Hi everyone...
This is to inform all the winners of SQuieeez series, held by IIT Roorkee, during Cognizance'08
Some of you might have received the prizes by now, or will be getting it in the coming week.. Have fun!! :-)
Sunday, April 27, 2008
Sunday, March 23, 2008
solutions to sQUIeeeZ part-4
Ans1-
28
Ans 2-
From conditions Prince has at least 3 balls and a number of balls from this sequence:
3,6,9,12,15,18,21,24.....
Also Vipul has at least 4 balls and a number of balls from this sequence:
4,8,12,16,20,24....
Moreover Keshav has at least 5 balls and a number of balls from this sequence:
5,7,9,11,13,15,17.......
So the total number of balls is at least 12 and at most 24. Also if the number of balls is even, Prince must have an odd number of balls; if the total number of balls is odd, Prince must have an even number of balls. Hence by proper logic it is revealed that, the total number of balls can not be 13 .Also total number can not be 12,14,15,16,17 because then the total number of balls each had would be known. Also the total number cannot be 18,20,21,22,23,24 because then no number of balls could be known for anybody. So total is 19.
When total is 19. Prince must have an even number of balls and from the sequences, this number must not be greater than 10. So Prince must have 6 balls. Then Vipul and Keshav together must have 13 balls, then Vipul must have either 4 or 8 balls .but not definite one.
Hence speaker is Prince.
28
Ans 2-
From conditions Prince has at least 3 balls and a number of balls from this sequence:
3,6,9,12,15,18,21,24.....
Also Vipul has at least 4 balls and a number of balls from this sequence:
4,8,12,16,20,24....
Moreover Keshav has at least 5 balls and a number of balls from this sequence:
5,7,9,11,13,15,17.......
So the total number of balls is at least 12 and at most 24. Also if the number of balls is even, Prince must have an odd number of balls; if the total number of balls is odd, Prince must have an even number of balls. Hence by proper logic it is revealed that, the total number of balls can not be 13 .Also total number can not be 12,14,15,16,17 because then the total number of balls each had would be known. Also the total number cannot be 18,20,21,22,23,24 because then no number of balls could be known for anybody. So total is 19.
When total is 19. Prince must have an even number of balls and from the sequences, this number must not be greater than 10. So Prince must have 6 balls. Then Vipul and Keshav together must have 13 balls, then Vipul must have either 4 or 8 balls .but not definite one.
Hence speaker is Prince.
Saturday, March 15, 2008
squieeez part 3 solutions
Ans1-
We show more precisely that the game terminates with
one player holding all of the pennies if and only if n=2^(m) +1 or n = 2^(m)+2 for some m. First suppose
we are in the following situation for some k>=2.(Note: for us, a “move” consists of two turns, starting
with a one-coin pass.)
– Except for the player to move, each player has k coins;
– The player to move has at least k coins.
We claim then that the game terminates if and only if the number of players is a power of 2. First suppose
the number of players is even; then after m complete rounds, every other player, starting with the player who
moved first, will have m more coins than initially, and the others will all have 0. Thus we are reduced to the
situation with half as many players; by this process, we eventually reduce to the case where the number of players
is odd. However, if there is more than one player, after two complete rounds everyone has as many coins
as they did before (here we need m>=2), so the game fails to terminate. This verifies the claim.
Returning to the original game, note that after one completeround, players remain, each with 2 coins
except for the player to move, who has either 3 or 4coins.Thus by the above argument, the game terminates
Ans 2 –
Assumption: Two of the three can be present on same day.
First we need to identify with what day this month can start with:
Monday: It is possible. Then Prince will be the guy who goes every 2 days, Vipul every 3 days and Methi every 7 days. No other combination will be possible.
Tuesday: Can’t start with Tuesday. Because no combination will be possible as no two guys go on Tuesday, Wednesday and Thursday necessary for satisfying every 2 and 3 day conditions.
