Playing with Chances (Tuklas Vol. 16, No. 2 - October 25, 2014)

PLAYING WITH CHANCES

One popular card game in the Philippines is Pusoy Dos. The goal of the game is simple: be the first player to use all of his or her cards. A player can either use a single card, pair, three of a kind, or a five-card combination, depending on the current play. Each player must play a card that has the same number of cards as the previous player, but should have a higher rank than the previous. The rank of single, pair and three cards are based on its face value and suit. What is interesting to look at is the ranking of five-card hands. Have you ever wondered why they are ranked as such?

The allowable five-card combinations, ranked in increasing order, are: straight, flush, full house, four of a kind, straight flush, and royal flush. Notice that they are similar to poker hands. At first glance, it would seem that the combinations are ranked based on the chances of a player forming that combination from any five randomly chosen cards from his or her hand. Now, let us try to put a number on that chance.

Probability
The probability of an event refers to a measure of the chances that the said event will occur. The most common definition of probability of an event \(E\) occurring is the ratio between the number of times the event can occur and the number of times an outcome (let us call it \(S\)) can happen. In formula, \(P(E) = \frac{n(E)}{n(S)}\), where \(n(\cdot)\) refers to the number of times it can occur.

In this case, \(E\) refers to the different five-card combinations mentioned above, while \(S\) refers to any set of five cards. For easier computation, let us assume that we are choosing five cards from a standard deck of cards (meaning we have 52 cards), rather than just the 13 cards in our hand. Notice that regardless of \(E\), your \(S\) is fixed. That is, the number of combinations you can do with five cards is always \({52 \choose 5}= 2,598,960\).

Royal Flush
To form a royal flush, your five cards must be '10', 'J', 'Q', 'K', 'A', all with the same suit. Since there are four suits, then the number of times to get a royal flush is 4. Then, the probability of getting a royal flush is \(\frac{4}{2,598,960}\), or a mere \(0.0002\%\).

Straight Flush
Next, to form a straight flush, your five cards must have consecutive face values, all with the same suit. Counting the number of straight flushes is similar to counting the number of times you can choose your lowest card, since the next four cards depends on the first card. However, we should be careful not to form a royal flush. After choosing the face value, we can now choose what the suit of the five cards will be. Since there are 9 cards to choose from for the lowest (`10' is excluded since this will form a royal flush) and 4 suits to choose from, the number of times we can get a straight flush is \({9\choose 1} \times {4\choose 1} = 36\). Then, the probability of getting a straight flush is \(\frac{36}{2,598,960}\) or \(0.0014\%\).

Four of a Kind
Forming a four of a kind is straightforward. You need to have all four suits of a particular card, then one other card. There are 13 values to choose from for the 'four' cards, and 12 values (since the value of the `four' cards cannot be used) for the `one' card. After choosing the value of the 'one,' its suit will be chosen. Then, the number of times we can get a four of a kind is \({13\choose 1}\times {12 \choose 1} \times {4\choose 1} = 624\). Then, the probability of getting a four of a kind is \(\frac{624}{2,598,960}\) or \(0.024\%\).

Full House
Forming a full house is similar to forming a four of a kind. Instead of having all four suits for a particular card, you just need three. The remaining two cards, however, must also have the same value. There are 13 values to choose from for the `three' cards, and 12 values for the 'two' cards. After choosing the values of the 'three' and 'two,' we are now going to choose the suits for the 'three' and 'two' separately. Then, the number of times we can get a full house is \({13\choose 1}  \times {4\choose 3} \times {12 \choose 1} \times  \times {4\choose 2} = 3,744\). Then, the probability of getting a full house is \(\frac{3,744}{2,598,960}\) or \(0.1441\%\).

Flush
Next, to form a flush, your five cards must have the same suit, regardless of their face values. However, we must be careful not to count the straight and royal flushes. We have four suits to choose from, and 13 values for the five cards. Then, the number of times we can get a flush is \({4 \choose 1}\times {13 \choose 5} - \left(36 + 4\right) = 5,108\). Then, the probability of getting a flush is \(\frac{5,108}{2,598,960}\) or \(0.1965\%\).

