Hence we have the following equations:

z=4, 3x+2y=8;

z=3, 3x+2y= 16;

z=2, 3x+2y=24;

z=1, 3x+2y=32.

And it will be found that all the solutions required are 2, 1, 4 4, 2, 3; 2, 5, 3; 6, 3, 2; 4, 6, 2; 2, 9, 2; 10, 1, 1; 8, 4, 1; 6,7,1; 4, 10, 1; and 2, 13, 1.

Ex. 2. Find the positive integral solutions of the equation

6x2 - 13xy+6y2= 16.

;

We have (3x-2y) (2x-3y) = 16; hence, as x and y are integers, 3x-2y must be an integer, and must therefore be a factor of 16. Thus one or other of the following simultaneous equations must be held good

2x-3y=8...... .(iv);

3x-2y= ±1, 2x-3y=16.. ..(v).

Whence we find that 5x must be ±(48-2), ±(24 −4), ±(12-8), +(6-16) or

(3−32).

Hence the only integral values of x are 4 and 2, the corresponding values of y being 2 and 4.

1710

must be an integer, and therefore 7x - 5 must be a

Hence 7x-5 factor of 1710. Hence, if x is a positive integer, the only possible values of 7x-5 (since 7x-5 cannot be a multiple of 5) are 342, 171, 114, 57, 38, 19, 18, 9, 6, 3, 2, and 1. It will be found that, of these values of 7x-5, only the values 114, 9 and 2 give integral values of x, the values being 17, 2 and 1 respectively. If x=17, y is negative; if x=2, y=3; and if x=1, y=17. Thus the only positive integral solutions are x = =2, y=3 and x=1, y=17.

EXAMPLES XL.

Find all the positive integral solutions of the equations:

(1) 7x+15y=59.

(3) 7x+9y=100.

(2) 8x+13y = 138.

(4) 15x+71y= 10653.

2. Find the number of positive integral solutions of

2x + 3y = 133 and of 7x+11y = 2312.

3. Find the general integral solutions of the equations

(1) 7x-13y = 15.

(2) 9x-11y=4.

(3) 119x-105y = 217. (4) 49x-69y=100.

4. Find the positive integral solutions (excluding zero) of the equations

(i) 2xy - 3x + 2y = 1329.

(ii) x2 - xy + 2x − 3y = 11. (iii) 2x+5xy-12y = 28.

(iv) 2x2 - xy - y2 + 2x + 7y = 84.

6. Shew that integral values of x, y and z which satisfy the equation ax + by + cz = d, form three arithmetical progressions.

7. Divide 316 into two parts so that one part may be divisible by 13 and the other by 11.

8. In how many ways can £1. 6s. 6d. be paid with half-crowns and florins?

9. In how many ways can £100 be made up of guineas and crowns?

10. In how many ways can a man who has only 8 crown pieces pay 11 shillings to another who has only florins?

11. Find the greatest and least sums of money which can be paid in eight ways and no more with half-crowns and florins, both sorts of coins being used.

12. Find all the different sums of money which can be paid in three ways and no more with four-penny pieces and threepenny pieces, both sorts of coins being used.

13. Find all the numbers of two digits which are multiples of the product of their digits.

14. Two numbers each of two digits, and which end with the same digit, are such that when divided by 9 the quotient of each is the remainder of the other. Find all the sets of numbers which satisfy the conditions.

15. A man's age in 1887 was equal to the sum of the digits in the year of his birth: how old was he?

then the number of solutions in positive integers (including zero) of the equation a,x,+ax, + ... + ax = m, is A, a, a, ..., a being all integers.

2

n

The number of solutions of the equations x + 2y = n is }{2n + 3+ (-1)"},

At an entertainment the prices of admission were 1s., 2s. and £5, and the total receipts £1000; shew that there are 1005201 ways in which the audience might have been made up.

17. The money paid for admission to a concert was £300, the prices of admission being 5s., 3s. and 1s.; shew that the number of ways in which the audience may have been made up is 1201801.

CHAPTER XXX.

PROBABILITY.

399. THE following is generally given as the definition of probability or chance :

Definition. If an event can happen in a ways and fail in b ways, and all these ways are equally likely to occur, then the probability of its happening is

b

α

a+b

and the pro

bability of its failing is

To make the above definition complete it is necessary to explain what is meant by 'equally likely'. Events are said to be equally likely when we have no reason to expect any one rather than any other. For example, if we have to draw a ball from a bag which we know contains unknown numbers of black and white balls, and none of any other colour; we have just as much reason to expect a black ball as a white; the drawing of a black ball and of a white one are thus equally likely. Hence, as either a black ball or a white ball must be chosen, the probability of drawing either is, for there are two equally likely cases, in one of which the event happens and in the other it fails. Again, if we have to draw a ball from a bag which we know contains only black, white and red balls, but in unknown proportions, we have just as much reason to expect one colour as to expect either of the others, so that the drawing of a black, of a white and of a red ball

are all equally likely; and hence the probability of drawing any particular colour is, for there are three equally likely cases, and any particular colour is drawn in one case and is not drawn in the other two cases.

Another meaning may however be given to 'equally likely'; for events may be said to be equally likely when they occur equally often, in the long run. For example, if a coin be tossed up, we may know that in a very great number of trials, although the number of 'heads' is by no means necessarily the same as the number of 'tails', yet the ratio of these numbers becomes more and more nearly equal to unity as the number of trials is increased, and that the ratio of the number of heads to the number of tails will differ from unity by a very small fraction when the number of trials is very great; and this is what is meant by saying that heads and tails occur equally often in the long run.

Now, if each of the a ways in which an event can happen and each of the b ways in which it can fail occur equally often, in the long run, it follows that the event happens, in the long run, a times and fails b times out of every a+b cases. We may therefore say, consistently with the former definition, that the probability of an event is the ratio of the number of times in which the event occurs, in the long run, to the sum of the number of times in which events of that description occur and in which they fail to occur.

Thus, if it be known that, in the long run, out of every 41 children born, there are 21 boys and 20 girls, the probability of any particular birth being that of a boy is

21 41'

Again, if one of two players at any game win, in the long run, 5 games out of every 8, the probability of his winning any particular game is

5

8'

We may remark that, in the great majority of cases, including all the cases of practical utility such as the data used by Assurance Companies, the only way in which probability can be estimated is by the last method, namely, by finding the ratio of the actual number of times the event