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09 May 2018, 00:36
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (01:50) correct
42%
(01:46)
wrong
based on 196
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How many zeros will the decimal equivalent of \(\frac{1}{2^{11} * 5^7}+\frac{1}{2^7 *5^{11}}\) have after the decimal point prior to the first non-zero digit?
(A) 6
(B) 7
(C) 8
(D) 11
(E) 18
(A) 6
(B) 7
(C) 8
(D) 11
(E) 18
09 May 2018, 10:36
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Bunuel
\(\frac{1}{2^{11} * 5^7}+\frac{1}{2^7 *5^{11}}=\frac{1}{2^7 * 5^7}(\frac{1}{2^4}+\frac{1}{5^4})\)
\(=>\frac{1}{10^7}[\frac{(5^4+2^4)}{5^4*2^4}]\)
\(=>\frac{625+16}{10^{11}}=\frac{641}{10^{11}}\)
\(=>0.641*10^3*10^{-11}=0.641*10^{-8}\)
therefore there will be \(8\) zeros before the first non zero number.
Option C
09 May 2018, 14:31
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Bunuel
Number of zeros?
One way to determine the number of zeros before the first nonzero digit:
get a fraction with an integer over a power of 10
• The number of digits in the integer = the number of decimal places the integer uses
• The exponent on 10 = total # of decimal places
If we have, for example: \(\frac{23}{10^{6}}\)
• \(10^6\) tells us that there are six decimal places total[/size]
• two of which are taken by 23: [size=90]\(.000023\)
Here, there are 4 zeros before the first nonzero digit.
For the prompt's terms, I think calculating them separately is easy. The key:
equalize the number of powers ("copies") of 2 and/or 5 in the denominator
FIRST TERM
\(\frac{1}{2^{11}* 5^7}\)
There are eleven powers of 2 but only seven powers of 5
We need four more 5s, such that denominator becomes
\(2^{11}*5^{7}*5^{4}=\)
\(2^{11}*5^{7+4}=2^{11}*5^{11}=10^{11}\)
Multiply numerator and denominator by \(5^4\)
\((\frac{1}{2^{11}* 5^7}* \frac{5^4}{5^4})=\)
\(\frac{5^4}{2^{11}*5^{11}}=\frac{625}{10^{11}}\)
There are 11 decimal places
625 takes 3 places
(11-3) = 8 zeros* before the first nonzero digit
\(.00000000625\)
SECOND TERM
The second fraction needs 4 more powers (copies) of 2 to yield a denominator with a power of 10
\((\frac{1}{2^7 *5^{11}}*\frac{2^4}{2^4})=\)
\(\frac{2^4}{2^{11}*5^{11}}=\frac{16}{10^{11}}\)
\(\frac{16}{10^{11}}=.00000000016\)
The sum of 625 and 16 = 641
"641" and "625" use the same number of decimal places
The number of zeros before the first nonzero digit is still 8
If in doubt, sum the decimals
\(.00000000625\)
\(.00000000016\)
-----------------------
\(.00000000641\)
There are 8 zeros before the first nonzero digit
Answer C
*You can also just count the number of zeros. Write a decimal point, then make 11 slots to the right. Write 625 in the last three slots. Fill empty slots with zeros, and count them.