Straight
Lastly, to form a straight, your five cards must have consecutive face values. Counting the number of straights is similar to counting the number of times you can choose your lowest card, since the next four cards depends on the first card. After choosing the face value, we can now choose what the suit of each of the five cards will be. We have 10 cards to choose from for the lowest, and 4 suits for each  card. However, we should be careful not to count the straight and royal flushes. Hence, the number of times we can get a straight is given by \({10\choose 1} \times {4\choose 1}\times {4\choose 1}\times {4\choose 1}\times {4\choose 1}\times {4\choose 1}  - \left(36 + 4\right) = 10,200\). Then, the probability of getting a straight is \(\frac{10,200}{2,598,960}\) or \(0.3925\%\).

Playing the Game
After doing all the computations, our guess is indeed valid. The five-card combination rankings are based on the probability of getting them. Since royal flushes have the smallest probability, then they are ranked highest. Same is true for straights. Since they have the highest probability of getting formed, they are ranked lowest.

Now, the next time you play five cards in Pusoy Dos, it might help to think of the chances of your opponents having a higher set of five cards, given you already know that there are 13 cards not in his hand. This may be your secret to winning.

ABOUT THE AUTHOR:
Jeric Briones is an Instructor at the Ateneo de Manila University. He obtained his Master of Applied Mathematics major in Mathematical Finance degree at the same university in 2013.

OLYMPIAD CORNER

from the Asian Pacific Mathematics Olympiad, 2014

Problem: For a positive integer \(m\) denote by \(S(m)\) and \(P(m)\) the sum and product, respectively, of the digits of \(m\). Show that for each positive integer \(n\), there exist positive integers \(a_{1}, a_{2}, \ldots, a_{n}\) satisfying the following conditions: \[ S(a_{1}) < S(a_{2})<\cdots <S(a_{n}) \]\[ S(a_{i}) = P(a_{i+1}), i=1,2,...n \] (We let \(a_{n+1}=a_{1}\))

Solution: 

Let \(k\) be a sufficiently large positive integer. For each \(i=2,3,...,n\), the digits of \(a_{i}\) are such that the number 2 appears exactly \(k+i-2\) times, the number 1 appears exactly \(2^{k+i-1}-2\left( k+i-2\right)\) times, and no other number appears. Then\[ S\left( a_{i}\right)  = 2\times \left( k+i-2\right) +1\times \left[ 2^{k+i-1}-2\left( k+i-2\right) \right] =2^{k+i-1} \]\[ P\left( a_{i}\right)  = 2^{k+i-2} \]Observe that\[ S\left( a_{i}\right) =2^{k+i-1}<2^{k+i}=S\left( a_{i+1}\right) \]and\[ S\left( a_{i}\right) =2^{k+i-1}=P\left( a_{i+1}\right) \]for \(i=2,3,..,n-1\).

Next, set the digits of \(a_{i}\) so that the number 2 appears exactly \(k+n-1\) times, the number 1 appears exactly \(2^{k}-2\left( k+n-1\right)\) times, and no other number appears. Then\[ S\left( a_{1}\right)  = 2\times \left( k+n-1\right) +1\times \left[2^{k}-2\left( k+n-1\right) \right] =2^{k} \]\[ P\left( a_{1}\right)  = 2^{k+n-1} \]Observe that\[ S\left( a_{1}\right) =2^{k}<2^{k+1}=S\left( a_{2}\right) \]and\[ S\left( a_{1}\right) =2^{k}=P\left( a_{2}\right) \]\[ S\left( a_{n}\right) =2^{k+n-1}=P\left( a_{1}\right) =P\left( a_{n+1}\right) \]Moreover, such a choice of \(a_{1}\) is possible if we take \(k\) to be large enough to satisfy \(2^{k}>2\left( k+n-1\right)\). Therefore, the numbers \(a_{1},a_{2},...,a_{n}\) that are chosen based on the above conditions satisfy the given requirements. 

PROBLEMS

4. Is there a triangle where the three altitudes have lengths \(1\), \(\sqrt{5}\), and \(1+\sqrt{5}\)? 
5. Let \(0<a<b<c<d\) be odd integers such that \(ad=bc\) and \(a+d=2^{k}\), \(b+c=2^{m}\), for some integers \(k\) and \(m\). Determine the value of \(a\).
6. Show that the interval \(\left[0,1\right]\) cannot be partitioned into two disjoint sets \(A\) and \(B\) such that \(B=A+a\) for some real number \(a\).

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions may ONLY be submitted online via vantonio@ateneo.edu. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 4:00 PM November 8, 2014.

Stuffing Things in Boxes (Tuklas Vol. 16, No. 1 - October 18, 2014)

STUFFING THINGS IN BOXES

Suppose you and four other friends are dining at a restaurant. Looking at the menu, you see that there are four different rice meals that can be ordered. Is it possible for each of you to order a unique rice meal?

