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ashiima
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we have in denominator 2^11 * 5^17
as we know 2*5 =10 which is a number very easy to be handle
here we have 11 powers of 2 and 17 powers of 5
multiply both numerator and denominator by 2^6 so that we now have exp in denominator as
2^17 * 5^17 = 10^17

in numerator we have only 2^6 = 64

so we have only 2 non zero digits in this decimal
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ashiima
If 1/(2^11 * 5^17) is expressed as a terminating decimal, how many non-zero digits will the decimal have?

A) 1
B) 2
C) 4
D) 6
E) 11

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ashiima
If 1/(2^11 * 5^17) is expressed as a terminating decimal, how many non-zero digits will the decimal have?

A) 1
B) 2
C) 4
D) 6
E) 11

1/(2^11 * 5^17)
= 1/(10^11*5^6)
= 10^6/(10^17*5^6)
= 2^6/10^17
= 64/10^17

So decimal will have 2 non-zero digits 6 & 4

Answer B
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If \(\frac{1}{(2^{11})(5^{17})}\) is expressed as a terminating decimal, how many nonzero digits will the decimal have?

A One
B Two
C Four
D Six
E Eleven


\(\frac{1}{(2^{11})(5^{17})}\)

Since number of zeros is determined by the combination of 2's and 5's, let us make both of them equal in the denominator by multiplying both numerator and denominator by \(2^6\)

\(\frac{1 * 2^6}{(2^{17})(5^{17})} = \frac{2^6}{10^{17}} = \frac{64}{10^{17}}\)

So, we can see that we will have 15 zeros followed by 2 non zero units (64)

Answer - B
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