – so the answer is 10 zeros.
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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
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Bunuel wrote:
If .002^4 is represented in its shortest possible decimal form, how many 0’s to the right of the decimal point does it contain?

A. 0
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.


.002^4= 2/1000^4 or 16/10^12. When you divide 16 by 10^12, two zeros will go in moving decimal point to the left of 1, and there will be 10 zeros left (of the total 12 that 10^12 has). Answer D.
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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
Expert Reply
Bunuel wrote:
If .002^4 is represented in its shortest possible decimal form, how many 0’s to the right of the decimal point does it contain?

A. 0
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.


We know that 2^4 = 16
Hence the last 2 digits would be 16.
Also, there are 3 digits in .002.
Hence (3 digits )^4 = 12 digit number
Out of these 12 digits, last 2 digits are 16
Hence there are 10 zeros after decimal.

Hence option D.

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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
Expert Reply
Hi All,

If anyone is confused by some of the posts in this thread, it's because they are in response to the original prompt (which had a 'typo' in it). The current prompt is correct as written (we're meant to deal with [.002]^4, not .0024).

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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
Bunuel wrote:
If .002^4 is represented in its shortest possible decimal form, how many 0’s to the right of the decimal point does it contain?

A. 0
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.


\(0.002^4\) = \((\frac{2}{1000})^4\)

= \(\frac{2^4}{10^{3*4}}\)

Now the fun begins -

\(2^4 = 16\) and 10^12 = 1000000000000 ( 12 trailing zero's)

So, 16 will take up 2 places of 12 digits and this we will be having only 10 zeroe's...

Hence answer will be (D)
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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
Bunuel wrote:
If .002^4 is represented in its shortest possible decimal form, how many 0’s to the right of the decimal point does it contain?

A. 0
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.


Simplification is the key word

{2/1000}^4

16/{10^12}

how many 0’s to the right of the decimal point does it contain => 10
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Re: If .0024 is represented in its shortest possible decimal form, how man [#permalink]
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