Page images


[ocr errors]

And, putting =p, we shall have x=17p+7.

Which value of x, being, substituted in the second frac.

17p+7-13 17p- 6 dion, gives

wh. 26


But it is obvious that is also =wk.

And consequently


26 26 26




p+18 Where, by rejecting p, there remains

= wh. = m

26 Therefore p=261–18; Whence, if r be taken =1, we shall have p=8. And consequently r=17p+7=17X8+7=143, the number sought.

2. It is required to find the least whole number, which, being divided by 11, 19, and 29, shall leave the remainders 3, 5, and 10, respectively.

Let x= the number required. 1-3 2-5 -10 Then


shole numbers. 11 19




And, putting


Ep, we shall have x=11p+3. Which value of x, being, substituted in the second frac

11p-2 tion, gires =wh.





Or 11p-2


And, by rejecting p, there will remain 3p-4mh.


1 Whis

18p-24 Also by mulen 3p-4




18p-- 5
Or, by rejecting the i, =wh.


[ocr errors]


is likewise =wh.

19 Whence 19p._ 1&p=5_p+5

= wh, which put =


Then we shall have p=191-5, and x=11(19r-5)+3=209r-52. And if this value be substituted for x in the third fraction, there will arise 2098-62


29 Or, by neglecting 7r-2, we shall have the remaining

6r - 4
part of the expression wh.

But, by multiplication,
674 30r-20 r20
X 5=

=rt =wh.


-20 Or, by rejecting , there witl remain = whis

29 which put =s.

Then ps298+-20; where, by taking s=0, we shall have r=20.

And consequently x=209r -52=209 X 20-52=4128% the number required,



[ocr errors]

3. To find a number, which, being divided by 6, shall leave the remainder 2, and when divided by 13, shall leave the remainder 3.

Ans. 68 4. It is required to find a number, which being divided by 7, sball leave 5 for a remainder, and if divided by 9, the remainder shall be 2.

Ans. 110 5. It is required to find the least whole number, which, being divided by 39, shall leave the remainder 16, and when divided by 56, the remainder shall be 27.

Ans. 1147 6. It is required to find the least whole number, which, being divided by 7, 8, and 9, respectively, shall leave the remainders 5, 7, and 8.

Ans. 1727 7. It is required to find the least whole number, which, being divided by each of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8, 9, shall leave no remainders.

Ans. 2520 8. A person receiving a box of oranges, observed, that, when he told them out by 2, 3, 4, 5, and 6 at a time, he had none remaining ; but when he told them out by 7 at a time, there remained 5; how many oranges were there in the box ?

Ans. 180



[ocr errors]

This branch of Algebra, which is so called from its inventor, Diophantus, a Greek mathematician of Alexandria in Egypt, who flourished in or about the third 'century after Christ, relates chiefly to the finding of square and cube numbers, or to the rendering certain compound expressions free from surds; the method of doing which is by making such substitutions for the unknown quantity, as will reduce the resulting equation to

a simple one, and then finding the value of that quantity in terms of the rest. (p)

These questions are so exceedingly curious and abstruse, that nothing less than the most refined Algebra, applied with the utmost skill and judgment, can surmount the difficulties which attend them. And, in this respect, no one has extended the limits of the analytic art further than Diophantus, or discovered greater knowledge and penetration in the application of it.

When we consider bis work with attention, we are

(0) That Diophantus was not the inventor of Algebra, as has been generally imagined, is obvious; since his method of apply. ing it is such, as could only have been used in a very advanced state of the science ; besides which, he no where speaks of the fundamental rules and principles, as an inventor certainly would have done, but treats of it as an art already sufficiently known; and seems to intend, not so much to teach it, as to cultivate and improve it, by solving such questions as, before his time, had been thought too difficult to be surmounted.

It is highly probable, therefore, that Algebra was known among the Greeks, long before the time of Diophantus ; but that the works of preceding writers have been destroyed by the ravages of time, or the depredations of war and barbarism.

His Arithmetical Questions, out of which these problems were mostly collected, consisted originally of thirteen books; but the first six only are now extant ; the best edition of which is that published at Paris, by Bachet, in the year 1670, with Notes by Fermat. In this work, the subject is so skilfully handled, that the moderns, notwithstanding their other improvements, have been able to do little more than explain and illustrate his method. Those who have succeeded best in this respect, are Vieta, Kersey, De Billy, Ozanam, Prestet, Saunderson, Fermat, and Euler; the last of whom, in particular, has amplified and illustrated the Diophantine Algebra in as clear and satisfactory a manner as the subject seems to admit of.

The reader will find a methodical abstract of the several me. thods made use by these writers, with a variety of examples to illustrate them, in the first and second volumes of my Treatise of Algebra, before quoted

at a loss which to admire most, his wonderful sagacity, and the peculiar artifices he employs, in forming such positions as the nature of the problems required or the more than ordinary subtility of bis reasoning upon them.

Every particular question puts us upon a new way of thinking, and furnishes a fresh vein of any analytical treasure, which cannot but prove highly useful to the mind, in conducting it through other difficulties of this kind, whenever they may occur, but, also, in enabliog it to encounter, more readily, those that may arise in subjects of a different nature.

The following method of resolving these questions will be found of considerable service ; but no general rule can be given, that will suit all cases ; and therefore the solution must often be left to the ingenuity and skill of the learner.

[ocr errors]



1. Put for the root of the square or cube required, one or more letters such that, when they are involved, either the given number, or the highest power of the unknown quantity, may vanish from the equation ; and then if the unknown quantity be only of one dimension, the problem will be solved by reducing the equation.

2. But if the unknown quantity ke still a square, or a higher power, some other new letters must be assumed to denote the root; with which proceed as before ; and so on, till the unknown quantity is but of one dimension ; when, from this, all the rest may be determined.


1. To divide a given square number (100) into two such parts, that each of them may be a square num

ber. 9)

(9) If x-10 had been made the side of the second square, in the following solution of this question, instead of 2-1%; tbe

« PreviousContinue »