Hi Usamashake. what a great start:
fork=4
spoon=4
knife=8
plate=12
135 tiles
Also a big high too And rian: :wink:
"Well, you got the number right."
Mark clears the fog of confusion with:
12 miles
Special mention: Mark's answer.
PRODIGITIOUS
Update:
The answer is 13.
You are in line for an ?'Emmy'
Stock answer:
Note that if any number n in our sequence ends in a 3, this implies 3 divides n. However, then n+10 also ends in a 3, yet is clearly not divisible by 3 (since 3 does not divide 10). The same trick works for numbers ending in 6 and 9, thus you can only have one of each in any consecutive sequence with the desired property. It follows that any such sequence has length at most 13, and indeed that it must go *0, *1, ..., *9, %0, %1, %2. Since I told you the answer was not 12, it has to be 13!
Consider sequences of the form (1)^n 000, (1)^n 001, ..., (1)^n 012, where for example (1)^3 000 = 111000. The freebies are the numbers ending in 00, 01, 02, 04, 05, 08, 10, 11, and 12. Let n be a multiple of 9, say 9*m; this gives us 03, 06, and 09 by the old trick of a number is divisible by 3 (9) if and only if the sum of its digits is divisible by 3 (9). The last number left is 07; we need that (10^{9m}-1)/9 is divisible by 7. This happens only if 10^{9m} = 1 (mod 7); by Fermat's little theorem this is ensured if 9*m is a multiple of 6 (and indeed, 10 is a primitive root modulo 7, so this is required). Let m=2k, then, so n=18k.
So we're done, the answer is 13, and one such sequence runs from 111,111,111,111,111,111,000 to 111,111,111,111,111,111,012.
The three of us made some bets:
First, Waldo won from Molly as much as Waldo had originally.
Next, Molly won from Spike as much as Molly then had left.
Finally, Spike won from Waldo as much as Spike then had left.
We ended up having equal amounts of money.
I started with 50 cents.
Who am I
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