A quick guess could be no. How would we solve this? If we pair off each rice meal to a person, we see that by the time we reach the fifth person there is no longer a unique rice meal that that person can order. 1

This example illustrates a very powerful technique with interesting applications. In fact, the underlying math involved is so simple that hardly anyone can see it. This technique is called the pigeonhole principle. First formally introduced by the German mathematician Peter Gustave Lejeune Dirichlet under the name Schubfachprinzip 2, we can state it roughly as follows:

If \(n+1\) objects are put into \(n\) boxes, then at least one box contains two or more objects.

In our first example, the objects refer to the diners and the boxes refer to the kinds of meals. Since there are five diners and only four kinds of meals, one diner must have the same kind of meal as one of the other four. Albeit simple, this technique can be used to prove some interesting and surprising mathematical results.

Let us first consider an easy variation of what is called the birthday problem.

Given a set of \(n\) randomly chosen people, what is the probability that some pair of them will have the same birthday?

The problem arises when one is called to find the value of \(n\) for different situations. While this is an interesting problem for another time, the pigeonhole principle actually guarantees that a pair will always exist for \(n \geq 367\), simply because there are \(366\) unique days in a year (including February 29). Here, the days are the boxes and the people are the objects.

Suppose we now make those \(n\) people shake hands. The pigeonhole principle can be used to show that there is always a pair of people who will shake hands with the same number of people. To illustrate this, we determine what the boxes or holes should be. Note that each person can shake hands with \(0\) up to \(n-1\) people. If we let our holes be equivalent to the number of people a person has shaken hands with, then one of the "\(0\)" or "\(n-1\)" holes must necessarily be empty. Why is this the case? Note that if the "\(0\)" hole is nonempty, then there is a person who has not shaken hands with anyone. Hence, the "\(n-1\)" hole must be empty. 3 This implies that we have at most \(n-1\) holes to fill in with \(n\) people, guaranteeing at least one pair that would have shaken hands with the same number of people.

The previous example highlights one of the mildly frustrating consequences of using the pigeonhole principle. Its highly intuitive concept makes it harder to apply than other more methodical techniques. The difficulty lies in trying to determine what the holes should represent; once that is overcome, everything else becomes easier.

Consider, for instance, this charming problem.

Show that given any set \(A\) of \(13\) distinct real numbers, there exist \(x, y \in A\) such that \[ 0 < \frac{x-y}{1+xy} \leq 2 - \sqrt{3}. \]

An elegant solution to this problem is as follows. We recognize the middle expression as equivalent to the difference of tangents identity. Thus, we set \(x = \tan(u)\) and \(y = \tan(v)\) for some angles \(u,v\) considered modulo \(\pi\) (by the periodicity of the tangent function). We have \[ \frac{x-y}{1+xy} = \tan(u-v), \]implying that we are interested in closing the gap between the angles \(u\) and \(v\) (as implied by \(u-v\)). Specifically, we set the gap to be smaller than \(\frac{\pi}{12}\). This observation is supported by noting that the \(13\) distinct real numbers imply that we divide the angles in \([0,\pi]\) into 12 equal sectors. Moreover, we have \(\tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}\). By the pigeonhole principle, we are done.

ABOUT THE AUTHOR:
Jesus Lemuel L. Martin, Jr. is an Instructor at the Ateneo de Manila University. He obtained his B.S. in Applied Mathematics major in Computational Science at the Ateneo de Manila University in 2010 and his M.S. in Mathematics at the same university in 2013.

ENDNOTES:

[1] In short, person number five will have to copy someone else's order, or get an appetizer instead if uniqueness is such an issue.

[2] Roughly translated as "drawer principle" or "shelf principle", it is sometimes called Dirichlet's box principle or Dirichlet's box principle.
[3] Equivalently, if the "n-1" hole is nonempty, then there is a person who has shaken everyone else's hand, making the "0" hole empty.

ENDNOTES:

- Bogomolny, Alexander. Pigeonhole Principle. Retrieved from http://www.cut-the-knot.org/do_you_know/pigeon.shtml.

- Chen, Beifang and Wang, J. "The Pigeonhole Principle." (2005) Retrieved from https://www.math.ust.hk/~mabfchen/Math391I/Pigeonhole.pdf.

- Chen, Chuan-Chong and Koh, K.-M. Principles and Techniques in Combinatorics (1992).

