How many zeros will the decimal equivalent of 1/2^11*5^7 + 1/2^7*5^11

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.