forum.virtualchallengemeets.com

I do the olympiad stuff posted out of boredom. :P It certainly doesn't help me...I'm a senior, so it's not really doing me any good. Oh well...hahaha.

there is always the putnam...

Something tells me I won't make the Putnam team at Caltech. :P

anyone can take it....

Number 44: In the decimal number 65xy2y, the letters x and y represent digits. If the number is divisible by 72 then x is:
A) 7
B) 6
C) 4
D) 3
E) 1


If it's divisible by [unparseable or potentially dangerous latex formula], it's divisible by [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula].

Numbers divisible by 8 have the last three digits divisible by [unparseable or potentially dangerous latex formula]... so [unparseable or potentially dangerous latex formula] must be [unparseable or potentially dangerous latex formula].

now for divisibility by 9:

[unparseable or potentially dangerous latex formula] a multiple of nine

[unparseable or potentially dangerous latex formula].... x must be 6 so that they sum to 27



47.. thought it said positive integers

I doubt Dr. White will put any more questions about dates on there since he's been informed that the TI-89 has a date function.

But in general, any particular date one year, will fall one day later the next year (with the exception of a leap year, in which case it's two days later.) Or equivalently, a particular day will fall the date earlier the next year. [unparseable or potentially dangerous latex formula]

So if April 23, 2005 is a Saturday, then 5 years later it will be 6 days earlier (1 extra one because of the leap year). So April 17th. However, this is now the 3rd Saturday, so you need to add 7...so April 24th.

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If a number is divisible by 72, it's divisible by 9 and 8. Since you only look at the last 3 digits for divisibility by 8, you can quickly think about all 9 cases and see which one is divisible by 8. That tells that digit y is 4. So to find x, we check for divisibility by 9. The sum of the digits (mod 9) excluding x is 3...so x must be 6 for the number to be divisible by 9.

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[unparseable or potentially dangerous latex formula].
For x to be an integer, y must be a multiple of 4. Since x0...or x<0, y>0. So the only solutions have the form x,y<0. To see the limit for solutions, check y=x, or [unparseable or potentially dangerous latex formula]. So all x,y must be less than -40. However, for any y that is a multiple of 4, but less than -40...x will be an integer smaller than y, since the limit for equality is at -40. So there are an infinite number of solutions.