puzzle time

[mostsmooth]

im no math whiz, so i dont know if this is hard or easy (actually i think its probably on the easy side)
also, i probably couldnt solve this myself very quickly, but i do know the answer(unlike some other puzzle posters /images/graemlins/grin.gif):
What is the smallest integer greater than 0 that can be written entirely with zeros and ones and is evenly divisible by 225?
im just curious to see how people go about solving this

[mostsmooth]

btw, if this is too easy, ill find a harder one. im bored

[JFB37]

1100

[kyro]

[ QUOTE ]
1100

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at first i was like "is it really that easy and i couldn't figure it out?" and then i realized your calculator was broken.

[asofel]

[ QUOTE ]
1100

[/ QUOTE ]

[ QUOTE ]
evenly divisible by 225?

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fwiw i'd just write up a few lines of code and let it figure it out because i'm lazy and busy...

[ethan]

In white: <font color="white">11,111,111,100</font>

I'm pretty sure this is correct.

[kyro]

<font color="white"> good call, i think i figured out the "method" after i saw the answer. 225 is 1/4 of 900, so what number consisting of all 1s is divisible by 9? tack on two zeros and you're set. I THINK this is how you'd go about it but i'm not positive.</font>

[nickey009]

The way I figured it out was to put "Number divisble by 225" into google.

[ethan]

[ QUOTE ]
good call, i think i figured out the "method" after i saw the answer. 225 is 1/4 of 900, so what number consisting of all 1s is divisible by 9? tack on two zeros and you're set. I THINK this is how you'd go about it but i'm not positive.

[/ QUOTE ]

Pretty much, although I went at it from the opposite direction. The number has to be divisible by 5, meaning it has to end in 0 or 5. So, the units digit is 0, meaning the number will also be divisible by 450. The rest of the digits will be the smallest {0, 1} number divisible by 45. Again, it has to be divisible by 5 and thus ends in 0, which is the tens digit. From there we need the smallest {0, 1} number divisible by 9, which is going to be 111,111,111.

[Rick Diesel]

I thought about it for about 30 seconds and realized that the number must contain at least nine 1's and end in 00, so I will go with 11,111,111,100. I got to this by figuring out that the number must end in 00, so it must also be divisible by 900 (4*225). Any number that is divisible by 900 must have characters that add up to 9 or some multiple of 9. Since it can only contain 1s and 0s, there must be at least 9 1s and end in 00.

Rick Diesel

[Jack of Arcades]

00110010 00110010 00110101

[Nomad84]

[ QUOTE ]
I thought about it for about 30 seconds and realized that the number must contain at least nine 1's and end in 00, so I will go with 11,111,111,100. I got to this by figuring out that the number must end in 00, so it must also be divisible by 900 (4*225). Any number that is divisible by 900 must have characters that add up to 9 or some multiple of 9. Since it can only contain 1s and 0s, there must be at least 9 1s and end in 00.

Rick Diesel

[/ QUOTE ]

Same logic. I knew it was divisible by 9 and by 25. Ending in 00 takes care of the divisible by 25 requirement. Also knew that to be divisible by 9, the digits must add up to 9. Since they are all 1s and 0s, I also quickly arived at the answer 11,111,111,100.

[GuyOnTilt]

<font color="white">It has to be divisible by 9 and end in 00, so 11,111,111,100 is probably the answer.</font>

Edit: Replace "probably" with "for sure is." Since 900 is divisible by 225, all other numbers ending in 00 divisible by 9 have to be as well, meaning that number is for sure the lowest possible positive solution.

GoT

[Duke]

I'd just write 255 in binary, since you can write any integer completely with 1's and 0's that way.

~D