OLYMPIAD CORNER

from the Belarus National Contest, 2000

Problem: Let \(M\) be the intersection point of the diagonals \(AC\) and \(BD\) of a convex quadrilateral \(ABCD\). The bisector of angle \(ACD\) hits ray \(BA\) at \(K\). If \(MA\cdot MC+MA\cdot CD=MB\cdot MD\), prove that \(\angle BKC=\angle CDB\).

Solution: 

Let \(N\) be the intersection of lines \(K\) and \(BD\). By the Angle Bisector Theorem applied to triangle \(MCD\),\(\frac{CD}{DN}=\dfrac{MC}{MN}\), or \(CD=\dfrac{MC\cdot DN}{MN}\). We then have \[ MB\cdot MD=MA\cdot MC+MA\cdot \dfrac{MC\cdot ND}{MN}=(MA\cdot MC)\cdot\frac{MD}{MN}, \]or \(MA\cdot MC=MB\cdot MN\). 

Because \(M\) lies inside quadrilateral \(ABCN\), the Power of a Point Theorem implies that \(A\), \(B\), \(C\), and \(N\) are concyclic. Hence, \(\angle KBD=\angle ABN=\angle ACN=\angle NCD=\angle KCD\), implying that \(K\), \(B\), \(C\), and \(D\) are concyclic. Thus, \(\angle BKC=\angle CDB\), as desired.

PROBLEMS

1. There are given \(2014\) sets, each containing \(40\) elements. Every two sets have exactly one element in common. Prove that all \(2014\) sets have exactly one element in common.
   
2. Let \(a, b, c\) be positive real numbers such that \(abc=1\). Prove the inequality \[ \frac{a}{a+b^{4}+c^{4}}+\frac{b}{b+c^{4}+a^{4}}+\frac{c}{c+a^{4}+b^{4}}\leq1. \]
3. Circles \(C_{1}\) and \(C_{2}\) with centers \(O_{1}\) and \(O_{2}\) respectively intersect each other at \(A\) and \(B\). Ray \(O_{1}B\) intersects \(C_{2}\) at \(F\) and ray \(O_{2}B\) intersects \(C_{1}\) at \(E\). The line parallel to \(EF\) and passing through \(B\) intersects \(C_{1}\) and \(C_{2}\) at \(M\) and \(N\), respectively. Prove that \(MN=AE+AF\).

We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions may ONLY be submitted online via vantonio@ateneo.edu. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 4:00 PM October 25, 2014.

PEM 2014 Class Lists

 PROGRAM FOR EXCELLENCE IN MATHEMATICS 2014

GRADE 7

Schedule: Sat, 9:00-12:00NN

Room: SEC A 202

Last Name	First Name	MI	School
Abrahan	Carmen		Assumption College, San Lorenzo
Bolosan	Ara Marinoli	B.	Valenzuela City Science HS
Cabalar	Andrea	A.	Valenzuela City Science HS
Cruz	Remus Japhet	D.C.	Pateros Catholic School
Cruz	Romulus Jathniel	D.C.	Pateros Catholic School
Dacanay	Valerie		St. Joseph's College of QC
Dayrit	Gabrielle		Assumption College, San Lorenzo
De Los Angeles	Lauren Angelica		St. Paul University QC
Demition	Andrew	DR.	Valenzuela City Science HS
Espinocilla	Kate		St. Joseph's College of QC
Jayme	Anya Patricia	J.	Saint Pedro Poveda College
Lanuza	Gwyneth Lee	D.	St. Paul University QC
Lazaro	Yves Gabriel	M.	Philippine Science High School
Munar	Marissa		St. Theresa's College, QC
Ong	Dion Stephan	J.	Ateneo HS
Ong	Elizabeth		Philippine Institute of QC
Padua	Mikhail Anthony	G.	Manila Science High School
Parra	Leibniz Charisse		St. Paul College, Pasig
Pineda	Leanne Ysabelle	W.	Saint Pedro Poveda College
Siazon	Darelle Jade	T.	Valenzuela City Science HS
Taguinod	Juan Roy	B.	Valenzuela City Science HS
Tenorio	Rallon Christopher	A.	Pateros Catholic School
Tsoi	Sha Sha		Philippine Institute of QC
Tuazon	Marcia		St. Theresa's College, QC
Villareal	Alexa Sofia		St. Paul College, Pasig

