DOOR #100 (WINTER 2013)
Puzzle:
You have 100 doors in a row that are all initially closed. You make 100 passes by the doors starting with the first door every time. The first time through, you visit every door and toggle the door (if the door is closed, you open it; if it’s open, you close it). The second time you only visit every second door (door #2, #4, #6, etc.). The third time, every third door (door #3, #6, #9, etc. ), and so on, until you only visit the 100th door. What state are the doors in after the last pass? Which are open and which are closed?
Solution:
After the first pass every door is open. On the second pass you only visit the even doors (2,4,6,8, etc.) so now the even doors are closed and the odd ones are opened. The third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc. You can figure out that for any given door, e.g., door #42, you will visit it for every divisor it has. So 42 has 1 and 42, 2 and 21, 3 and 14, 6 and 7, so on pass 1 you will open the door, pass 2 you will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. For every pair of divisors the door will just end up back in its initial state. You might think that every door will end up closed? Well what about door #9?
9 has the divisors 1 and 9, 3 and 3. But 3 is repeated because 9 is a perfect square, so you will only visit door #9 on pass 1, 3, and 9, etc. leaving it open at the end. Only perfect square doors will be open at the end.
(SOURCE: techInterview)
FINAL SCORE (WINTER 2013)
Puzzle:
Jennifer took a test that had 20 questions. The total grade was computed by awarding 10 points for each correct answer and deducting five points for each incorrect answer. Jennifer answered all 20 questions and received a score of 125. How many wrong answers did she have?
Solution:
She had five wrong answers.
If Jennifer had answered all 20 questions correctly, she would have scored 200. Since she only scored 125, she must have lost 75 points. Since each incorrect answer results in a total loss of 15 points (10 for not getting it correct and 5 for answering incorrectly) she must have missed 5 questions. 5 x 15 = 75, 200 - 75 = 125.
(SOURCE: malini-math)
Calendar Confusion (Fall 2013)
Puzzle:
Three days ago, yesterday was the day before Sunday. What day will it be tomorrow?
Solution:
Thursday; three days ago was Sunday, so today is Wednesday and tomorrow will be Thursday.
(SOURCE: mathisfun)
Happy Birthday (Fall 2013)
Puzzle:
When asked about his birthday, a man said: “The day before yesterday I was only 25 and next year I will turn 28.” This is true only one day in a year: When was he born?
Solution:
He was born on December, 31st and spoke about it on January, 1st.
(SOURCE: malini-math)
1,000 LOCKERS (Summer 2013)
Puzzle:
A high school has a strange principal. On the first day, he has his students perform an odd opening day ceremony. There are 1,000 lockers and 1,000 students in the school. The principal asks the first student to go to every locker and open it. Then he has a second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?
Solution:
31. The only lockers that remain open are perfect squares (1, 4, 9, 16, etc.) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be "changed" an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed. So the number of open lockers is the number of perfect squares less than or equal to 1,000. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. Thirty two squared is greater than 1,000, and therefore out of range. So the answer is 31..
(SOURCE: Puzzle #2 at http://www.rinkworks.com/brainfood/p/math1.shtml)SOUND SERIES PATTERN (Summer 2013)
Puzzle:
Can you figure out the logic used to order the following words: gun, shoe, spree, door, hive, kicks, heaven, gate, line, den?
Solution:
Each word rhymes with its numeric position in the list. (e.g. 'gun' rhymes with 'one', etc.)
(SOURCE: http://dailybrainteaser.blogspot.it/2011/07/sound-series-pattern.html)the right number (spring 2013)
Puzzle:
A man has two bankcards, each with a four digit number. The first number is four times the second. The first number is the reverse of the second. What is the first number?
Solution:
You could use trial and error. The smaller number must start with a 1 or a 2, otherwise the bigger number would have five digits. Therefore the bigger number must end with 1 or 2. Actually, it has to be 2, because the bigger number must be a multiple of four. Consequently the numbers are 2**8 and 8**2. The digit after the 2 has to be small, because four times the number is less than 10.
(SOURCE: www2.eng.cam.ac.uk/~tpl/maths/puzzles.html)loose change (spring 2013)
Puzzle:
There are 12 coins. One of them is false; it has a different weight than the others. It is not known, if the false coin is heavier or lighter than the other coins. How do you find the false coin if you only have three chances to weigh the coins on a simple scale?
Solution:
Divide the coins into three groups of four, say A, B, and C. Weigh A and B. Case I: A and B are equal. Now weigh three coins of A and C, lets call them A1 and C1: (a) If they are equal, then the remaining coin in C (not in C1) is the odd one; (b) If they are different, say A1 weighs more than C1, weigh two coins in C1. If these weigh the same, the remaining coin in C1 is the odd man out. If they are different then the lighter coin in C1 is the odd man out.
(SOURCE: lokvani.com)