So you have a mashed potato mountain left over from Thanksgiving. Solution? Add more potatoes. And other things. The goal here is to turn the mashed potatoes into a delicious dip, so add whatever else you like (gravy? cranberry sauce?) to give it that extra kick!
Ingredients (Serves 1-2):
1 small bag (~15) sweet potato chips (or other dipping tool of your choice)
Bread stuffing is a traditional holiday season recipe, usually used to fill poultry like turkey on Thanksgiving and Christmas. Typical ingredients include bread/croutons, onions, celery, sage, butter, salt, and pepper, though a quick google search demonstrates the vast range of possibilities. One problem with preparing for bountiful holiday feasts is, of course, leftovers! What do you do with tupperware stuffed with stuffing when you've run out of stuff to stuff it into? Here is one easy appetizer answer, made in 5 minutes with a microwave. (A baking pan in an oven would work too.)
Bread Stuffing Meteors (Serves 1):
Bread stuffing (~1/2 cup)
Parmesan cheese flakes (or your favorite cheese)
Warm marinara sauce (or your favorite sauce)
Olive oil
Roll bread stuffing with hands or spoons into three bite-size to meatball-size balls.
Spread olive oil very lightly over a plate, roll around the bread stuffing until lightly coated, and microwave for four 25sec intervals. Flip or roll each ball to a new side between each interval, since the top will dry faster than the bottom. In the end, the balls should be slightly chewier and should not break apart under light pressure.
Place three dollops of sauce around one half of a small plate, and spread to make 'meteor tails'. Place one ball at the center of each meteor shape, and top each with a flake of cheese.
Microwave until the cheese has melted, and enjoy! :)
Portobello mushrooms are are large, meaty, and are fantastic when marinated and grilled. Once cooked, they work well in everything from burgers and sandwiches to pasta and salad. They are low-calorie, low-fat, and packed with great vitamins and minerals. Now, every superhero needs a sidekick, and portobello and mozzarella are a match made in heaven, along with PB&J and tomatoes & bread. They even sound like they're meant to be together! I've included two recipes below featuring marinated portobello and melted mozzarella--one appetizer and one main course. :)
Marinated Portobello Mushroom
There are endless potential ingredients and methods of marinading. This is just a simple version that I could mix up quickly in my college dining hall. The portobello there is sliced, which speeds up the marination, but keeping it whole may make more sense for burgers.
1 portobello mushroom, sliced or whole
1 tsp each of olive oil and balsamic vinegar
1/2 tsp each of minced basil and garlic
Salt/pepper/cayenne to taste
In a bowl, microwave the portobello in a microwave for 2 30sec intervals, pressing with a fork and draining excess liquid after each. Mix in the rest of the ingredients to coat the mushroom evenly, and microwave for 4 30sec intervals, stirring between each.
Portobello Mozzarella Bruschetta (Serves 1-2)
1 chopped marinated portobello mushroom
2 slices french or sourdough bread
2 slices mozzarella cheese
Olive oil
Pepper and chopped scallion to taste
Brush french or sourdough bread with olive oil and toast until golden. Top with portobello and mozzarella, and melt in a microwave or oven. Garnish with pepper and scallions.
Portobello Burger (Serves 1)
1 whole marinated portobello mushroom
Sliced mozzarella cheese
Sliced tomato
Sliced red onion
Crisp lettuce
Burger bun or bread roll
Other burger condiments to taste
On a plate, microwave mozzarella on top of the portobella to melt. In a burger bun or bread roll, place in order lettuce, tomato, portobello, mozzarella, and red onion. Add whatever condiments you like. Enjoy!
Fried food and fresh fruits--scrumptious separately and delicious together! This dessert is essentially a variant of fruit pie; instead of baked crust filled with cooked fruit and topped with whipped cream, we have a fried tostada 'crust' topped with fresh berries, using the whipped cream to hold everything together. This recipe can be changed dramatically to suit your tastes and available ingredients. For example, yogurt or ice cream can be used instead of whipped cream, and you can use whatever fruits you have on hand! :)
Ingredients:
Tostada chips
Whipped cream
Sliced strawberries
Raspberries
Blueberries
Julienned mint
Honey
Steps:
Warm the tostadas in a microwave or oven, and arrange artistically on a plate.
Dollop generously with whipped cream and top with berries, or alternate between whipped cream layers and fruit layers.
These are the hardest set of flipping riddles I know, but when you get them, you'll light up for the rest of the day! Answers (along with a
hint for the hardest one) can be found below each question, but feel free
to ask for hints or let me know if you have the solution! :D
Easier: I
have 1000 numbered lights turned off, all in a row, each with
associated light switches. I first flip all the lights (turning them all
on). I then flip just the even numbered lights (2, 4, etc). I then flip
just every 3rd light (3, 6, etc). I continue this through 1000
iterations (for the last one, I just flip the 1000th light). At the end,
how many lights are on?
The first light will be flipped just once, on the first run. How many times will the 12th light be flipped? On the runs 1, 2, 3, 4, 6, and 12. Notice that these are the factors of 12, or all the positive integers that divide 12 evenly. The fun thing about factors is that they come in pairs! (1 * 12 = 2 * 6 = 3 * 4 = 12). So it would seem that every number besides 1 is going to be flipped an even number of times, and therefore end up off. But wait! What if a factor is paired with itself? For example, 4's factors are 1, 2, and 4 (2 is paired with 2!). That means that 4 will be flipped 3 times and end up on. If a number has a factor that is paired with itself, that means the number is a square. How many squares are there between 1 and 1000? It turns out that 322, or 210 as many will know it, is 1024, which is just over 1000. So since there are 31 perfect squares under 1000, there will be 31 lights on at the end!
