A Pattern in Powers of Ten and Their Binary Equivalents

http://www.exploringbinary.com/a-pattern-in-powers-of-ten-and-their-binary-equivalents/


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:

Some Powers of Ten and Their Binary Equivalents
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?

Dingbat

One Response to “A Pattern in Powers of Ten and Their Binary Equivalents”

  1. James Says:

    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.

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