or

.....

.. whole number next less than rn+ +

&c.

rn+a

pm-1

m

rn p
p"-p-1p-

a
m-19

rn rn
Р p

that is, the whole number next less than

p-1'p"-p"-(p-1)

mth operation.

=number of cards before A in the pack, after the

Hence, if be the whole number next greater than rnp

dent, when is not an integer, that the card A will be restricted to

cannot make a difference of 1 in the value of

cannot make a difference of 1 in the above ex

Hence at the mth and every succeeding operation the card A is restricted to the qth place from the top.

If

rnp

p-1

be an integer, that is, rn be divisible by p-1, then it is evident that the number of cards before A may be either q-1 or q-2; and therefore A must be restricted to the qth or q-1th place.

SINGLE AND DOUBLE POSITION.

MANY problems are readily solved by means of the Arithmetical Rules called "Single Position", and "Double Position", without the application of Algebra; but the Rules themselves require algebraic proof.

(1) The Rule of 'SINGLE POSITION' is applied to those cases only, in which the required quantity is some multiple, part, or parts, of some other given quantity; that is, if a represent the required value, a and b known quantities, the cases for 'Single Position' are such as produce an equation of the form

ax=b.

Thus, if it be required to find such a value x that axx=b, suppose s to be the value, and instead of b, we find axs-b', then we have

which points out the Rule:-namely, Suppose some value (s) to be the one sought, and having operated upon it as the question directs, let the result (b) be noted. Then the true value is equal to the true result (b) divided by the false one (b') and multiplied by the supposed value (s).

Ex. Find the number which being added to the half and fourth of itself will produce 14.

(2) The rule for DOUBLE POSITION' is applied to those cases in which the required quantity is not a multiple, part, or parts, of a given quantity, but furnishes an equation of the form

ax+b=cx+d.

By transposition this equation becomes

(a-c)x+b-d=0.........(1).

Now suppose s to be the value of x, which by substitution does not satisfy the equation, but gives

(a-c)s+b-d=e......

then subtracting (1) from (2),

(a−c)(s-x)=e.

(2),

Again, supposes to be the value of x, which by substitution and subtraction, as before, gives

which proves the common Rule, namely, Make two suppositions (s and s') for the required quantity; treat each of them in the manner pointed out by the question; and note the errors (e and e'); then the required quantity will be found by dividing the difference of the products es', e's by the difference of the crrors e, e'.

Ex. What number is that which, upon being increased by 10, becomes three times as great as it was before?

1st. Suppose the number to be 20, (s)

Ex. 1. IN a pyramidal pile of shot of which the base is a square, the number in one side of the base is given, find the whole number n the pile.

Let n be the given number in one side of the base, then n2 is the number in the base;

n-1 is the number in a side of the next superincumbent square,
(n−1)2
..the whole square.

Similarly (n-2) is the number in the next square; and so on, until the squares are diminished to a single shot.

Hence the whole number in the pile is equal to the sum of the series n2+ (n−1)3+ (n − 2)3+&c....+13,

or 12+22+32+...+n', which by Art. 295, Ex. 4, is equal to

Ex. 2. In a pyramidal pile of shot of which the base is an equilateral triangle, the number in a side of the base is given, find the whole number in the pile.

Let n be the given number in a side of the base,

then n-1 is the number in a side of the next superincumbent triangle,

and so on, until we get to a single ball; that is, there are n triangles, which, beginning at the other end of the series, may be represented thus,

&c.

so that the whole number of shot in the pile, being the sum of the shot in these » triangles, will be the sum of the series

1+3+6+10+15+21+ &c. to n terms.

1st. Let n be even; then the series, taking pairs of term

n+1
2

or 1′+32+52+72+&c. to terms; which is equal to

Ex. 3. To find the number of shot in a pile of which the base is a rectangular parallelogram, whose sides contain a given number.

Let and represent the given number of balls in the length and breadth respectively of the base;

The next layer will be a paralleloand .. (b−1)(l-1) is the number number is (b-2)(1-2): and so on,

then bl=number of balls in the base. gram whose sides are b-1, and l-1; of balls in this layer. In the next the until we come to (b-b-1)(l-b-1), or 1x(l-b+1); the last parallelogram being reduced to a straight line of l-b+1 balls.

Hence the whole number of balls in the pile

=bl+(b−1)(l−1)+(b−2) (l−2)+......to b terms,
=bl+bl+bl+&c. to b terms-(1+2+3+&c.+b−1)(b+1)

OBS. To find the number of balls in an incomplete pile it is only necessary to find the number in the pile which is wanting to complete the given one, and subtract that number from the number in the given pile supposed complete.