Wednesday: It is possible. Then Vipul will be the guy who goes every 2 days, Methi every 3 days and Prince every 7 days. No other combination will be possible.
Thursday: Not possible with same reason as for Tuesday.
Friday: Not possible with same reason as for Tuesday.
Saturday: Not possible with same reason as for Tuesday.
Sunday: Not possible with same reason as for Tuesday.
Lets first analyze rest question for Monday.
P ----- Prince V ------ Vipul K ----- Keshav
Date range Mon Tue Wed Thu Fri Sat Sun
1-7 p V, p K , p v p
8-14 V , p p K , v p
15-21 V , p p v K , p V , p
22-28 p v p k V , p
29-31 p v p
Since there is no day possible on which all three come together. Hence Monday is not possible.
Lets do it for Wednesday.
Date Mon Tue Wed Thu Fri Sat Sun
1-5 ------ ------ v V , k v
6-12 P , k v V , k v k
13-19 P , v V , k v k v
20-26 p V , k v k v
27-31 P , v , k v k v
As is clear from the table, all the three will meet on 27th of this month on Monday. You can also verify that all the three will not be present together on same day in the last turn of previous month
We show more precisely that the game terminates with
one player holding all of the pennies if and only if n=2^(m) +1 or n = 2^(m)+2 for some m. First suppose
we are in the following situation for some k>=2.(Note: for us, a “move” consists of two turns, starting
with a one-coin pass.)
– Except for the player to move, each player has k coins;
– The player to move has at least k coins.
We claim then that the game terminates if and only if the number of players is a power of 2. First suppose
the number of players is even; then after m complete rounds, every other player, starting with the player who
moved first, will have m more coins than initially, and the others will all have 0. Thus we are reduced to the
situation with half as many players; by this process, we eventually reduce to the case where the number of players
is odd. However, if there is more than one player, after two complete rounds everyone has as many coins
as they did before (here we need m>=2), so the game fails to terminate. This verifies the claim.
Returning to the original game, note that after one completeround, players remain, each with 2 coins
except for the player to move, who has either 3 or 4coins.Thus by the above argument, the game terminates
Ans 2 –
Assumption: Two of the three can be present on same day.
First we need to identify with what day this month can start with:
Monday: It is possible. Then Prince will be the guy who goes every 2 days, Vipul every 3 days and Methi every 7 days. No other combination will be possible.
Tuesday: Can’t start with Tuesday. Because no combination will be possible as no two guys go on Tuesday, Wednesday and Thursday necessary for satisfying every 2 and 3 day conditions.
Wednesday: It is possible. Then Vipul will be the guy who goes every 2 days, Methi every 3 days and Prince every 7 days. No other combination will be possible.
Thursday: Not possible with same reason as for Tuesday.
Friday: Not possible with same reason as for Tuesday.
Saturday: Not possible with same reason as for Tuesday.
Sunday: Not possible with same reason as for Tuesday.
Lets first analyze rest question for Monday.
P ----- Prince V ------ Vipul K ----- Keshav
Date range Mon Tue Wed Thu Fri Sat Sun
1-7 p V, p K , p v p
8-14 V , p p K , v p
15-21 V , p p v K , p V , p
22-28 p v p k V , p
29-31 p v p
Since there is no day possible on which all three come together. Hence Monday is not possible.
Lets do it for Wednesday.
Date Mon Tue Wed Thu Fri Sat Sun
1-5 ------ ------ v V , k v
6-12 P , k v V , k v k
13-19 P , v V , k v k v
20-26 p V , k v k v
27-31 P , v , k v k v
As is clear from the table, all the three will meet on 27th of this month on Monday. You can also verify that all the three will not be present together on same day in the last turn of previous month
squieeez part 2 solutions
Ans: 1
Let 5 digits are X1, X2, X3, X4 and X5.
