How Many Non-Zero Digits?
2012-03-01
How many nonzero digits does `1/(2^13 * 5^19)` have when written as a decimal?
[Spoiler Below]
`1/(2^13*5^19) = 1/(2^13*5^13*5^6) = 1/((2*5)^13*5^6) = 1/(10^13*5^6) = 10^-13 * 1/(5^6)`
note I can multiple by 1 whenever I want without changing the value, and `1 = (2^6/2^6)`, continuing:
` = 10^-13 * (2^6/2^6)*1/5^6`
` = 10^-13 * 2^6/(2^6*5^6)`
` = 10^-13 * 2^6/(2*5)^6`
` = 10^-13*2^6/10^6 = 10^-13 * 10^-6 * 2^6 = 2^6 * 10^-19`
` = (2^3)^2 * 10^-19 = 8^2 *10^-19 = 64 * 10^-19`
the `10^-19` is only going to change the place of the decimal, not the number of non-zero digits, so the answer is simply 2, the number of digits in `64`.
[Spoiler Below]
`1/(2^13*5^19) = 1/(2^13*5^13*5^6) = 1/((2*5)^13*5^6) = 1/(10^13*5^6) = 10^-13 * 1/(5^6)`
note I can multiple by 1 whenever I want without changing the value, and `1 = (2^6/2^6)`, continuing:
` = 10^-13 * (2^6/2^6)*1/5^6`
` = 10^-13 * 2^6/(2^6*5^6)`
` = 10^-13 * 2^6/(2*5)^6`
` = 10^-13*2^6/10^6 = 10^-13 * 10^-6 * 2^6 = 2^6 * 10^-19`
` = (2^3)^2 * 10^-19 = 8^2 *10^-19 = 64 * 10^-19`
the `10^-19` is only going to change the place of the decimal, not the number of non-zero digits, so the answer is simply 2, the number of digits in `64`.