perplexus.info :: Numbers : Deck of Cards Almanac

Deck of Cards Almanac (Posted on 2007-09-11)

If you assign the values 1 through 13 for each playing card denomination Ace through King, if you add up the values of all the cards you get 364. In one year, on Sunday, January 1, Susan noted this in the sense that the 364 points were the same as the 364 days remaining in the year that have not arrived yet.

Then, the following Sunday and every Sunday thereafter, she removed one card from the deck, and either counted or calculated the number of points remaining in the deck. As a result, on Sunday, December 31, after she discarded the last card, there were zero points left in the pack, and zero days left to the year.

In fact, every alternate Sunday, starting January 1, then January 15, etc., she made the point count remaining in the pack equal to the number of days remaining still to be seen in the year. But it's the other Sundays that make the problem interesting. On those Sundays, starting with January 8, she made sure that the number of points remaining in the deck after the removal was either a palindrome, a perfect square or a perfect cube (including allowing fitting into more than one of these categories, possibly at times even also matching the count of remaining days in the year).

Each time she discarded the first king of a given color, the very next week she discarded the other king of the same color.

On which four dates did she discard the sixes?

Submitted by Charlie

Rating: 3.0000 (1 votes)

Solution: (Hide)

A program finds 16 sequences of discards that work:
3 11 7 7 12 2 9 5 5 9 2 12 8 6 4 10 10 4 6 8 2 12 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
3 11 7 7 12 2 9 5 5 9 2 12 8 6 4 10 10 4 6 8 12 2 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
3 11 7 7 12 2 9 5 5 9 2 12 8 6 10 4 10 4 6 8 2 12 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
3 11 7 7 12 2 9 5 5 9 2 12 8 6 10 4 10 4 6 8 12 2 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
3 11 7 7 12 2 9 5 5 9 12 2 8 6 4 10 10 4 6 8 2 12 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
3 11 7 7 12 2 9 5 5 9 12 2 8 6 4 10 10 4 6 8 12 2 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
3 11 7 7 12 2 9 5 5 9 12 2 8 6 10 4 10 4 6 8 2 12 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
3 11 7 7 12 2 9 5 5 9 12 2 8 6 10 4 10 4 6 8 12 2 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
11 3 7 7 12 2 9 5 5 9 2 12 8 6 4 10 10 4 6 8 2 12 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
11 3 7 7 12 2 9 5 5 9 2 12 8 6 4 10 10 4 6 8 12 2 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
11 3 7 7 12 2 9 5 5 9 2 12 8 6 10 4 10 4 6 8 2 12 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
11 3 7 7 12 2 9 5 5 9 2 12 8 6 10 4 10 4 6 8 12 2 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
11 3 7 7 12 2 9 5 5 9 12 2 8 6 4 10 10 4 6 8 2 12 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
11 3 7 7 12 2 9 5 5 9 12 2 8 6 4 10 10 4 6 8 12 2 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
11 3 7 7 12 2 9 5 5 9 12 2 8 6 10 4 10 4 6 8 2 12 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
11 3 7 7 12 2 9 5 5 9 12 2 8 6 10 4 10 4 6 8 12 2 8 6 5 9 11 3 7 7 3 11 9 5 1 13 13 1 10 4 3 11 4 10 12 2 6 8 1 13 13 1
The differences occur in four isolated pairs, so there's one basic solution with two alternative swaps of adjacent week at four spots, leading to these 2^4=16 sequences. Adding dates and having separate columns for the alternatives gives:

Date discard remaining alternative count disc.remaining Jan 8 3 361 11 353 Jan 15 11 350 3 350 Jan 22 7 343 Jan 29 7 336 Feb 5 12 324 Feb 12 2 322 Feb 19 9 313 Feb 26 5 308 Mar 5 5 303 Mar 12 9 294 Mar 19 2 292 12 282 Mar 26 12 280 2 280 Apr 2 8 272 Apr 9 6 266 Apr 16 4 262 10 256 Apr 23 10 252 4 252 Apr 30 10 242 May 7 4 238 May 14 6 232 May 21 8 224 May 28 2 222 12 212 Jun 4 12 210 2 210 Jun 11 8 202 Jun 18 6 196 Jun 25 5 191 Jul 2 9 182 Jul 9 11 171 Jul 16 3 168 Jul 23 7 161 Jul 30 7 154 Aug 6 3 151 Aug 13 11 140 Aug 20 9 131 Aug 27 5 126 Sep 3 1 125 Sep 10 13 112 Sep 17 13 99 Sep 24 1 98 Oct 1 10 88 Oct 8 4 84 Oct 15 3 81 Oct 22 11 70 Oct 29 4 66 Nov 5 10 56 Nov 12 12 44 Nov 19 2 42 Nov 26 6 36 Dec 3 8 28 Dec 10 1 27 Dec 17 13 14 Dec 24 13 1 Dec 31 1 0
The remaining count columns have been split into two, for alternate weeks, for more easily seeing the ones that match the remaining days of the year from those that need to be palindromes, squares or cubes.
In all cases, the 6's were discarded April 9, May 14, June 18 and November 26.
The program was written to call recursively a subroutine that chose the next card that resulted in a palindrome/square/cube, and the following card, that had to be 14 minus the first. In the case of an ace, that had to be followed by two kings and another ace, in succession, so in all the recursion level got to 26 - 2 = 24. (Each pair of two kings do not add up to 14 themselves, and so one of each pair must be preceded by an ace and the other followed by an ace. The two such pairs of kings then use up all the aces.)
DECLARE SUB playCd (cNo!)
CLEAR , , 9999
DIM SHARED nLine(364), lMax

