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Posers and Puzzles

Posers and Puzzles

  1. Standard member Mathurine
    sorozatgyilkos
    24 Nov '06 15:09
    An absent-minded maths professor was telling his eccentric colleague about losing some of his paperwork.
    "I can't find my special number lists," he complained.
    "What was so special about them?" replied his colleague.
    "Well," he explained, "I call a list of numbers 'special' if it has the following 7 properties:"
    (1) All of the numbers on the list are positive integers, base 10.
    (2) There is more than one number on the list.
    (3) All of the numbers on the list have the same number of digits.
    (4) No two of the numbers on the list are equal.
    (5) The smallest number on the list is equal to the total number of digits of all the numbers on the list.
    (6) The largest number on the list is equal to the sum of all of the digits of all of the numbers on the list.
    (7) The number of items on the list is equal to the largest number divided by the smallest number.
    "There must be thousands of special lists," replied his colleague.
    "Hmm, I'm not sure about that, but I was only interested in four of them."
    "What four?"
    "What for? Well, I divide the sum of the numbers on each list by 100 and the four remainders comprise the four-number combination to my office safe. The safe is locked and I can't remember the combination without them!"
    "No, I meant which four!"
    "Well obviously I can't remember the lists, but I do recall their properties. First, all of the numbers on a list that lie strictly between the smallest and largest numbers I will call the middle-numbers. Now all of the middle-numbers on the first list are primes, and all of the middle-numbers on the second list are composite. Furthermore, the two largest middle-numbers on the third list do not appear on the first or second lists. Finally, the largest middle-number on the fourth list does not appear on any of the other three lists, and the smallest middle-number of the fourth list is not equal to the smallest middle-number of any of the other three lists."
    "Hmm, let me think about that. By the way, what do you have in your safe?"
    "I can't remember! But I usually lock up important papers that I am currently working on."
    The next day the eccentric professor approached the absent-minded professor while waving a handful of papers. "I found your lists!" he shouted.
    "Where in the world did you find them? I've looked everywhere!"
    "I found them in your safe!"

    Questions:
    (1) What is the combination to the safe?
    (2) How many possible "special" lists are there?
  2. 25 Nov '06 02:35
    Montilly is the drug capital of the world
  3. 29 Nov '06 08:19
    this is an unbelievable puzzle.....i'll give it a shot.
  4. 01 Dec '06 12:03 / 1 edit
    i tried and i had numbers and variables EVERYWHERE. i started to get somewhere, but got lost in my own thinking.....maybe some other time.
  5. Standard member TheMaster37
    Kupikupopo!
    01 Dec '06 15:18 / 1 edit
    (5) Would mean all numbers had that amount of digits, including the number itself. This is not possible if the number is larger than 1.

    The smallest number is therefore 1. And with that all numbers are single digit numbers.

    Once you have that, there is no possible combination of digits to make list 1.
  6. Subscriber BigDoggProblem
    The Advanced Mind
    01 Dec '06 21:09 / 1 edit
    Tough puzzle here, but I have solved question 1).

    Special list rule 6 (S6) implies the list numbers can't be 1-digit [Two one-digit numbers added together yields a sum greater than either of the two]. S3 and S7 imply that the numbers can't be 3 or more digits [The lowest candidate for small number is 102, which results in a large number of 102*34, which is larger than 3 digits.]

    The combination of rules S3 and S7 severly limit candidate pairs of small and large numbers. Only (10,50), (12,72) and (14,98) are worth considering. Odd small numbers aren't possible thanks to S3, while 16 * 8 = 104, which is too many digits.

    I made an excel spreadsheet of two-digit numbers from 10 to 98, and sorted it by sum of the digits. Using this list and rule S6, I eliminated (10,50) and (12,72) as potential lists. The highest possible sum of all digits with (10,50) is 43, which falls short of the 50 required by S6. The best you can do with (12,72) is 68, which falls 4 short.

    List 1 is 14, 59, 67, 79, 89, 97, 98. I used the numbers with the highest possible sum of digits (SOD), and this is the only way the digits sum to 98 using only prime middle numbers.

    List 2 is 14, 69, 78, 87, 88, 96, 98. Again, only the composite numbers with the highest possible SOD will achieve a tally of 98.

    List 3 is 14, 79, 88, 89, 94, 95, 98. I started with 94 and 95, the highest numbers not already used on the other two lists, then supplemented with high SOD numbers from the spreadsheet.

    List 4 is 14, 78, 79, 88, 89, 93, 98. I started with 93, since everything above that is ruled out. High-SOD numbers 89, 88, 79 were again a natural choice. Of the next-highest SOD numbers (69, 78, 87, 96), only 78 is still available [69 has been used as List 2's smallest number, 96 is higher than the largest middle number, 87 is larger than 79, which must remain to keep SOD high].

    Summing the lists, and taking the remainder of sum/100 yields:

    The combination to the safe is 3, 30, 57, 39.
  7. Subscriber BigDoggProblem
    The Advanced Mind
    02 Dec '06 01:03 / 5 edits
    Question 2) looks tedious enough to justify a bit of programming. The answer is 557 possible 'special' lists. Here is the C++ code:

    #include(stdio.h) //RHP doesn't allow "less than" symbol

    int sod(int num);

    void main(void) {
    const int small=14, large=98;
    int n, o, p, q, r;
    int lists=0;

    for(n=small+1; n[less than]large-4; n++) {
    printf("considering n = %d\n",n);
    for(o=n+1; o[less than]large-3; o++) {
    for(p=o+1; p[less than]large-2; p++) {
    for(q=p+1; q[less than]large-1; q++) {
    for(r=q+1; r[less than]large; r++) {
    if(sod(small)+sod(n)+sod(o)+sod(p)+sod(q)+sod(r)+sod(large)==98) {
    lists++;
    printf("Valid list found: %d %d %d %d %d %d %d\n",small, n, o, p, q, r, large);
    }
    }
    }
    }
    }
    n_loop:
    }

    printf("%d valid lists found\n", lists);
    }

    int sod(int num) {
    return(int(num/10) + num%10);
    }
  8. Standard member Mathurine
    sorozatgyilkos
    02 Jan '07 15:20
    Originally posted by BigDoggProblem
    Tough puzzle here, but I have solved question 1).