GRADE 8
Schedule: Sat, 9:00-12:00NN
Room: SEC A 203

Last Name	First Name	MI	School
Abio	Marian Louise		St. Joseph's College of QC
Aguilar	Elisha Rynne	T.	De La Salle Zobel School
Ampong	Mark Louis	A.	Pateros Catholic School
Astada	Darielle		St. Theresa's College, QC
Castro	Mara Gabrielle	S.	St. Paul College of Parañaque
Catindig	Jayson Dwight	S.	Ateneo HS
Concepcion	Danielle Andrea Rose	S.	Manila Science High School
Cruz	Julia Martina	P.	Saint Pedro Poveda College
Cummings	Angele	L.	De La Salle Zobel School
Dela Cruz	Veronica Anna	S.	St. Paul College of Parañaque
Domingo	Gellie		Manila Science High School
Magsajo	Theresa Denise		St. Paul College, Pasig
Mallari	Cristina Beatrice		St. Paul College, Pasig
Medrero	Janna Ysabelle		De La Salle Zobel School
Mesa	Kiara Anne		St. Joseph's College of QC
Ortega	Gerard Francis	M.	Ateneo HS
Quinio	Giselle		Assumption College, San Lorenzo
Ramirez	Anna Marie		St. Paul University QC
Ramos	Kristine		St. Theresa's College, QC
Regino	Alyssa Bianca	D.	Saint Pedro Poveda College
Sacro	Jacinta		Assumption College, San Lorenzo
Sy	Kyla Nicole		Philippine Institute of QC
Tan	Paolo Caleb		Philippine Institute of QC
Villa	Gabriel	C.	Pateros Catholic School

GRADE 9-A
Schedule: Sat, 9:00-12:00NN
Room: SEC A 204

Last Name	First Name	MI	School
Carreon	Alexine		St. Theresa's College, QC
Chow	Mary Florence		Assumption College, San Lorenzo
Chua	Khyla		ICA Greenhills
Corpuz	Frederick Matthew	T.	Ateneo HS
Evan	Daniella Kristin		De La Salle Zobel School
Geraldez	Beatrice Martha		St. Theresa's College, QC
Hocson	Andrea Betina		De La Salle Zobel School
Magno	Christopher Joseph	G.	Ateneo HS
Mendoza	Bettina		Colegio de Sta. Rosa - Makati
Recio	Anne Kaila		Philippine Institute of QC
Santiago	Martin Rodolfo	B.	Ateneo HS
Santos	Maria Antoinette Gabrielle		Assumption College, San Lorenzo
Sava	Maria Mikaela Joyce	C.	Pateros Catholic School
Tolentino	Carolina Lim		St. Theresa's College, QC
Ueda	Fumiya Kyle		De La Salle Zobel School
Wang	Cindy		Colegio de Sta. Rosa - Makati
Wong	Samantha		Philippine Institute of QC
Yamane	Christine Marie	S.	Pateros Catholic School

GRADE 9-B
Schedule: Sat, 9:00-12:00NN
Room: SEC A 116

Last Name	First Name	MI	School
Alcovendaz	Ma. Jennica		St. Paul College, Pasig
Azuma	Ruka	D.	St. Paul College of Parañaque
De Guzman	Kristen Kyra		St. Paul University QC
Fernandez	Katrine Chanel		St. Paul University QC
Flores	Lorenzo Jaime	Y.	Philippine Science High School
Flores	Anina-Lei		St. Paul College, Pasig
Gutierrez	Jonathan Mari		St. Joseph's College of QC
Mactal	Maria Katrina	L.	Saint Pedro Poveda College
Mape	Mary Zenaida		Saint Pedro Poveda College
Merano	Dolores		St. Paul College of Parañaque
Navarro	Julianna		St. Paul College, Pasig
Niño	Francis		St. Paul University QC
Ordoñez	Marie Mel	C.	St. Paul College of Parañaque
Pajarillaga	Melissa Shane		St. Joseph's College of QC
Santiago	Rafael	D.	Philippine Science High School
Villanueva	Camille Ann		St. Paul College, Pasig

