3, will leave 2 remainder; but if divided by 4, will leave 3 remainder ? Ans. 11.

4. Required the least whole number, which being divided by 6, 5, and 4, will leave 5, 2, and 1, for the respective remainders? Ans. 17.

5. To find the least whole number, which being divided by 3, 5, 7, and 2, there shall remain 2, 4, 6, and 0, respectively. Ans. 104.

6. Required the least whole number, which being divided by 16, 17, 18, 19, and 20, will leave the remainders 6, 7, 8, 9, and 10, respectively?

6. Any equation involving two different powers only of the unknown quantity, may be reduced by substitution to the form of an indeterminate equation, involving two variable quantities. Hence, all commensurate quadratic equations, commensurate cubics wanting one term, commensurate biquadratics wanting two terms, &c. may be resolved by this method. It will be proper for the convenience of reference, to premise the following table of roots and powers

Roots 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Squares 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. Cubes 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728.

EXAMPLES.-1. Let x2+4x=32 be given, to find x.

First, by transposition and division,

32-x2 4

х

= Secondly,

1

it is plain, that whatever equimultiples of 4 and 1 be taken, the fractions whose terms are constituted of these equimultiples re

12 16 20

= =

and to one another, that is, =

&c. Wherefore, thirdly, if the quantity

4 8 1

32-x2

3 4 5 be made equal to either of these fractions, which (after transposing the known quantity 32) will give the resulting numerator equal to the square of the denominator, that denominator will be the value

See on this subject, Dodson's Mathematical Repository, Vol. I. Emerson's Algebra, Simpson's Algebra and Select Exercises, Vilant's Elements of Mathematical Analysis, &c.

20 ==

5

32-x2 4 8

=

12 16

16

be taken, we

4

&c. here it is plain, that if the fraction

shall have 32-x2=16, or x2=(32—16=) 16, whence x=4.

6

2. Given x2+6x=40, to find x.

By transposition and division, as before, we have

12 18 24

2

=; and if the latter fraction be taken, we shall

3 4

v2=64, or v=8,

Whence { 24-v=16, or v=8; whence y=2, as before.

8. Given y3-7y=36, to find y. y3-36 7 14 21 28

Here

y

Whence {

- = =

=

1 2 3

y3 (36+28) 64,

and y=4, the answer.

7. INDETERMINATE PROBLEMS.

1. How must tea, at 7 shillings per pound, be mixed with tea at 4 shillings per pound, so that the mixture may be worth 6 shillings per pound?

Let x=the number of pounds at 7 shillings, then 7x=their value; y=the number at 4 shillings, then 4y=their value.

Whence by the problem 7x+4y=(6.x+y=)6x+6y, or x= 2y, or 1xx=2xy x:y:21. there must be twice as much

in the mixture at 7 shillings, as there is at 4 shillings.

*These problems are of the kind which belong to the rule of Alligation.

2. Twenty poor persons received among them 20 pence; the men had 4d. each, the women d. each, and the children d. each; what number of men, women, and children, were relieved?

Let x=the number of men, y=the number of women, z=the number of children; then by the problem x+y+z=20, and (4x+ ży++z=20, or) 16x+2y+z=80: subtract the first equation

y

from this, and 15x+y=60, or y=(60-15x=)4-x.15, or 4-x

=

15 30 45

1 2

3

&c. but by the problem y < 20 ·:: y=15; and since

4—x=1 ·.· x=3, and z=(20—x—y=)20—18==2.

3. How many ways can 1007. be paid in guineas and crownpieces ?

Let x=the number of guineas, y=the number of crowns.
Then by the problem 21x+5y=(100×20=)2000.

21

21

20+y

21

and x=(

-=W=p, y=21p-20; let p=1, then y=1 crown,

2000-5 y

21

=) 95 guineas: and if (21) the coefficient of

x, be continually added to the value of y, and (5) the coefficient of y, continually subtracted from that of x, the corresponding values of x and y will be as follows, viz.

x=95, 90, 85, 80, 75, 70, 65, 60, 55, 50, 45, 40, 35, 30, 25, 20, 15, 10, 5, 0.

y=1, 22, 43, 64, 85, 106, 127, 148, 169, 190, 211, 232, 253, 274, 295, 316, 337, 358, 379, 400. Ans. 19 ways.

4. To divide the number 19 into three parts, such that seven times the first part, four times the second, and twice the third, being added together, the sum will be 90.

Let the parts be x, y, and z; then by the problem x+y+z= 19, and 7x+4y+2z=90; from the first x=19-y-z, this value being substituted for x in the second, it becomes (133-7 y-7z+ 4y+2x=)133-3y-5 z=90; or (3 y=43-5 z, or) y=

43-5% 3

1±2=W=p · : z=3p−1; if p be taken=1, then z=2, y=

3

=-52

3

=)11, and x=(19-y-z=) 6; if p=2, then will z=5, =6, and x=8; if p=3, then z=8, y=1, and x=10: these are the possible values in whole numbers.

5. How many ways is it possible to pay 100l. in guineas at
shillings each, and pistoles at 17 shillings each? Ans. 6.
6. If 27 times A.'s age be added to 16 times B.'s, the sum will
1600; what is the age of each? Ans. A.'s 48, B.'s 19.
7. A Higler's boy, sent on a market day

With eggs, fell down and smash'd them by the way;
The news reach'd home, and Master, in a rage,
Vow'd him a whipping, bridewell, or the cage:
""Tis through your negligence the eggs are lost,
"So pay me if you please the sum they cost."
The boy, since nought avail his tears and prayers,
Fetches his leathern bag of cash down stairs;
The cash a year's hard earnings had put in,
But much he wish'd to sleep in a whole skin.
"How many were there, Master?" In a doubt,
The Higler calls his wife to help him out;

Says she, "I counted them by twos, threes, fours,

Fives, sixes, sev'ns, before he left these doors;

"And one, two, three, four, five, and nought, remain'd

8. Is it possible to pay 100l. with guineas and moidores only? Ans. It is impossible.

9. A, who owes B a shilling, has nothing but guineas about him, and B has nothing but louis d'ors at 17 shillings each; how, under these circumstances, is the shilling to be paid? Ans. A must give B 13 guineas, and receive 16 louis d'ors change.

10. With guineas and moidores the fewest, which way
Three hundred and fifty-one pounds can I pay?