There are only two pentuplets given by :
(M,X,R(X),P,Q) = (10,684,486,6,9) and (20,342,243,6,9), which satisfies all the conditions of the problem.
EXPLANATION:
From conditions of the problem, keeping in mind that the conference commenced between P o'clock and P+1 o'clock and the conference concluded between Q o'clock and Q+1 o'clock, we obtain:
(A- B/12) = 5Q and (B - A/12) = 5P where A=M*X/143 and
B=M*R(X)/143
or, 12*M*X -M*R(X) = 143 *( 60Q) AND 12*M*R(X)-M*X = 143 *( 60P).
Solving the above system of simultaneous equations ,we obtain:
X = 60(12Q+P)/M and R(X)= 60(12P+Q)/ M........(*)
Since, 11 >=Q > P >=1 with the proviso that P and Q are whole number , substituting (P,Q) = (1,2), (1,3),……..,(10,11) in turn, we obtain a total of 55 distinct quadruplets corresponding to (P,Q,X,R(X)) which are furnished below:
(P,Q,M*X,M*R(X))=(1,2,1500,840),(1,3,2220,900),(1,4,2940,960),(1,5,3660,1020),(1,6,4380,1080),(1,7,5100,1140),(1,8,5820,1200),(1,9,6540,1260),(1,10,7260,1320),(1,11,7980,1380),(2,3,2280,1620),(2,4,3000,1680),(2,5,3720,1740),(2,6,4440,1800),(2,7,5160,1860),(2,8,5880,1920),(2,9,6600,1980),(2,10,7320,2040),(2,11,8040,2100),(3,4,3060,2400),(3,5,3780,2460),(3,6,4500,2520),(3,7,5220,2580),(3,8,5940,2640),(3,9,6660,2700),(3,10,7380,2760),(3,11,8100,2820),(4,5,3840,3180),(4,6,4560,3840),(4,7,5280,3300),(4,8,6000,3360),(4,9,6720,3420),(4,10,7440,3480),(4,11,8160,3540),(5,6,4620,3960),(5,7,5340,4020),(5,8,6060,4080),(5,9,6780,4140),(5,10,7500,4200),(5,11,8220,4260),(6,7,5400,4740),(6,8,6120,4800),(6,9,6840,4860),(6,10,7560,4920),(6,11,8280,4980),(7,8,6180,5520),(7,9,6900,5580),(7,10,7620,5640),(7,11,8340,5700),(8,9,6960,6300),(8,10,7680,6360),(8,11,8400,6420),(9,10,7740,7080),(9,11,8460,7140),(10,11,8520,7860)
-----------------(##)
Now, for any pair of the form (ks, kt) where P=ks and Q=kt, with s and t being relatively prime to each other and k being a positive whole number, we observe that; since P is less than Q, so s is less than t; and consequently, the minimum possible value of (s,t) is (1,2). Consequently, the maximum value of k is 5, whenever(P,Q)=(ks,kt) = (5,10)
Accordingly, by (*), for any given quadruplet (ks,kt,X,R(X)); the g.c.d of M*X and M*R(X) is 60*k, where k = 1,2,3,4,5.
Hence, M must be equal to a divisor of 60*k (including M).
When, g.c.d.(M*X, M*R(X))=60,corresponding to k=1, then:
M=1,2,3,4,5,6,10,12,15,20,30,60.---------(#)
When, g.c.d.(M*X, M*R(X))=120,corresponding to k=2, then:
M= values given in (#) and 8,24,40 ,120.
When, g.c.d.(M*X, M*R(X))=180, corresponding to k=3, then:
M=values given in (#) and 9, 18, 36, 45, 90, 180.
When, g.c.d.(M*X, M*R(X))=240, corresponding to k=4, then:
M=values given in (#) and 8,16,24,40,48,80,120,240.
When, g.c.d.(M*X, M*R(X))=300,corresponding to k=5, then:
M=values given in (#) and 25, 50, 75, 100, 150, 300.
In the light of the foregoing, checking individually for each of the 55 quadruplets in (##), we observe that only the quadruplet
(P,Q,M*X,M*R(X))=(6,9,6840,4860) generates two distinct pentuplets given by:
(M,X,R(X),P,Q) = (10,684,486,6,9) and (20,342,243,6,9)
None of the other 54 quadruplets in (##) generate any pentuplet corresponding to (M,X,R(X),P,Q) which satisfies all the conditions of the problem.
Consequently, there are only two pentuplets given by (M,X,R(X),P,Q) = (10,684,486,6,9) and (20,342,243,6,9), which satisfies all the conditions of the problem.
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