OPEN "deckcaln.txt" FOR OUTPUT AS #2

PRINT : PRINT

FOR i = 1 TO 364
  ok = 0
  n$ = LTRIM$(STR$(i))
  good = 1
  FOR j = 1 TO LEN(n$) / 2
    IF MID$(n$, j, 1) <> MID$(n$, LEN(n$) + 1 - j) THEN good = 0: EXIT FOR
  NEXT
  IF good THEN
    ok = 1
  ELSE
    tst = INT(SQR(i) + .5)
    IF tst * tst = i THEN ok = 1
    tst = INT(i ^ (1 / 3) + .5)
    IF tst * tst * tst = i THEN ok = 1
  END IF
  IF ok THEN
    PRINT i; : nLine(i) = 1
    IF i > 26 THEN
     IF nLine(i - 26) = 1 THEN PRINT "<==*** (\\"; i - 26; \\")\\";   END IF
NEXT
PRINT

DIM SHARED cct(13), played(52), ctRem
FOR i = 1 TO 13: cct(i) = 4: NEXT
ctRem = 364

playCd 1

CLOSE

DIM soln(1, 52)

OPEN "deckcaln.txt" FOR INPUT AS #1
DO
  LINE INPUT #1, l$
  FOR week = 1 TO 52
    soln(sNo, week) = VAL(MID$(l$, week * 3 - 1, 2))
  NEXT
  IF sNo = 0 THEN
   sNo = sNo + 1
  ELSE
   ctmm = 0
   FOR i = 1 TO 52
    IF soln(0, i) <> soln(1, i) THEN ctmm = ctmm + 1
   NEXT
   IF ctmm = 8 THEN EXIT DO
  END IF
LOOP UNTIL EOF(1)
CLOSE

OPEN "deckcal2.txt" FOR OUTPUT AS #2

DIM moLen(12), moName$(12)
DATA 31,28,31,30,31,30,31,31,30,31,30,31
FOR i = 1 TO 12: READ moLen(i): NEXT
DATA Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec
FOR i = 1 TO 12: READ moName$(i): NEXT

mo = 1: da = 8
remain1 = 364: remain2 = 364
FOR week = 1 TO 52
  PRINT moName$(mo); " ";
  PRINT #2, moName$(mo); " ";
  PRINT USING "## "; da;
  PRINT #2, USING "## "; da;
  PRINT USING " ##"; soln(0, week);
  PRINT #2, USING " ##"; soln(0, week);
  remain1 = remain1 - soln(0, week)
  remain2 = remain2 - soln(1, week)
  IF week MOD 2 = 1 THEN PRINT " ";
  IF week MOD 2 = 1 THEN PRINT #2, " ";
  PRINT USING " ###"; remain1;
  IF week MOD 2 = 0 THEN PRINT " ";
  PRINT #2, USING " ###"; remain1;
  IF week MOD 2 = 0 THEN PRINT #2, " ";
  IF soln(1, week) <> soln(0, week) THEN
    PRINT USING " ##"; soln(1, week);
    IF week MOD 2 = 1 THEN PRINT " ";
    PRINT USING " ###"; remain2;
    IF week MOD 2 = 0 THEN PRINT " ";
    PRINT #2, USING " ##"; soln(1, week);
    IF week MOD 2 = 1 THEN PRINT #2, " ";
    PRINT #2, USING " ###"; remain2;
    IF week MOD 2 = 0 THEN PRINT #2, " ";
  END IF
  PRINT
  PRINT #2,
  da = da + 7
  IF da > moLen(mo) THEN da = da - moLen(mo): mo = mo + 1
NEXT

CLOSE

SUB playCd (cNo)
  IF cNo > lMax THEN lMax = cNo
  IF cct(1) THEN
   IF nLine(ctRem - 1) AND nLine(ctRem - 27) THEN
     played(cNo) = 1
     played(cNo + 1) = 13
     played(cNo + 2) = 13
     played(cNo + 3) = 1
     cct(1) = cct(1) - 2
     cct(13) = cct(13) - 2
     ctRem = ctRem - 28

     IF cNo = 49 THEN
      FOR j = 1 TO 52
       PRINT USING " ##"; played(j);
       PRINT #2, USING " ##"; played(j);
      NEXT
      PRINT
      PRINT #2,
     ELSE
      playCd cNo + 4
     END IF

     cct(1) = cct(1) + 2
     cct(13) = cct(13) + 2
     ctRem = ctRem + 28
   END IF
  END IF
FOR i = 2 TO 12
  IF cct(i) THEN
   IF nLine(ctRem - i) THEN
     played(cNo) = i
     played(cNo + 1) = 14 - i
     cct(i) = cct(i) - 1
     cct(14 - i) = cct(14 - i) - 1
     ctRem = ctRem - 14

     IF cNo = 51 THEN
      FOR j = 1 TO 52
       PRINT USING " ##"; played(j);
       PRINT #2, USING " ##"; played(j);
      NEXT
      PRINT
      PRINT #2,
     ELSE
      playCd cNo + 2
     END IF

     cct(i) = cct(i) + 1
     cct(14 - i) = cct(14 - i) + 1
     ctRem = ctRem + 14
   END IF
  END IF
NEXT
END SUB

From Enigma No. 1455, Deck calendar, by Susan Denham, New Scientist, 11 August 2007.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Thoughts	K Sengupta	2023-04-16 22:23:00
re: Solution	Penny	2007-09-12 10:07:18
Solution	Penny	2007-09-11 20:44:44
Card Removal Order (Spoiler)	Guest	2007-09-11 20:14:48
Solution	Guest	2007-09-11 15:18:24

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