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If (10^4 * 3.456789)^8 is written as a single term, how many [#permalink]
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11 Mar 2013, 22:19
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If (10^4 * 3.456789)^8 is written as a single term, how many digits would be to the right of the decimal place? A. 2 B. 12 C. 16 D. 32 E. 48
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Last edited by Bunuel on 12 Mar 2013, 01:09, edited 1 time in total.
Edited the question.



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Re: If [m](10^4 x 3.456789)^8[/m] is written as a single term, h [#permalink]
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11 Mar 2013, 22:52
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emmak wrote: If \((10^4 X 3.456789)^8\) is written as a single term, how many digits would be to the right of the decimal place? a) 2 b) 12 c) 16 d) 32 e) 48 \((10^4 X 3.456789)^8 = (10^4 X 10^{6} X 3456789)^8 = (10^4 X 10^{6} X 3456789)^8 = (10^{2} X 3456789)^8\) = \(3456789^8 X 10^{16}\) Answer is C
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Re: If (10^4 * 3.456789)^8 is written as a single term, how many [#permalink]
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12 Mar 2013, 01:13
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Re: If (10^4 * 3.456789)^8 is written as a single term, how many [#permalink]
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04 Sep 2013, 10:30
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emmak wrote: If (10^4 * 3.456789)^8 is written as a single term, how many digits would be to the right of the decimal place?
A. 2 B. 12 C. 16 D. 32 E. 48 First of all split the terms (10^4)^8 * (3.456789)^8 > (10)^32 * (3.456789)^8 10^32 will have 32 zeros after 1. (3.456789)^8 > Here we should understand 2 things 1) Any decimal when squared its decimal places doubles e.g. 1.2^2 = 1.44 2) x^2 means (x^2)^2)^2 So here when 3.456789 squared its decimal places would become 12. Squaring the resultant figure again would give 24 decimal places AND squaring the resultant figure for the final time would give 48 decimal places. So we will have the figure as (x.abcd....48 times)*(1000000.....32 times). The 32 zeros would shift the decimal sign of x.abcd... to the right side by 32 places. So beyond decimal point we will have 48  32 = 16 places. Hence C
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Re: If (10^4 * 3.456789)^8 is written as a single term, how many [#permalink]
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11 Oct 2013, 11:40
In general, a fraction is something upon 10. ex : 2.34 = 234/100 3.4 = 34/10 3.6789= 36789/10000
So when you square such numbers, you basiaclly square numerator and denominators both. Say (2.34)^2 = (234/100)^2 = 234^2 / 100 ^2
so you end up having some number upon 10000, in this case. Generally speaking you increase number of decimal digits by a multiplication factor of 2. If its cubing then multiplication factor is 3, and decimal will have 6 digits
So if you had one place of decimal, say 1.1 after squaring you shall have somenumber.2 places of decimal.
To our question 10^4 * 3.456789 becomes (34567.89)^8.
Sow we have (3456789/100)^8
or (3456789)^8 / 100^8
= some number / 10^16 or 16 places of decimal.



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Re: If (10^4 * 3.456789)^8 is written as a single term, how many [#permalink]
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25 Oct 2013, 03:05
(34567.89)^8 Whenever you multiply a decimal by itself, you get the number of decimals times the power for the numbers on the right 0.2^2 = 0.04, 0.2^3=0.008 So, 2 numbers to the right of the decimal times power 8. This is how I got the answer but if there's any flaw in the logic please let me know.



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Re: If (10^4 * 3.456789)^8 is written as a single term, how many [#permalink]
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26 Feb 2014, 18:46
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(10^4 * 3.456789)^8 Just resolve the power to move decimal point 4 digit ahead = (34567.89)^8 There are 2 digits after decimal in the above term & the power is 8, so digits after decimal would be 8 x 2 =16 = Answer = C
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If (10^4 * 3.456789)^8 is written as a single term, how many [#permalink]
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13 Mar 2015, 14:21
Well, what I did was to do the multiplication in the brackets and end up with this: 3,4567.89
Now, we want to raise 3,4567.89 to the eighth power. Instead of doing this, I tried with 0.01. I noticed that 0.01^2, is adding two more digits: 0.0001.
So, ading 2 more digits until we reach the eighth power will lead to 14 digits. Plus the 2 we already have (.89) sums up to 16 digits. ANS C



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Re: If (10^4 * 3.456789)^8 is written as a single term, how many [#permalink]
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31 Jan 2016, 17:29
i got to (34567.89)^8
now, we have 2 digits after the decimal point. if we multiply two decimals, the number of digits after the decimal point are the sum of the digits after the decimal point of the decimals. thus, we can easily eliminate A and B. C looks good, rest are way too much.



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