    Special list rule 6 (S6) implies the list numbers can't be 1-digit [Two one-digit numbers added together yields a sum greater than either of the two]. S3 and S7 imply that the numbers can't be 3 or more digits [The lowest candidate for small number is 102, which results in a large number of 102*34, ...[text shortened]... ing the remainder of sum/100 yields:

    [b]The combination to the safe is 3, 30, 57, 39.
    [/b]
    Congratulations, BDP...

    Nifty C++ coding, btw!

    Answers:
    (1) the safe combination is: 3,30,57,39.
    (2) there are 557 possible special lists.

    Complete Solution to (1):

    For any special list, let S = the smallest number, L = the largest number, N = the number of items on the list, and D = the number of digits in each number on the list (all equal by (3)). Also, for any number X, let d(X) = the number of digits in X.

    Then by (5) S = N*D, and by (7) L = N*S, so L = N*N*D. Now by (6) the most L can be is when all digits are 9, so L < 9*D*N (not = by (2) and (4)). So N*N*D < 9*D*N, so N < 9. Also by (2) and (4), S < L, so N*D < N*N*D, so 1 < N. So 1 < N < 9. Therefore D < N*D < 9*D. So S < 9*D. This means d(S) 4 (since S = 2*N has 2 digits) and N < 8 (since L = 2*N*N has 2 digits). So N is 5, 6, or 7.

    If N=5, S=10, L=50, and the list has 3 middle numbers. The list with the largest digital sum would be {10,39,48,49,50}, with a digital sum of 43, but L = 50, which contradicts (6), so N is not 5. If N=6, S=12, L=72, and the list has 4 middle numbers. The list with the largest digital sum would be {12,59,67,68,69,72} which has a digital sum of 68, but L = 72, which contradicts (6), so N is not 6. This means N can only be 7. So S=14, L=98, and the list has 5 middle numbers. (There are many such lists.) Since the digital sum of the list must be L which is 98, and the digital sum of S and L adds to 22, the digital sum of the 5 middle digits must be 98-22=76.

    Now the middle numbers range downward from 97 to 15. Calculate the digital sum of each and sort downward by digital sum, getting the following range (with digital sum in parentheses):
    89(17),97(16),88(16),79(16),96(15),87(15),78(15),69(15),95(14),86(14),77(14),68(14),59(14),94(13),etc. Note that the largest digital sum is 17, and there is one 17, three 16s, four 15s, etc.
    Now there are only 10 combinations of digital sums which add to 76:
    (1) 17,16,16,16,11
    (2) 17,16,16,15,12
    (3) 17,16,16,14,13
    (4) 17,16,15,15,13
    (5) 17,16,15,14,14
    (6) 17,15,15,15,14
    (7) 16,16,16,15,13
    (8) 16,16,16,14,14
    (9) 16,16,15,15,14
    (10) 16,15,15,15,15
    Note that the smallest possible digital sum is 11.
    Now list the possible middle numbers in decreasing order (with digital sum in parentheses):
    97(16),96(15),95(14),94(13),93(12),92(11),89(17),88(16),87(15),86(14),85(13),84(12),83(11),
    79(16),78(15),77(14),76(13),75(12),74(11),69(15),68(14),67(13),66(12),65(11),59(14),58(13),
    57(12),56(11),49(13),48(12),47(11),39(12),38(11),29(11)
    Note that there are only 8 primes on this list: 97(16),89(17),83(11),79(16),67(13),59(14),47(11),29(11)
    Furthermore, note that among the digital sums of the remaining composite numbers there is no 17, only one 16, and all four 15s.

    Now on the professor's first list all middle numbers are primes. There is only one combo that fits, combo (3), and there is only one possible list: {14,59,67,79,89,97,98}
    On the second list all middle numbers are composite, and amazingly there is only one combo that fits, combo (10), and only one possible list: {14,69,78,87,88,96,98}
    Now note that the first two lists contain all of the numbers with digital sums of 15,16,and 17.
    On the third list the two largest middle numbers do not appear on the first two lists, and so (by checking the combos) must have digital sums of 14,13 or 14,14. By checking the pattern of 15s, 16s, and 17 on the main list we discover that the 2 largest numbers must be 95(14) and 94(13), which fits only combo (3), and there is only one possible list: {14,79,88,89,94,95,98}
    Finally, on the fourth list the largest middle number does not appear on the first 3 lists, and again the patterns show it can only be 93(12), fitting only combo (2). Checking the patterns on the main list, while remebering that 93 must be the largest, yields only one value each for the 16s and 17, and these must be 89(17),88(16),79(16), with 3 possible values for the 15: 87,78,69. If the value of the 15 is 69, it is the smallest middle number on both the fourth and second lists, which is not allowed. If the 15 is 87, then 79 is the smallest middle number on both the fourth list and the third list, which is not allowed. So this leaves only 78 for the 15 value, and thus the list must be {14,78,79,88,89,93,98}

    So the sums of the numbers on the 4 lists are: 503,530,557,539, which makes the safe combination 3,30,57,39.