Some Miscellaneous Questions

Excerpts

• Some Miscellaneous Questions

  The first of these is to place eight queens on a chess-board so as to command the fewest possible squares.

• Some Miscellaneous Questions

  Let $a$ stand for $a_1$, or $a_2$, $b$ for $b_1$ or $b_2$, and so on.

• Some Miscellaneous Questions

  Thus there are seven possible triads, such as $abc$, $ade$, $afg$, $bdf$, $beg$, $cdg$, and $cef$.

• Some Miscellaneous Questions

  It has been asserted that, if $k = 2$, the number of ways in which two kings can be placed on a board so that they may not occupy adjacent squares is $\frac{1}{2}(n-1)(n-2) (n^2 + 3n-2)$.

• Some Miscellaneous Questions

  Now a cyclical permutation of $n$ letters is equivalent to $n-1$ simple interchanges; accordingly an odd cyclical permutation is equivalent to an even number of simple interchanges.

• Some Miscellaneous Questions

  Hence, if it requires $x$ transfers of simple discs to move a tower of $n-1$ discs, then it will require $2x +1$ separate transfers of single discs to move a tower of $n$ discs.

• Some Miscellaneous Questions

  Proceeding in this way we see that with a tower of $n$ discs it will require $2^n-1$ transfers of single discs to effect the complete transfer.

• Some Miscellaneous Questions

  Denote the rings which are on the bar by the digits $1$ or $0$ alternately, reckoning from left to right, and denote a ring which is off the bar by the digit assigned to that ring on the bar which is nearest to it on the left of it, or by a $0$ if there is no ring to the left of it.

• Some Miscellaneous Questions

  In the case of an ordinary chess-board the determinant is of the $8$th order, and therefore contains $8!$, that is, $40320$ terms, so that it would be out of the question to use this method for the usual chess-board of $64$ cells or for a board of larger size unless some way of picking out the required terms could be discovered.

• Some Miscellaneous Questions

  Would that English writers were in the habit of inventing equally interesting origins for the puzzles they produce!

• Some Miscellaneous Questions

  The problem is to dispose them so that for seven consecutive days no girl will walk with any of her school-fellows more than once.

• Some Miscellaneous Questions

  The suffixes $1$ and $2$ are called complementary.

• Some Miscellaneous Questions

  Then, if the suffixes are permuted cyclically, we obtain six other arrangements which satisfy the conditions of the problem:

• Some Miscellaneous Questions

  Is it possible to place the eight queens so as to leave more than eleven cells out of check? I have never succeeded in doing so, nor in showing that it is impossible to do it.

• Some Miscellaneous Questions

  Suppose for instance that a pack of $n$ cards is shuffled, as is not unusual, by placing the second card on the first, the third below these, the fourth above them, and so on.

• Some Miscellaneous Questions

  Let the number $n$ of the pack be divided into $p, q, r, \ldots$ such cycles, whose sum is $n$; then the l.c.m. of $p, q, r, \ldots$ is the utmost number of shufflings necessary before all the cards will be brought back to their original places.

• Some Miscellaneous Questions

  For example, if we are told that in figure i the card is in the third row, it must be either $9$, $10$, $11$, or $12$: hence, if we know in which row of figure ii it lies, it is determined.

• Some Miscellaneous Questions

  This depends on the fact that the number of homogeneous products of two dimensions which can be formed out of four things is $10$. Hence the homogeneous products of two dimensions formed out of four things can be used to define ten things.

• Some Miscellaneous Questions

  Request a spectator to note a card, and remember in which pile it is. After finishing the deal, ask in which pile the card is.

• Some Miscellaneous Questions

  The reason is that after the first deal you know it is one of sixty-four cards. In the next deal these sixty-four cards are distributed equally over the four piles, and therefore, if you know in which pile it is, you will know that it is one of sixteen cards.

• Some Miscellaneous Questions

  Hence, if $n-1$ is expressed in the ternary scale of notation, $x$, $y$, $z$ will be determined, and therefore $a$, $b$, $c$ will be known.

• Some Miscellaneous Questions

  If the $k$th card has the number $k$ on it---which event is called a *hit*---the player takes up the card and begins counting afresh.

• Some Miscellaneous Questions

  I believe that these arrangements by sentences are known, but I am not aware who invented them

Equations

• Some Miscellaneous Questions

  2^{64}-1

  The number of single-disc transfers needed to move the Tower of Hanoi with sixty-four discs is 2 to the 64th power minus 1.

• Some Miscellaneous Questions

  2^n-1

  A tower of n discs needs 2 to the power n minus 1 single transfers, so eight discs need 255.

• Some Miscellaneous Questions

  2x +1

  Moving a tower of n discs takes twice the x transfers needed for n-1 discs, plus one, by the recursive method.

