Math is Fun
Here's another math problem to wrestle with while counting down to the New Year.
Find the smallest positive integer (base 10), composed entirely of the digits 0 and 1 that is divisible by 225.
Friday, December 29, 2006
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7 comments:
All numbers are divisible by 225, silly!
evenly divisible by 225....billy.
Based on the fact that I don't even know how to begin to solve this, I'm going to guess it has something to do with using calculus to determine a minimum (something I can't even remember how to do, without looking it up).
I think I could write a Java program to determine the answer, through brute force, faster than I could ever figure this out mathematically. Sigh.
Heh - look at the last problem - that will give you a hint. First take the number 225 and prime factor it. This is a good idea with any number-theory problems - kinda like putting fractions into lowest-common-denominators. We have 5*5*3*3 or 5^2*3^2.
From here, I'm stuck. All I know for sure is that the last number has to be a 2 because 5*2 ends in 0. 225*52 is a reasonable guess (11700), but that is a simple multiplication and not a proof, which is what is needed.
But 11700 has that nasty "7" in it, Chris. Did I misunderstand what the question was asking?
Chris is on the right track...
Notice that 225 = 9*25.
Think about what properties multiples of both those number have... (see previous post about divisibility rules too...)
Since 225 = 9x25, we know the number in question has to end in 00.
Also, you can quickly determine if any positive integer is divisible by 9 by adding up the (base 10) digits in that number.
Using those two facts, the smallest positive base 10 integer that (a) consists of only the digits 0 and 1, and (b) is equal to 225 * k, where k is also an integer, must be:
11,111,111,100
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