Thursday, April 06, 2006

Winx, Bionicles, McSocks, and Math

Just an update with 4 random things.
a) I know of the Winx club, and the other day I watched a whole episode. They're quite sexy =)
b) Lego is expensive, but sometimes I buy the bionicles. It's like a bundle of joy, not too expensive, and it's fun. It's one of those rewarding experiences, when you follow the nicely illustrated instructions. It makes you feel good when you successfully complete a piece. I like bionicles =)
c) We went to McDonald's the other day, and there was a sign that says that you can buy socks at McDonalds.

and lastly, as part of this math party thing, I think I'll put up bi-daily puzzles from the math contests. So, here is a question which I haven't solved yet:

Fermat (Grade 11) 2004, #22, Calculators permitted:
If k is the smallest positive integer such that (2^k)(5^300) has 303 digits when expanded, then the sum of the digits of the expanded number is
a) 11 b) 10 c) 8 d) 7 e) 5

good luck! and hope you have fun. I'll try to solve it too.

4 Comments:

Blogger Unknown said...

BAKA! Man, I went through all these steps, found out that 5^300 has 201 digits, and then found out some pattern with the 2^ whatever... and then I took a step back, and then I got the answer really easily!!!

AHHH!!!!!

3:42 PM  
Blogger Unknown said...

..doh, almost made a mistake. heh. okay, I have the real answer now!!

3:44 PM  
Blogger Unknown said...

... doh again... okay, now I'm sure I have the right answer!

And there. Now we have 3 comments here, everyone's all curious and stuff. BOOYAH!! I shall post a new math problem on Saturday or so, give you time to work with this one.

3:46 PM  
Blogger Ambrose said...

Um, the format for comments can be a bit weird, since I'm not sure if you want people to reply with: "I have the answer."

I suppose that for now I'll do a "spoiler" thing and hope that others don't scroll too quickly or something.

Hm, perhaps people want hints or something, which I'll give one:
2*5 = 10.


==== ANSWER SPOILER =====














The answer is A, 11.


Since there are no rules I'll just post my solution as well, muhahaha



==== SOLUTION SPOILER ======

















The key to answering this is to realize that 2*5 = 10, and that 10 = + 1 digit in a number. Thus, for every 2 there is in 2^k, you can "match it up" with (up to 300) 5's to "add digits". Thus, for k < 300, the number of digits of the expanded number is k + number of digits of 5^(300-k). It is evident that 303 digits cannot be accomplished with k < 300 (you have 300 digits if k = 300, and as k decreases, so does the number of digits, since you do not always "gain a digit" with exponents of 5's, i.e. 5^4 = 625 only 3 digits). Thus, k > 300, and then the question becomes pretty easy. If k > 300, then 300 2's are matched with 300 5's to add 300 digits to the remaining number. This can be done if 2^(k-300) is a 3 digit number. Since the question specifies that k is the smallest integer that does this, we now only need to find the first exponent of 2 that has 3 digits, which would be 2^7= 128. Since the 300 10's (2*5) add zeroes, they add nothing to the sum of the expanded number's digits. Thus the sum of the digits is simply 1+2+8=11, yeilds answer A.

11:18 PM  

Post a Comment

<< Home