Imágenes de página
PDF
ePub

x the third number. Thus the sum of the first and second
= 3x, and the sum of the three = 4x = 100.
Hence x = 25, and the sum of the first two = 75.
Let y be the first'.

[ocr errors]

Therefore sum of second and third

- 41, 5y = 100 and y = 20.

The required parts are 20, 55, 25.

21. To find three numbers such that the greatest exceeds the middle number by a given fraction of the least, the middle exceeds the least by a given fraction of the greatest, but the least exceeds a given fraction of the middle number by a given number.

Necessary condition. The middle number must exceed the least by such a fraction of the greatest that, if its denominator' be multiplied into the excess of the middle number over the least, the coefficient of x in the product is greater than the coefficient of in the expression for the middle number resulting from the assumptions made3.

Suppose greatest exceeds middle byof least, middle exceeds least by of greatest, and least exceeds

of middle by 10. [Diophantus assumes the three given fractions or submultiples to be one and the same.]

+10 the least. Therefore middle = 3x, and greatest

[blocks in formation]

1 As already remarked (p. 52), Diophantus does not use a second symbol for the second unknown, but makes åpμbs do duty for the second as well as for the first. 844 'Denominator," literally the "number homonymous with the fraction," i.e. the denominator on the assumption that the fraction is, or is expressed as, a submultiple.

3 Wertheim points out that this condition has reference, not to the general solution of the problem, but to the general applicability of the particular procedure which Diophantus adopts in his solution. Suppose X, Y, Z required such that X- Y=Z|m, Y-Z=X/n, Z-a Ylp. Diophantus assumes Z=x+a, whence Y=px, X=n (px-x-a). The condition states that np-n>p. If we solve for x by substituting the values of X, Y, Z in the first equation, we in fact obtain

or

m {(np-n-p) x − na} = x+a,
x (mnp - mn - mp − 1) = a (mn+1).

In order that the value of x may be positive, we must have mnp>mn+mp+1, that is,

.

[blocks in formation]

[Another solution1.

Necessary condition. The given fraction of the greatest must be such that, when it is added to the least, the coefficient of x in the sum is less than the coefficient of x in the expression for the middle number resulting from the assumptions made1.

Let the least number be x + 10, as before, and the given

fraction; the middle number is therefore 37.

Next, greatest = middle + (least)=3x+31.

[blocks in formation]

=2fx + 11f.

Therefore 12}, and

=

the numbers are, as before, 45, 37, 22.]

22. To find three numbers such that, if each give to the next following a given fraction of itself, in order, the results after each has given and taken may be equal.

Let first give of itself to second, second of itself to

third, third of itself to first.

[ocr errors]

Assume first to be a number of 's divisible by 3, say 3r, and second to be a number of units divisible by 4, say 4.

Therefore second after giving and taking becomes x + 3. Hence the first also after giving and taking must become x+3; it must therefore have taken x+3−2x, or 3-x; 3-x must therefore be of third, or third = I5 - 5.

[blocks in formation]

23. To find four numbers such that, if each give to the next following a given fraction of itself, the results may all be equal. itself to second, second of itself

of

Let first give
to third, third
itself to first.

of itself to fourth, and fourth of

Assume first to be a number of x's divisible by 3, say 3x, and second to be a number of units divisible by 4, say 4.

1 Tannery attributes this alternative solution, like the others of the same kind, to an ancient scholiast.

• Wertheim observes that the scholiast's necessary condition comes to the same thing as Diophantus' own.

The second after giving and taking becomes + 3. Therefore first after giving to second and receiving of fourth-x+3; therefore fourth

=6(x+3-2x)= 18-6x.

But fourth after giving 3-x to first and receiving of
third +3; therefore third = 3ar-60.

Lastly, third after giving 6-12 to fourth and receiving
I from second = x + 3.

That is, 24-47=x+3, and x=.

The numbers are therefore 80, 4, 4, 4;

or, after multiplying by the common denominator, 150, 92, 120, 114.

24. To find three numbers such that, if each receives a given fraction of the sum of the other two, the results are all equal.

Let first receive of (second + third), second of (third + first), and third of (first+second).