GRADE 10-A
Schedule: Sat, 9:00-12:00NN
Room: SEC A 205

Last Name	First Name	MI	School
Abrilla	Denise Michelle	A.	Colegio de Sta. Rosa - Makati
Bacila	Ysabelle Coelyn		Colegio de Sta. Rosa - Makati
Bautista	Jason Angelo	N.	Ateneo HS
Bruce	Gabrielle Nikhaela		Assumption College, San Lorenzo
Capistrano	Alessandro Raphael	G.	Lourdes School of Mandaluyong
Ching	Jaymi Mae		Jubilee Christian Academy
Co	Mason Alonzo	C.	Jubilee Christian Academy
Cua	Sonny	S.	Philippine Cultural College - Mla
Del Mundo	Jo Adrian	P.	Ateneo HS
Eala	John Patrick	B.	Ateneo HS
Fernandez	Jervie Jan	M.	Lourdes School of Mandaluyong
Flores	Luis Miguel	C.	Lourdes School of Mandaluyong
Gopaldas	Shireen		Assumption College, San Lorenzo
Labao	Richard Braulio	J.	Lourdes School of Mandaluyong
Lazaro	Lorenzo Thomas		Ateneo HS
Libunao	Philip Neri	R.	Ateneo HS
Moises	Ricardo Sebastian	L.	Pateros Catholic School
Ong	Richmond Lance		Philippine Cultural College - Mla
Ordona	Janzrel Kyle	D.G.	Pateros Catholic School
Ratilla	Gabrielle		Philippine Institute of QC
Saret	Lemuel Gavin	G.	Manila Science High School
See	Danielle Joyce	L.	Jubilee Christian Academy
Sia	Ma. Angelika		Philippine Cultural College - Mla
Young	Ezekiel		Jubilee Christian Academy
Yu	Janine Michelle		Philippine Institute of QC

GRADE 10-B
Schedule: Sat, 9:00-12:00NN
Room: SEC A 117

Last Name	First Name	MI	School
Buenaseda	Justine		St. Joseph's College of QC
Chu	David Anthony	T.	St. Stephen's High School
Co	Miko Johnson	O.	St. Stephen's High School
Cunanan	Sylvester Walt		St. Paul University QC
De Leon	Bret Michaels	D.	Valenzuela City Science HS
Fernandez	Kimberly Danielle		St. Paul University QC
Herrera	Arzeliz Dyrene		Valenzuela City Science HS
Hipolito	Kristen Rae		St. Theresa's College, QC
Holgado	Jeanine		St. Paul University QC
Llorin	Gewell		St. Paul College, Pasig
Macalalad	Nadine		St. Paul College, Pasig
Mape	Christine Mary Rose	R.	Saint Pedro Poveda College
Onglao	Ma. Andrea Kristina		St. Paul College, Pasig
Petalver	Arlyn Emmelle Graham	O.	Valenzuela City Science HS
Rafael	Dwayne Joshua	P.	Valenzuela City Science HS
Ramos	Jek Joel		St. Paul University QC
Reyes	Trisha Anne		St. Paul College, Pasig
Reyes	Irish Claire	R.	Valenzuela City Science HS
Rojo	Abigail Anne	O.	St. Paul College of Parañaque
Santos	John Dave	L.	Valenzuela City Science HS
Suyat	Neil Joshua	B.	St. Paul College of Parañaque
Tagle	Jazzrine	A.	Valenzuela City Science HS
Tan	Hans Markson	C.	St. Stephen's High School
Wong	Kent Louie	Y.	St. Stephen's High School
Yao	Kim Lauren	W.	Saint Pedro Poveda College
Yu	Jade Kathleen	F.	Valenzuela City Science HS

TEACHERS
Schedule: Sat, 9:00-12:00NN
Room: SEC A 124

Last Name	First Name		School
Angeles	Marvin John		Jubilee Christian Academy
Aniban	Diana Grace	B.	Manila Science High School
Bagcal	Maria Angelica	A.	Jubilee Christian Academy
Capada	Precilla		St. Paul College, Pasig
Cosme	Dariel Dene	G.	De La Salle Zobel School
Delizo	Loreto	D.	Saint Pedro Poveda College
Herezo	Frances Ida	S.	Saint Pedro Poveda College
Manzano	Orlando	M.	De La Salle Zobel School
Maranan	Donna Cristy	S.	St. Paul College, Pasig
Mendoza	Paula Mae		ICA Greenhils
Olos	Joanne		Saint Pedro Poveda College
Ortilla	Francis Thomas		St. Theresa's College, QC
Policarpio	Mark Julius		St. Paul College, Pasig
Ramirez	Julie Anne	O.	Manila Science High School
Rosario	Jay-Ar	S.	UNO HS
Trasmonte	Mary Cris	M.	Jubilee Christian Academy
Verdan	Kenneth	F.	Jubilee Christian Academy
Villalon	Louise Joy		ICA Greenhils