• Some Miscellaneous Questions

  \frac{1}{3}(2^{n+1}-1)

  For an odd number n of Chinese rings, disconnecting them from the bar takes one third of (2 to the power n+1 minus 1) steps.

• Some Miscellaneous Questions

  \frac{1}{3}(2^{n+1}-2)

  For an even number n of Chinese rings, disconnecting them from the bar takes one third of (2 to the power n+1 minus 2) steps.

• Some Miscellaneous Questions

  \tfrac{1}{3}(2^{2n+2} - 1)

  The number of steps to take off the first 2n+1 Chinese rings is one third of (2 to the power 2n+2 minus 1).

• Some Miscellaneous Questions

  \tfrac{1}{3}(2^{2n+1}-2)

  The number of steps to take off the first 2n Chinese rings is one third of (2 to the power 2n+1 minus 2).

• Some Miscellaneous Questions

  2^{2n}

  If the first two rings are taken off or put on in one step, the count for an even number of rings becomes 2 to the power 2n.

• Some Miscellaneous Questions

  2^{2n-1}-1

  If the first two rings are taken off or put on in one step, the count for an odd number of rings becomes 2 to the power 2n-1 minus 1.

• Some Miscellaneous Questions

  4x

  Once the first m rings are off, the additional steps needed to take off the (m+3)th and (m+4)th rings equal 4x, where x is the step count for the previous pair.

• Some Miscellaneous Questions

  1 + 4 + 4^2 + \dotsb + 4^n

  The steps needed to take off the first 2n+1 rings are the sum of the powers of 4 from 4 to the power 0 up to 4 to the power n.

• Some Miscellaneous Questions

  (1+2^1 + 2^2 + \ldots + 2^{2n})

  Putting on a set of 2n+1 rings in the binary numbering requires the sum of the powers of 2 from 2 to the power 0 up to 2 to the power 2n steps.

• Some Miscellaneous Questions

  (2 + 2^3 + \ldots + 2^{2n-1})

  Putting on a set of 2n rings in the binary numbering requires the sum of the even powers of 2 starting at 2 up to 2 to the power 2n-1 steps.

• Some Miscellaneous Questions

  mn-3

  For a rectangular box with both m and n even, a right-angle rotation is equivalent to mn-3 simple interchanges, so it changes a solvable position into an unsolvable one and vice versa.

• Some Miscellaneous Questions

  n-1

  A cyclical permutation of n letters is equivalent to n-1 simple interchanges.

• Some Miscellaneous Questions

  y=(p^m-1)/(p-1)

  If p is a prime and x = p^m, a = p, b = 2, then the greatest number of ways y is (p^m - 1)/(p - 1) (Kirkman's theorem).

• Some Miscellaneous Questions

  \frac{1}{2}(n-1)(n-2) (n^2 + 3n-2)

  The number of ways to place two kings on an n-by-n board (n^2 cells) so that they do not occupy adjacent squares is said to be this expression (an assertion the book reports, not proves).

• Some Miscellaneous Questions

  \frac{1}{6}(n-1) (n-2) (n^4+ 3n^3-20n^2-30n + 132)

  The number of ways to place three kings on an n-by-n board so that no two occupy adjacent squares is said to be this expression (reported as an assertion, not proved).

• Some Miscellaneous Questions

  x_1=\frac{1}{2}(2p+x_0+1)

  One shuffle of a pack of 2p cards moves the card in place x_0 (x_0 odd) to place x_1.

• Some Miscellaneous Questions

  x_1=\frac{1}{2}(2p-x_0 + 2)

  One shuffle of a pack of 2p cards moves the card in place x_0 (x_0 even) to place x_1.

• Some Miscellaneous Questions

  2^{m+1}x_m=(4p+1)(2^{m-1} \pm 2^{m-2} \pm \dotsb \pm 2 \pm 1) \pm 2x_0 + 2^m \pm 1

  After m shuffles of a pack of 2p cards, the card starting in place x_0 ends in place x_m, where the signs are an ambiguity of sign.

• Some Miscellaneous Questions

  3 (2m+1)(3m+1)

  The number of fundamental Anstician arrangements is this expression in m.

• Some Miscellaneous Questions

  y=n(2n-1)

  For x = 2n girls walking in rows of 2 with every pair together once, the greatest number of ways y is n(2n-1).

• Some Miscellaneous Questions

  z=2n-1

  For x = 2n girls walking in rows of 2 with every pair together once, the number of days z is 2n-1.

• Some Miscellaneous Questions

  y=35

  For x = 15 girls, a = 3, b = 2, the greatest number of rows y is 35.

• Some Miscellaneous Questions

  y=\frac{3}{2}(x-1)/x

  For x = 5 x 3^m girls, a = 3, b = 2, the book gives y as this expression. FLAG: as printed it gives 1.4 at x = 15, where the text states y = 35 (x(x-1)/6 gives 35); possible erratum in the book, recorded not corrected.

