Solutions
369 Smorgasbord
364 Paradox
339 Brad Pitt
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337
336 Dick Cheney
335 Divine intervention
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333 Jason Segel
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331
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329
328 Chamomile
327 I'll be damned
326 Raise the stakes
325 Emoji
324
323 Project runway
322 Pisces
321 Catastrophe
320 Season finale
319 I'll pencil you in
318 Illegal Aliens
317 Euthanasia
316 Safety in numbers
315 Scar face
314 Movers and Shakers
313 Wisconsin
312 Back to the drawing board
311 Red White and Blue
310 No chance in hell
309 Contact Solution

308 Mayonnaise

307 Tie Breaker

306 Alfalfa

305 Dodge Charger

304 George Washington

303 Overnight Delivery

302 World Series

301 HBO

300 Station Wagon

299 You'll be sorry

298 Settle a score

297 Meteor shower

296 Pigeon toed

295 Freaky Friday

294 Automatic Weapons

293 Third time's the charm

292 MTV

291 West Palm Beach

290 Long time no see

289 Amazeballs

288 Frankincense

287 USB port

286 Cover charge

285 Go for broke

284 Hedge Bets
283 Miley Cyrus
282 Hugh Grant
281 Ooze and Oz
280 Ponytail

279 Quiznos
278 Bartender
277 Yogi Bear
276 Little rock Arkansas
275 Jheri curl
274 Running behind schedule
273 Boulder Colorado
272 Polariod
271 Search Party
270 Page Turner
269 Run for office
268 Plot Twist
267 Bloomingdales
266 Twenty thousand leagues under the sea
265 Monologue
264 Range rover
263 Nestle quick
262 Pumpernickel
261 Baskin robbins
260 Spreadsheet
259 Austin Powers
258 Masturbate
257 Even Stevens
256 Tit for tat
255 Goalkeeper
254 Mixed Martial Arts
253 Butt dial
252 Dim Sum
251 Boarding school
250 German Engineering
249 Cher Crop
248 Steer clear
247 Vanity Fair
246 Pickup artist
245 Dane Cook
244 Pinky swear
243 Mix and match.
242 Ceiling fan
241 Sausage links
240 Cross dresser
239 Barnes and Nobles
238 Made in China
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236 Hostle takeover
235 Instagram
234 Xbox
233 Porcupine
232 Bruno Mars
231 Pandora's box
230 50 cent
229 Dramatic pause
228 Make up for lost time
227 Goldie Hawn
226 General Mills
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Alternating T-Shirt Colors

This is a puzzle from Microsoft Research Page: "Ten friends walk into a room where each one of them receives a hat.  On each hat is written a real number; no two hats have the same number.  Each person can see the numbers written on his friends' hats, but cannot see his own.  The friends then go into individual rooms where they are each given the choice between a white T-shirt and a black T-shirt.  Wearing the respective T-shirts they selected, the friends gather again and are lined up in the order of their hat numbers.  The desired property is that the T-shirts colors now alternate.The friends are allowed to decide on a strategy before walking into the room with the hats, but they are not otherwise allowed to communicate with each other (except that they can see each other's hat numbers).  Design a strategy that lets the friends always end up with alternating T-shirt colors."

My solution is based on induction. 

Assume we have a numbered sequence 

1 2 3 4 5 6 7 8 ...

We may assign alternating ranks to to stand for the alternating colors for each number. 

1 0 1 0 1 0 1 0 ...

Now if we permutate 2 arbitrary numbers once with exactly the same rank, for example

1 2 7 4 5 6 3 8...

We can see any number beside the chosen 2 numbers will observe one permutation. For the 2 numbers themselves, because the chosen 2 numbers have the same rank, there were always odd number of digits between them (4, 5, 6) . Once the permutation happened, the numbers that are switched will be observing odd number of permutations. In conclusion,  the parity of the number of permutation will be preserved by an observation made by inside numbers and outside numbers alike.

Next if we permuate 2 arbitrary numbers of different ranks, for example

1 2 6 4 5 3 7 8

For an outside observer (those numbers which are not permutated), an odd number of permutation is observed, yet an even number of permutation is observed by the numbers which are permutated. We can extend this rule to multiple permutations of numbers of different ranks. In conclusion, the parity of the number of permutation will be preserved by an outsider's observation, and will be reverted by an insider's observation if its final position is in a different rank. 

Any other case can be reduced to the above 2 cases, for example

1 2 4 6 5 3 7 8 is a combination of 1 2 3  6 5 4 7 8 and then 1 2 4 6 5 3 7 8

Base on the above properties of permutations, we can conclude that the solution to the puzzle is: Each number observes the parity of the permutations. If it's odd then it switches its rank and if it's even then it retains its rank.