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Conference Time (Posted on 2005-12-09) Difficulty: 5 of 5
Considering a positive whole number X which contains more than one digit, let us define R(X) as the number obtained by reversing the digits of X. Neither X nor R(X) can contain any leading zeroes.

* A Conference commenced on a given day precisely at 10*X/ 143 minutes past P o'clock ( but before P+1 o'clock) and concluded at 10*R(X) / 143 minutes past Q o'clock ( but before Q+1 o'clock) on the same day ; where 11 >=Q > P >=1 with the proviso that P and Q are whole numbers. It was observed that the hour hand and the minute hand had exchanged places during the respective times of commencement and conclusion of the said conference.

* Determine X, R(X), P and Q and,hence or otherwise, find the precise times of commencement and conclusion of the conference.

  Submitted by K Sengupta    
Rating: 4.0000 (1 votes)
Solution: (Hide)
* X = 684; R(X)=486; P=6 and Q=9

* The Conference commenced at (47 + 119/143) minutes past 6 o'clock and concluded at (33 + 141/143) minutes past 9 o'clock.

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=10*X/143 and B=10*R(X)/143
or, 120X -10*R(X) = 143 *( 60Q) AND 120*R(X)-10X = 143 *( 60P).
Solving the above system of simultaneous equations ,we obtain:
X = 6(12Q+P) and R(X)= 6(12P+Q)........(i)

Now, under the provisions corresponding to the problem under reference , we know that 11 >=Q > P >=1 with the proviso that P and Q are whole numbers. Accordingly, the magnitude of Q is always greater than that of P. For example, (P,Q)=(1,2) may correspond to a valid choice of (P,Q) but (P,Q)=(2,1) is untenable. Consequently, in conformity with the above and having regard to relationship (i), we obtain a total of 55 possible quadruplets (P,Q,X,R(X)) which are furnished hereunder.

(P,Q,X,R(X))=(1,2,150,84),(1,3,222,90),(1,4,294,96),(1,5,366,102),(1,6,438,108),(1,7,510,114),(1,8,582,120),(1,9,654,126),(1,10,726,132),(1,11,798,138),(2,3,228,162),(2,4,300,168),(2,5,372,174),(2,6,444,180),(2,7,516,186),(2,8,588,192),(2,9,660,198),(2,10,732,204),(2,11,804,210),(3,4,306,240),(3,5,378,246),(3,6,450,252),(3,7,522,258),(3,8,594,264),(3,9,666,270),(3,10,738,276),(3,11,810,282),(4,5,384,318),(4,6,456,384),(4,7,528,330),(4,8,600,336),(4,9,672,342),(4,10,744,348),(4,11,816,354),(5,6,462,396),(5,7,534,402),(5,8,606,408),(5,9,678,414),(5,10,750,420),(5,11,822,426),(6,7,540,474),(6,8,612,480),(6,9,684,486),(6,10,756,492),(6,11,828,498),(7,8,618,552),(7,9,690,558),(7,10,762,564),(7,11,834,570),(8,9,696,630),(8,10,768,636),(8,11,840,642),(9,10,774,708),(9,11,846,714),(10,11,852,786)

Of the 55 quadruplets, only (P,Q,X,R(X))= (6,9,684,486) is in conformity with the definition of X and R(X).
Now, 6840/143 = (47 + 119/143) and
4860/143 = (33 + 141/143)

Consequently, the conference commenced at (47 + 119/143) minutes past 6 o'clock and concluded at (33 + 141/143) minutes past 9 o'clock.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(3): computer solution-spoilerMindRod2005-12-09 23:29:25
re(2): computer solution-spoilerCharlie2005-12-09 21:54:58
re: computer solution-spoilerMindRod2005-12-09 21:23:42
Hints/TipsPro ClockwiseMindRod2005-12-09 17:52:39
Solutioncomputer solution-spoilerCharlie2005-12-09 13:00:24
Questionre: Conference TimeCharlie2005-12-09 09:30:15
re: Conference TimeKELVIN2005-12-09 09:29:11
Conference TimeKELVIN2005-12-09 09:14:31
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