Karnataka Regional mathematical Olympiad-2006?

Question

Question:
Let A={n E N|5divides n}
B={n E N|both n and n/5 contain only odd digits}
C={n E N|both n and n/5 contain exactly 2006 dighits} where N denotes the set of all natural numbers.Find the number of elements in the intersection of A,B, and C

I got a weird answer: 3^2004. Is that correct?

Answer 1

I don't think 3^2004 is right, but I think you can figure it out. I doodled with this for a few minutes, and I think I can get you started.

A is the set A = {5, 10, 15, ...}

B is the set B = {5, 15, 35, 55, 75, 95, 115, 135, 155, 175, 195, 315, 335, ..., 975, 995, 1115, 1135, ..., 1975, 1995, 3115, ..., 7975, 7995, 9115, ..., 9975, 9995, 11115, ...}

The B set has a pattern that you can dig out.

Instead of jumping to a 2006-digit number, suppose we say C only has 6 digits, e.g., 100000 or 999999. (You'll try to generalize later.) Since n/5 must also be a 6-digit number, then n in C is between 500000 and 999999.

And, since you're looking for the intersection, why don't you define a new set D, containing only elements common to A, B, and C. (You can ignore A, since B is contained in A.) The set D will look something like this:

D = {555555, 555575, 555595, 555755, 555775,555795, 555955, 555975, 555995, ...}

Notice that dividing the D elements by 5, you get

{111111, 111115, 111119, 111151, 111155, 111159, 111191, 111195, 111199 ...}

You'll have to do a lot more work on this, including expanding it to 2006 digits, but there are some patterns in here, and with due diligence, I'm sure you can get it.

(And after seeing what's coming out of this, your 3^24 answer looks a bit better than it did at the beginning. Good luck.)

Answer 2

ok, honestly this is not easy but if you go through it in steps it isn't as bad as it sounds, well let us try to figure this out... i think i have it right:

the first one reads "Set A contains n, which is a element of the natural numbers, such that n is divisible by 5"

So the number needs to end in 0 or 5

next

"set B contains n, " " " ", such that n and n/5 contain only odd digits"

So n needs to end in 5 to make n/5 odd. Right?

"set C contains n, " " " " , such that n and n/5 contain exactly 2006 digits"

ouch! well i know that the numbers that are divisible by 5 have a repeating pattern after 100 right? So let's see if we can use this to our advantage.

exactly 2006 digits. well 10^3 has four digits and 10^6 has seven digits and 10^9 has 10 digits. So it looks like the pattern is that 10^n has n+1 digits right? So i want 10^2005. Now i still need to make more sense of this problem. Let me test something

for 10^5 i get 100,000 i know there are 10 such number in each 100 (namely: 5, 15, 25, 35, 45, 55, 65, 75, 85, 95), and there are 10^3 hundreds in 10^5 so we get 10*1000 (rules of counting... each of 6 moms has 2 children... how many children... 6*2) of these numbers, or 10^4

for 10^6 i know i get 10^4 hundreds and i will get 10^5 of these numbers

now we have our answer 10^2005 = is 10^2003 hundreds, and i will get 10^2004 of these numbers!

crap, i rethought the answer and i counted all the numbers that satisfy that are up to the digits you wanted. So to modify you need to find how many are in the intersection with one more than that digit and minus the number that i counted (which is up to the 2006 digits). My formula still works. So in 10 there is 10^0 satisfying A and B... namely 5. And up to 10^2 there are 10^1 satisfying A and B. So just use the formula for 2007 digits and minus up to 2006 and you will get the answer.

OK, i just needed to correct that. I am off to bed