Assume first=x, and for convenience' sake (Tоû πроxeíρov ěvekev) take for sum of second and third a number of units divisible by 3, say 3.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

The numbers, after multiplying by the common

denominator, are 13, 17, 19.

25. To find four numbers such that, if each receives a given fraction of the sum of the remaining three, the four results are equal.

Let first receive of the rest, second of the rest, third of rest, and fourth of rest.

[ocr errors]

Assume first to be x and sum of rest a number of units

divisible by 3, say 3.

Then

Now

sum of all = x+3.

first+(second + third + fourth)=x+1.

Therefore second + (third + fourth + first)=x+1,
3 times second + sum of all – 4x + 4,

whence

and therefore

Similarly
and

Adding, we have

and

second=x+}.
third =x+},
fourth-x+8.

4x+18=x+3,
x = ff.

The numbers, after multiplying by a common denominator, are 47, 77, 92, 101.

26. Given two numbers, to find a third number which, when multiplied into the given numbers respectively, makes one product a square and the other the side of that square.

Given numbers 200, 5; required number x.
Therefore 200r=(5x)3, and

x=8.

[ocr errors]

27. To find two numbers such that their sum and product are given numbers.

Necessary condition. The square of half the sum must exceed the product by a square number. ἔστι δὲ τοῦτο πλασματικόν. Given sum 20, given product 96.

2x the difference of the required numbers.
Therefore the numbers are 10+x, 10-x.
Hence 100-r2 = 96.

Therefore x = 2, and

the required numbers are 12, 8.

28. To find two numbers such that their sum and the sum of their squares are given numbers.

Necessary condition. Double the sum of their squares must exceed the square of their sum by a square. ἔστι δὲ καὶ τοῦτο πλασματικόν.

[ocr errors]

1 There has been controversy as to the meaning of this difficult phrase. Xylander, Bachet, Cossali, Schulz, Nesselmann, all discuss it. Xylander translated it by "effictum aliunde." Bachet of course rejects this, and, while leaving the word untranslated, maintains that it has an active rather than a passive signification; it is, he says, not something "made up" (effictum) but something "a quo aliud quippiam effingi et plasmari potest," from which something else can be made up," and this he interprets as meaning that from the conditions to which the term is applied, combined with the solutions of the respective problems in which it occurs, the rules for solving mixed quadratics can be evolved. Of the two views I think Xylander's is nearer the mark. πλασματικόν should apparently mean of the nature of a πλάσμα,” just as δραματικόν means something connected with or suitable for a drama; and wλáoμa means something

Given sum 20, given sum of squares 208.
Difference 2x.

Therefore the numbers are 10+x, 10-x.

Thus 200+ 2x2 = 208, and x = 2.

The required numbers are 12, 8.

29. To find two numbers such that their sum and the difference of their squares are given numbers.

Given sum 20, given difference of squares 80.

[blocks in formation]

30. To find two numbers such that their difference and product are given numbers.

Necessary condition. Four times the product together with the square of the difference must give a square. ἔστι δὲ καὶ τοῦτο πλασματικόν.

Given difference 4, given product 96.

2x the sum of the required numbers.

Therefore the numbers are x+2, x−2; accordingly x2-4=96, and x = 10.

The required numbers are 12, 8.

31. To find two numbers in a given ratio and such that the sum of their squares also has to their sum a given ratio.

Given ratios 3:1 and 5: I respectively.

Lesser number x.

Therefore 10r2 = 5.4x, whence x = 2, and

the numbers are 2, 6.

32. To find two numbers in a given ratio and such that the sum of their squares also has to their difference a given ratio.

Given ratios 3:1 and 10: 1.

Lesser number x, which is then found from the equation
Iar2 = 10. 2x.

Hence x=2, and

the numbers are 2, 6.

"formed " or "moulded." Hence the expression would seem to mean "this is of the nature of a formula," with the implication that the formula is not difficult to make up or discover. Nesselmann, like Xylander, gives it much this meaning, translating it "das lässt sich aber bewerkstelligen." Tannery translates wλaoμarików by “formativum.”

« AnteriorContinuar »