Since we are getting 10 different values, therefore we have all digits different. ( = 10)
So we Assume X1 < X2 < X3 < X4 < X5 (All different)
By given values of summation
X1 + X2 + X3 = 0 …………………………………….. (1)
X3 + X4 + X5 = 19……………………………………. (2)
X2 + X4 + X5 = 14 …………………………………… (3)
X1 + X2 + X4 = 3 …………………………………….. (4)
6(X1 + X2 + X3 + X4 + X5) = 90
Or X1 + X2 + X3 + X4 + X5 = 15……………………. (5)
=> X4 + X5 = 15
=> X3 = 4
=> X2 = -1
=> X1 = -3
=> X4 = 7
=> X5 = 8
Ans: 2
Let X be the number of hours, and Y be the number of minutes past the hour. When the hour hand is on a minute mark, the position of the hour hand is 5X + Y/12, and the position of the minute hand is Y. On the first occasion, Y = 5X + Y/12 + 6. Since X and Y are integers, therefore Y can only take one of the values in the set {0, 12, 24, 36, 48}, it can be determined that the only legal values for the equation are X = 1 and Y = 12. So the time is 1:12.
Similarly, the second occasion's equation is Y = 5X + Y/12 + 7. The only legal values here are X = 3 and Y = 24. So the time is 3:24.
Between 1:12 and 3:24, two hours and twelve minutes have elapsed.
Let 5 digits are X1, X2, X3, X4 and X5.
Since we are getting 10 different values, therefore we have all digits different. ( = 10)
So we Assume X1 < X2 < X3 < X4 < X5 (All different)
By given values of summation
X1 + X2 + X3 = 0 …………………………………….. (1)
X3 + X4 + X5 = 19……………………………………. (2)
X2 + X4 + X5 = 14 …………………………………… (3)
X1 + X2 + X4 = 3 …………………………………….. (4)
6(X1 + X2 + X3 + X4 + X5) = 90
Or X1 + X2 + X3 + X4 + X5 = 15……………………. (5)
=> X4 + X5 = 15
=> X3 = 4
=> X2 = -1
=> X1 = -3
=> X4 = 7
=> X5 = 8
Ans: 2
Let X be the number of hours, and Y be the number of minutes past the hour. When the hour hand is on a minute mark, the position of the hour hand is 5X + Y/12, and the position of the minute hand is Y. On the first occasion, Y = 5X + Y/12 + 6. Since X and Y are integers, therefore Y can only take one of the values in the set {0, 12, 24, 36, 48}, it can be determined that the only legal values for the equation are X = 1 and Y = 12. So the time is 1:12.
Similarly, the second occasion's equation is Y = 5X + Y/12 + 7. The only legal values here are X = 3 and Y = 24. So the time is 3:24.
Between 1:12 and 3:24, two hours and twelve minutes have elapsed.
Saturday, February 23, 2008
Queries
Hi all,
If u have any queries or doubts, post it in here as comments, so that we can try and answer them. You can also watch out this space to find similar queries which others may already have asked.
Have Fun!
Team Cognizance
Electronics & Computer Science Department
IITR
If u have any queries or doubts, post it in here as comments, so that we can try and answer them. You can also watch out this space to find similar queries which others may already have asked.
Have Fun!
Team Cognizance
Electronics & Computer Science Department
IITR
Monday, February 18, 2008
sQuieeez -1 13th Feb 08
Answers to the sQuieeez 1 problems are :
Answer 1:
3468
Starting at any of the N's, there are 17 different readings of NAK, or 68 (4 times 17) for the 4 N's. Therefore there are also 68 ways of spelling KAN. If we were allowed to use the same N twice in a spelling, the answer would be 68 times 68, or 4,624 ways. But the conditions were, "always passing from one letter to another." Therefore, for every one of the 17 ways of spelling KAN with a particular N, there would be 51 ways (3 times 17) of completing the NAK, or 867 (17 times 51) ways for the complete word. Hence, as there are four N's to use in KAN, the correct solution of the puzzle is 3,468 (4 times 867) different ways.
Answer 2:
Note that
= 2 + 1 = 3.
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