In my article “One Hundred Cheerios in Binary”, I made a comment about the decimal number 100, and its binary equivalent, 1100100:
“And will they wonder if the two sub strings of ‘100’ in the binary number have any significance?”
What I meant is if a novice might wonder if a decimal string made up of 1s and 0s must appear in its binary equivalent. Of course that’s not true in general, but it is true for nonnegative powers of ten — the trailing digits of the binary number will match the power of ten!
You can see the pattern in these examples:
| Power of Ten (in Decimal) | Power of Ten (in Binary) |
|---|---|
| 1 | 1 |
| 10 | 1010 |
| 100 | 1100100 |
| 1000 | 1111101000 |
| 10000 | 10011100010000 |
| 100000 | 11000011010100000 |
The pattern is easy to explain. A nonnegative power of ten is a multiple of a power of five and a power of two: 10n = 5n * 2n. A power of five always ends in ‘5’, so it’s odd — its binary representation always end in ‘1’. When you multiply by a power of two, you shift the power of five left by n bits, which adds n trailing 0s. So the binary representation ends with a ‘1’ followed by n 0s, which looks like the power of ten!
Cool, huh?
I know I’m 3 years late, but it is cool!
So, this should work for all bases which have exactly one “2” in their prime factorisation. Indeed, I don’t see why not, but I ought to do a demo, and the simplest choice is base 6 – senary, as it’s called. It does actually live up to its punnable name in terms of fraction representation, but that’s beyond the point.
(senary = binary)
1 = 1
10 = 110
100 = 100100
1000 = 11011000
As expected, all good! Also, the power of 2 in the prime factorisation can be used to tell you how many trailing zeros to expect. In my familiar base 2^2 * 3, for example, you get:
(dozenal = binary (comment))
10 ^ 0 = 1 (2 * 0 trailing zeros)
10 ^ 1 = 1100 (2 * 1 trailing zeros)
10 ^ 2 = 10010000 (2 * 2 trailing zeros)
10 ^ n has 2n trailing zeros.
This of course holds for odd bases (where the power of 2 is 0, and all powers are odd). But it only looks interesting for the “exactly one 2” bases.
Power of Five (in Pentimal) Power of Ten (in Binary)
1 1
10 101
100 11001
1000 1111101
10000 1001110001
100000 110000110101
well, lets analise this in powers of five to get a deeper understanding of the real question. first: they all end in 1 because its powers of 5.
5^0=1=1
5^1=5, 5-1=4, 4 is 100 + 001 is 101
5^2=25, 25-1=24, still ends in 4. that’s a 5 law. so no mater what we add 101
to the end (with exceptions) so, 20=16+4, the first exception. 16+4+4+1, 11001 is the result.
5^3=125, 125 is 3 less then 128, therefore 1111101, thus enacting the following:
it can’t be equated that easily. so lets think in binary.
101^10=11001, 11001=101+101+101+101+101=10100+101 soooo….. every power of 5 added you multiply 101 by 101 literally. there is the simplification at its root, your doing the 11 rule.
5^1=101
5^2=10100+101
5^3=(10100+101)00+(10100+101) or (1010+101)000+101
5^4=((10100+101)00+(10100+101))00+((10100+101)00+(10100+101)) or ((10100+1010+101)00+1010+101)00+101, this pattern grows further through a pattern.
((next tier)00+1010+101) – that in itself is a tier, if there is no next tier, then substitute 101. then, at the end add + 101 to the equation. powers of 5 indefinitely.
sorry, i made a mistake or 2,
5^1=101
5^2=(prev)00+101
5^3=(prev)00+101
get it? times 4 +5? that was my original thought.
now to improve:
5^1=101
5^2=(prev)00+(prev)=10100+101
5^3=(prev)00+(prev)=(10100+101)00+(10100+101) – the real solution. still, sorry about the above.
(i did it!!!)
then for decimal, just do the same and add the number of 0’s that is the power of 10.