Harder: There are 111 numbered lights turned off. You and I play a game where we take turns flipping lights. At least 1 and at most 10 flips must be made each turn. More lights must be flipped on than flipped off in any turn. The loser is the first one whose turn ends with all lights on. I let you choose who goes first. What should you choose, and what is your winning strategy?
If the loser is the one who turns the last light on, you can force me to lose if you leave me with just 1 light off. You can make sure that happens if there are anywhere between 2 and 11 lights left off on your turn. How can you make sure that there are that many lights? By making sure that I have 12 lights left off on my turn! As you can see, the number of lights you want off on my turn is going to start at 1 and go up by increments of 11, or any number (11n + 1). 111 = 11(10) + 1, so you want to give the first turn to me!
Hardest (first found on Dr. Miller's riddle page here): A
warden tells 50 numbered prisoners that he is giving them a chance to be
released, or executed--a fun game (by his twisted definition of fun)! They will not be able to communicate with each other after one
3-hour-long planning session before they are taken to their rooms and the game begins. There
is a special room containing two light switches numbered 1 and 2, which
can each be either up or down (on or off). They cannot be left in between, they are
not linked in any way. Their initial positions are unknown to the
prisoners. One at a time, a prisoner will be brought into the room. The
prisoner must flip one and only one switch. The prisoner is then
returned to
his cell. There is no fixed pattern to the order or frequency with
which prisoners visit the room, but at any given time, every prisoner is
guaranteed to visit again eventually. At any time, any prisoner may declare that all 50 of them have been
in room 0. If right, the prisoners all go free. If wrong, they are all
executed. If you were Person 1, what plan would you give your cohorts during the meeting?
Think about simpler versions of the riddle. Two switches where you are forced to flip at least one is similar to just one switch that you can choose not to flip. What would you do if there were just 2 people, and you knew the switch's starting position? Then see if you can remove assumptions to return to the original question.
You claim authority and says to the others, "Ok, listen up everyone! Here are the rules. The first two times you see Switch 1 down, flip it up. If Switch 1 is already up, or if you've already flipped it up twice, don't touch it and just flip Switch 2. If I ever see Switch 1 up, I'll flip it down. I'll flip Switch 2 otherwise. Once I see Switch 1 up 49 * 2 = 98 times, I'll know that each of you has flipped it at least once, though most if not all of you will have flipped it twice. Then I'll call the warden and we can get out of here."
These are less straightforward than the card-flipping riddles, but it all comes down to flipping in the end! Answers (along with a hint for the harder one) can be found below each question, but feel free to ask for hints or let me know if you have the solution! :D
Easier:
You are blindfolded with 72 coins on a table in front of you. Exactly 1/4 of them are heads up, though you don't know which they are. Separate the 72 coins into 2 piles, each containing the same number of heads up. You can flip as many coins as you like, though you won't be able to tell if they are heads or tails up.
This is a fun math trick that always blows my mind. There are 18 heads up. Say we take 18 coins from the 72 coin pile, and there are x heads up in the remaining 54. Then there are 18-x heads up in the 18-coin pile, and x tails. But then if we flip all of those 18 coins, there will be x heads up in that pile too, leaving x heads in each pile. :D
Harder (first read on the riddle page of Dr. Miller here):
You sit blindfolded in front of a square with a coin in each corner. You want to get all coins heads up or all tails up. You have no idea what the starting formation is, of course; they could even be all heads up to begin with. You may flip however many you want, then ask if you are done (this constitutes a turn). If you are not done, the square is then spun an undisclosed amount clockwise or counterclockwise. You then get another turn and so the game continues. Is there a strategy that is
guaranteed to work in a finite number of moves, and if so, what is that smallest number of moves you need?
Think about what possible states the four coins could be in. For example, there could be 1 head and 3 tails up. What other states could they be in? Do any of these states have a "guaranteed win" move?
This one is really tricky. The key is to enumerate all the possible states and work through them to see which you can solve fastest.
First, all the states: All 4 of one kind (4:0) 3 of one, and 1 of the other (3:1) 2 each, with same-sided coins next to each other (2:2 adjacent) 2 each, with same-sided coins diagonally opposite each other (2:2 diagonal) Next, let's define some flips: No flip (N) Just one coin (O) 2 adjacent coins (A) 2 diagonally opposite coins (D) The first thing to notice is that if we do a diagonal flip (D) from the 2:2 diagonal state, we win! This is good news. Even better, this has no effect on the 3:1 state or the other 2:2 state! Now, what happens if we try to solve the 2:2 adjacent state by using A? Either we win (yay!) or we get to the 2:2 diagonal state! Then we can do D again and win! Fortunately, neither of these 2 coin flips affect the 3:1 state. If we try to solve this state by using O, either we win or we get to one of the 2:2 states. Then we can do the same sequence of D, A, D to be sure of success! Since we also want to check the case where we're done to begin with, we'll check that first by flipping no coins. In sum, we have N, D, A, D, O, D, A, D. Just 8 flips and we're sure to win! :)
