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
A) 1
B) 2
C) 4
D) 6
E) 11
1/(2^11 * 5^17)
= 1/(2^11 * 5^11 * 5^6)----Since (a^(m+n) = (a^m) * (a^n))
= 1/(10^11 * 5^6)----Since ((a^m) * (b^m) = (ab)^m)
= 2^6/(10^11 * 5^6 * 2^6)----Multiply the numerator and denominator by 2^6
= 64/(10^11 * 10^6)
= 64/(10^17)
= 0.0000..64 (17 0's after the decimal)
Non-zero numbers are 6 and 4. Hence 2, Option B
In simple terms, multiply the numerator and denominator by a number which can convert the denominator into a multiple of 10(in this case)
i.e. 1/(2^11 * 5^17) = 2^6/10^17 = 64/10^17 = 0.000....64, Option B
