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What we do:	Upcoming Workshop:
The Math Help Centre is a free service offered by the Department of Mathematics for all first year students to get help with mathematics. Located at 412 Machray Hall, the help centre is staffed by graduate students with a passion for teaching. You can come to the help centre if you have specific questions you wish to ask one of our tutors, or you can just come and use it as a study space with help on hand should you get stuck. Either way, bring your notes and your brains and come get help with those problems you've been stuck on!	Laws of Logic and Rules of Inference, Sept 24, 25^th

What we do:

Upcoming Workshop:

The Math Help Centre is a free service offered by the Department of Mathematics for all first year students to get help with mathematics. Located at 412 Machray Hall, the help centre is staffed by graduate students with a passion for teaching. You can come to the help centre if you have specific questions you wish to ask one of our tutors, or you can just come and use it as a study space with help on hand should you get stuck. Either way, bring your notes and your brains and come get help with those problems you've been stuck on!

Laws of Logic and Rules of Inference, Sept 24, 25^th

Time	Monday	Tuesday	Wednesday	Thursday	Friday
9:00 - 10:00	Alex Elaine	Elaine Sergei	Alex M. Sadeghi	M. Sadeghi Waweru	Emmanuel Sarah
10:00 - 11:00	Alex Avleen Ruoxin	Aidin Emmanuel Sergei	Alan M. Sadeghi Ruoxin	Aidin Farzaneh M. Sadeghi	Yungang Ruoxin Sarah
11:00 - 12:00	Avleen Max Sergei	Aidin Babak Sahar	Deokro Ievgen Yungang	Alan Babak Farzaneh	Alan Avleen Sergei
12:00 - 1:00	Atika Sergei	Babak Tessa	Elaine Ievgen	Suraj Tessa	Alan Suraj
1:00 - 2:00	Aidin Sergei	Alex Irshad	Ian Yuanchao	Atika Olena	Yuanchao Ievgen
2:00 - 3:00	Kelsey Sooyeong	Irshad Sooyeong	Emmanuel Yuanchao	Atika Olena	Yuanchao Ievgen
3:00 - 4:00	M. Sadeghi Sooyeong	Emmanuel Olena	Yuanchao Olena	Sergei Olena	Emmanuel M. Shirazi
4:00 - 5:00	Kyrylo Sergei	Kyrylo Olena	Aidin Ian	Sergei Xiaohong	Sehaj Yungang
5:00 - 6:00	Kyrylo Sergei	Babak Deokro	Babak Avleen	Alan Xiaohong	Sehaj Yungang

In each of the following workshops, we will be tackling problems that aid understanding of the topic, and problems that are commonly found on midterm and final examinations. Each workshop starts with a (roughly) 20 minute demonstration, with time in the middle for students to attempt the problems and ask questions if/when they get stuck. The problems that caused students the most difficulty are then demonstrated for the last 20 minutes.
These workshops are totally free, and all first year students are welcome to attend as much or as little of the workshop as they wish. Comments or suggestions for these workshops are welcome -- just send an email to the MHC (MathHelpCentre@umanitoba.ca) and your comments will be considered.

Logic
Sept. 24th^th, 1:00pm-2:00pm.
Sept. 25th^th, 1:30pm-2:30pm.
This workshop will be relevant to students in MATH1240. In this workshop, we'll be looking at a large number of questions from old exams on the topic of logic.
SEATING IS LIMITED, SO ONLY STUDENTS WHO HAVE FILLED OUT THE ONLINE FORM MAY ATTEND.
Both of the above workshops will be the exact same, so each student may attend only one.

Three dogs, Beethoven, Lassie and Skip, are sitting at corners of an equilateral triangle of side 15m in a park. Simultaneously, Beethoven starts walking towards Lassie, Lassie starts walking towards Skip and Skip starts walking towards Beethoven. If all dogs are walking at the same speed of 1m/s, how long does it take for all dogs to meet?

Answer: 10 seconds

There are 63 coins, all identical in appearance, and all identical weight, except one coin that is slightly heavier that the others. Three men wish to identify the heavy coin from the 63 using a peculiar set of pan scales. These pan scales have three pans places symmetrically about a central pivot point. The only function of the scales is that if two pans carry a load of equal weight and the third pan carries a load of different weight, then the movement of the scales indicates whether the different weight is heavier or lighter than the other two weights.
Each of the three men must perform one weighing of a selection of coins without the other two men witnessing the result of the weighing. Beforehand, the men can plan a strategy, and after the weighings, they can regroup and compare their results. Is there a strategy such that when the men regroup, then can identify the heavy coin?

Answer: The first man takes 48 of the coins and place 16 on each pan. If all pans weigh the same, he concludes the heavy coin is one of the 15 remaning, so he puts the 48 aside, and leaves the 15 for the next man. If one of the pans is heavier than the others, the first man leaves the 16 coins on the heavy pan for the next man and puts the rest aside. The next man takes 12 of either the 15 or 16 coins depending on the result of the first weighing. He places 4 on each scale. If all pans weigh the same, he leaves the three (or 4) remaining coins for the next man to weigh. If one of the pans weighed more, he leaves the coins from that pan for the third man. The third man now has either three or four coins to weigh. If there are three, he places one on each pan and finds the heavy one. If there are four remaining, he places one on each pan. If one of the pans is heavier then the coin on that pan is the heavy one. If they weigh the same, then the one he didn't weigh is the heavy one.

A tennis tournament with n players is held. All the players are paired up and play one game. The winner goes through to the next round. If the number of players in any round is odd, then one randomly chosen player gets to skip the round and play in the next round. This system continues until someone wins the tournament. How many games are played in total in the whole tournament?

Answer: There is a one-to-one correspondence between the number of players who lose a game and the number of games played. Since each player loses exactly one game (except the winner), n-1 games are played.

Runner A, runner B and runner C all run a 100m race. Runner A finishes 10m ahead of runner B. Runner B finishes 10m ahead of runner C. How far ahead of runner C does runner A finish?

Answer: 19m. If runner A finishes 10m ahead of runner B, then runner B is running 9/10 the speed of runner A. Likewise, if runner B finishes 10m ahead of runner C, then runner C is running 9/10 the speed of runner B. Therefore if runner A is running at x m/s, runner B is running 9x/10 m/s and runner c is running 9/10*(9x/10) = 81x/100m/s. Therefore, runner C is running 81/100 times the speed of runner A. When runner A has finished the race (i.e., he has ran 100m), runner C has ran 81m, so he is 19m behind runner A.

Mathematics Help Centre