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rocdocmac

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Everything posted by rocdocmac

  1. YAY! second number: the last digit of powers of 3 runs through the cycle (3971), with exponent divisible by 4 ending in 1. 556677 is 1mod4 so it ends in 3.
  2. Earlier today (whilst attending his birthday party) I asked Andy his age, but he said that I can figure that out easily by myself! All I had to do was to add four numbers together. The first of these four numbers is the last digit in the figure obtained when 2 is raised to the power of 667788 (2667788). Similarly, I had to find the last digit in the evaluation of 3556677, 7445566, and 8334455 to complete the sum. If today is the 12th of March 2020, when was Andy born?
  3. Ah-ha! The first toss is irrelevant. Thus the second player has the best chance ...
  4. Having a stab at this to get the ball rolling ... Jane Cramer Janice Carter Jack Clark Jasper Clayton Jim Carver
  5. Also thought so, but no! It's in fact 13! Sticking to the method used to generate the next number, of course. But this is where the sequence stops. One can't use the resulting 13-digit number to determine the next one.
  6. The five polyominoes chosen with just a single example ...
  7. I wish to point out a trivial mistake in my OP ... The value at position C10 (1223) and the one at J11 (1226) have inadvertently not been swapped before posting, So both these values are in their correct placings. This means that only 14 swaps need to be done. No extra cycles, all straightforward X ↔ Y (one exchange per number with another).
  8. In the ten numerical sequences below, three numbers from each sequence have been moved to another sequence, i.e. 15 numbers have been swapped. Which 15 pairs of numbers should be exchanged to make each of the sequences follow a regular pattern? (A) 37, 85, 148, 232, 333, 442, 589, 744, 917, 1108, 1317, 1544 (B) 42, 84, 144, 219, 310, 417, 537, 679, 834, 1005, 1208, 1395 (C) 44, 95, 166, 253, 368, 493, 650, 821, 1012, 1223, 1454, 1705 (D) 70, 122, 189, 271, 366, 475, 598, 735, 890, 1051, 1230, 1423 (E) 123, 181, 252, 335, 430, 540, 656, 787, 930, 1085, 1283, 1431 (F) 149, 209, 280, 361, 452, 553, 691, 785, 916, 1057, 1192, 1369 (G) 160, 190, 220, 253, 288, 325, 364, 406, 448, 499, 540, 589 (H) 253, 320, 395, 478, 569, 666, 775, 886, 1013, 1144, 1252, 1430 (I) 257, 300, 351, 405, 465, 528, 595, 668, 741, 820, 903, 990 (J) 306, 371, 452, 519, 602, 664, 786, 887, 994, 1107, 1226, 1351 Example In the two sequences below, 48 and 52 should be swapped to make sense (A5 ↔ B4). (A) 20, 25, 32, 41, 48, 65. 80 (B) 15, 24, 35, 52, 63, 80, 99
  9. Phil1882 is right ... there is a discrepancy. The 6th term may possibly be 532 901 * 532 901 + 1 = 283 982 410 001.
  10. You've got all of them correct! Kindly tell all Braindenners how you managed that. My explanation to follow after that (if necessary).
  11. @mtngoat ... @flamebirde @Brainden
  12. Flamebirde ... there's an additional score (7/10) by Jim56 to consider!
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