 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Conference Time: Generalised Case (Posted on 2006-01-19) 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 M*X/ 143 minutes past P o'clock ( but before P+1 o'clock) and concluded at M*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 and M is a positive integer greater than 1.

#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 the total number of distinct choices of the pentuplet (M,X,R(X),P,Q) satisfying conditions of the problem.

NOTE:

(i)Any two choices of the pentuplet are defined to be distinct if they differ in the magnitudes of at least one of the five parameters (viz. M,X.R(X),P and Q).

(ii)It may be noted that Q is always greater than P. For example, (P=2,Q=3) may correspond to valid values for P and Q, but (P=3,Q=2) is not feasible.

 See The Solution Submitted by K Sengupta Rating: 2.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re(2): Computer solution Comment 4 of 4 | (In reply to re: Computer solution by Bob Smith)

At the outset, since 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 /143.
or, 12*m*x �m*r = 143 *( 60q) AND 12*m*r-m*x = 143 *( 60p).

Solving the above system of simultaneous equations ,we obtain:
m*x = 60(12q+p) and m*r= 60(12p+q) .---------------(*)

so, (p,q)= (1,7) in (*) gives (mx,mr) = (5100,1140). Hence, m=63 gives x=80.95238095238 and r =18.09523809524 ; so x and r are not  whole numbers. This is a contradiction.

(p,q)= (2, 5) in (*) gives (mx,mr) = (3720,1740). Hence, m=5 gives x= 744 and r =348 . This is not feasible since, these values are not in conformity with the definition of x and r.

(p,q)= (3,9) in (*) gives (mx,mr) = (6660,2700). Hence, m=8 gives x=832.5  and r =337.5 ; so x and r  are not whole numbers. This is a contradiction.

(p,q)= (6,9) in (*) gives (mx,mr) = (6840,4860).

Hence, m=10 gives x=684  and r =486. Also, m =20 gives x = 342 and r =243. These values are in conformity with all the conditions of the problem.

Hence, (p,q,m,x,r) = (6, 9, 10, 684, 486) ; (6,9, 10, 342, 243) constitute the only possible choices of pentuplets which satisfy all the conditions of the problem.

Edited on April 20, 2006, 9:43 pm

Edited on April 20, 2006, 9:47 pm
 Posted by K Sengupta on 2006-04-20 21:40:23 Please log in:

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