• Some Miscellaneous Questions

  z=\frac{1}{2}(x-1)

  For x = 5 x 3^m girls, a = 3, b = 2, the number of days z is half of (x-1).

• Some Miscellaneous Questions

  y=x(x-1)/p(p+1)

  If x = (p^2+p+1)(p+1) with p^2+p+1 having no divisor less than p+1, a = p+1, b = 2, then y = x(x-1)/(p(p+1)).

• Some Miscellaneous Questions

  y=x

  If x = p^3+p+1, a = p+1, b = 2, then the greatest number of ways y equals x.

• Some Miscellaneous Questions

  y=455

  For x = 15, a = 3, b = 3 (Sylvester's result), the greatest number of ways y is 455.

• Some Miscellaneous Questions

  z=91

  For x = 15, a = 3, b = 3, the number of days z is 91.

• Some Miscellaneous Questions

  y=84

  For x = 9, a = 3, b = 3, the greatest number of ways y is 84.

• Some Miscellaneous Questions

  z=28

  For x = 9, a = 3, b = 3, the number of days z is 28.

• Some Miscellaneous Questions

  y=7

  For x = 7, a = 3, b = 2 (Bills), the greatest number of ways y is 7.

• Some Miscellaneous Questions

  y=155

  For x = 31, a = 3, b = 2 (Bills), the greatest number of ways y is 155.

• Some Miscellaneous Questions

  y=66

  For x = 11, a = 5, b = 4 (Lea), the greatest number of ways y is 66.

• Some Miscellaneous Questions

  y=140

  For x = 16, a = 4, b = 3 (Lea), the greatest number of ways y is 140.

• Some Miscellaneous Questions

  (455)^7

  The total number of ways the school of fifteen girls can walk out in triplets for a week is (455)^7, so a chance arrangement satisfying Kirkman's condition is very improbable.

• Some Miscellaneous Questions

  15567,552000

  Power showed there are this many different solutions of the fifteen school-girls problem. FLAG: the printed number has a comma inside the digits (15567,552000); read here as 15567552000, which may be a typesetting error to check.

• Some Miscellaneous Questions

  \frac{1}{2}n(n+1)

  A pack of n(n+1) cards divided into couples contains one half of n(n+1) couples, which equals the number of homogeneous products of two dimensions that can be formed out of n things.

• Some Miscellaneous Questions

  m^m

  The pack size for Gergonne's generalization is m to the power m, dealt into m piles of m^(m-1) cards each.

• Some Miscellaneous Questions

  n=km^{m-1}-jm^{m-2} + \dotsb + bm - a + 1

  When m is even, after m deals into m piles, the selected card is the n-th card from the top, where n is fixed by the piles a, b, ..., k taken up after each deal.

• Some Miscellaneous Questions

  n=km^{m-1}-jm^{m-2} + \dotsb - bm + a

  When m is odd, the selected card is the n-th card from the top of the collected pack, with the signs of the terms alternating from the pile positions a, b, ..., k.

• Some Miscellaneous Questions

  64d-16c + 4b-a + 1

  In a pack of 256 cards dealt four times into four piles of 64, the selected card is the (64d-16c+4b-a+1)-th card from the top, where a, b, c, d are the pile positions taken up after each deal.

• Some Miscellaneous Questions

  64-16c + 4b-a + 1

  After the fourth deal of a 256-card pack, the selected card is the (64-16c+4b-a+1)-th card in the pile indicated as containing it.

• Some Miscellaneous Questions

  n = 9c-3b + a

  In a 27-card pack dealt three times into three piles of nine, the selected card is the n-th card from the top when the piles taken up are a, b, c.

• Some Miscellaneous Questions

  9-3b + a

  After the third deal, the selected card is the (9-3b+a)-th card from the top of the pile indicated as containing it.

• Some Miscellaneous Questions

  9(c-1) + (8-3b + a) + 1

  The selected card lies in the place 9(c-1)+(8-3b+a)+1 from the top of the pack when the pile indicated after the third deal is taken up c-th.

• Some Miscellaneous Questions

  n = 9c-3b + a= 14

  With the pile indicated always taken up in the middle (a = b = c = 2), the selected card is the 14th card from the top, as in the usual presentation of the trick.

• Some Miscellaneous Questions

  3p = 9c-3b

  Dividing the adjusted position by three gives a multiple of three equal to 9c minus 3b, which is used to solve for the pile positions.

• Some Miscellaneous Questions

  p = 3c-b

  The integer p equals 3c minus b, so b is the smallest positive number that added to p gives a multiple of three, and c is that multiple.

• Some Miscellaneous Questions

  9z + 3y + x = n-1

  Gergonne's equation rewritten with x = a-1, y = 3-b, z = c-1, so that n-1 expressed in the ternary scale gives x, y, z.

Problems

No exercises in this chapter.