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feries 12 +30 +48 + &c. thence arifing, will be a true arithmetical progreffion; whereof the common difference being 18, and the number of terms =

157

2

61

20880: to

48, the fum will therefore be given which adding 13485, the number of anfwers when q was lefs then 62, the aggregate 34365 will be the whole number of all the answers required.

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To determine how many different ways it is poffible to pay 1000l. without using any other coin than crowns, guineas, and moidores.

By the conditions of the problem we have 5x+21y + 27% = 20000; where taking zo, x is found 2, and from thence the leaft value 5

=4000

·4y

of yo (o being to be included, here, by the question): whence the greatest value of x is given

4000: Moreover, from the equation 5m + 21n 27, we have

m = 54n·

12

5

2

; from which n = 2, and m=

3 fo that the general values of x and y, given in the preceding problem, will here become 4000-219 + 3%, and 59-2z. Moreover, from the given equation, the greatest limit of z appears to be =

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20000

27

mz 4000 + 3 x 740

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190, expreffing the leffer limit of q, when the value of x, answering to fome interpretations of z, will become negative, while thofe of y ftill continue affirmative. To find the number of all thefe affirmative values, up to the greatest limit of q, let 0, 1, 2, 3, 4, 5, &c. be now wrote in the room of q (as in the margin). Whence it is evident that the faid number is compofed of the

O 3

feries

feries + 3+ 6 + 8 +

012

I

y =

95

22

Quot. N.Anf.

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325

-2% 7
70

Ι

3

6

8

11 + 13, &c. continued to 297 terms; which terms (fetting afide the first)being united in pairs, we fhall have the arithmetical progreffion 9 +19 + 29 &c. where the number of terms to be taken being 148, and common difference 10, the laft term will therefore be 1479, and the sum of the whole progreffion 110112: to which adding (1) the term omitted, we have 110113, for the number of all the answers, including those wherein the value of x is negative; which last must therefore be found and deducted.

22

2 10

22

315

4 20 22 10

II

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In order to this we have already found, that these negative values do not begin to have place till 9 is greater than 190 let, therefore, 191, 192, 193, &c. be subftituted, fucceffively, for q;

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x. Quot. N. Anf. from whence it will appear

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that the number of all the faid negative values is truly exhibited by the arithmeti cal progreffion 4 + 11+ 18 + 25, &c. continued to 296-190 terms; whereof the fum is 39379; which fubtracted from 110113, found above, leaves 70734, for the number of answers required.

موع

After the manner of these two examples (which illuftrate the two different cafes of the general folution, given in the preceding problem) the number of answers may be found in other equations, wherein there are three indeterminate quantities. But, in fumming up the numbers arifing from the different interpretations of 9, due regard must be had to the fractions exhibited in the third column expreffing the limits of z; because, to have a regular progreffion, the terms of the feries in the fourth column, exhibiting the number of answers,

muft

must be united by twos, threes, or fours, &c. according as one and the fame fraction occurs every fecond, third, or fourth, &c. term (the odd terms, when there happen any over, being always to be fet afide, at the beginning of the feries). And it may be obferved farther, that, to determine the fum of the progreffion thus arifing, it will be fufficient to find the firft term only, by an actual addition; fince, not only the number of terms, but the common difference alfo, will be known; being always equal to the common difference of the limits of z (or of the quotients in the faid third column), multiplied by the fquare of the number of terms united into one; whereof the reafon is evident. But all this relates to the cafes wherein the coefficients of the indeterminate quantities, in the given equations, are (two of them at least) prime to each other: I fhall add one example more, to fhew the way of proceeding when thofe coefficients admit of a coinmon measure.

PROBLEM XV.

Suppofing 12x+15y + 20% = 100001; it is required to find the number of all the answers in pofitive integers.

It is evident, by tranfpofing 20z and dividing by (3) the greatest common measure of x and y, that 4x + 51, and confequently it's equal 33333

6%

22 2

3

>

must be an integer, and therefore 2z 2 divifible by 3: but 3 is divifible by 3, and fo the difference of these two, which is z+2, must be likewife divisible by the same number, and confequently z + fome multiple of 3. Make, therefore, I + 3uz (u being an integer); then the given equation, by fubftituting this value, will become 12x + 15y + bou + 20 =100001; which, by divifion, &c. is reduced to 4x + 53+ 20u33327: wherein the coefficients of x and y are now prime to each other, and we are to find the number of all the variations, anfwering to the different interpretations of u, from o to the greatest limit, inclufive.

By proceeding, therefore, as in the aforegoing cafes, we haxe x = 8331 —y 8331-y-3; whence the leaft va

4

lue of y is given 3, and the greatest of x Moreover, from the equation 4m + 5n

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x=8328. 20, we have

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Therefore the general values of x and y (given in Problem 13) do here become 8328 59 5u, and 3 + 49; from the former of which the greatest limit of q is 8328 = 1665. Now, fince the value of y 5

given =

will here continue pofitive, in all fubftitutions for q and u (as no negative quantity enters therein); the whole number of anfwers will be determined by the values of x alone.

In order to this, let q be fucceffively expounded by 1665, 1664, 1663, &c. Quot. N. Anf. and it will thence appear

7

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that the faid number will be truly defined by 1666 terms of the arithmetical progreffion

+ 2 + 3

+ 4+ 5, &c. whereof the fum is found to be

When there are four indeterminate quantities in the given equation, the number of all the answers may be determined by the fame methods: for, any one of those quantities may be interpreted by all the integers, succeffively, up to its greateft limit (which is eafily determined); and the number of anfwers, correfponding to each of these interpretations may be found, as above; the aggregate of all which will confequently be the whole number of anfwers required: which fum, or aggregate, may, in many cafes, be derived by the methods given in Section 14, for fumming of feries's by means of a known relation of their terms. But this being a matter of more speculation than real use, I fhall now pafs on to other fubjects.

SEC

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SECTION XIV.

The Investigation of the fums of powers of Numbers in arithmetical progreffion.

B

ESIDES the two forts of progreffions treated of

3

in Section 10, there are infinite varieties of other kinds; but the most useful, and the best known, are thofe confifting of the powers of numbers in arithmetical progreffion; fuch as 1+ 22 + 32 + 42. ... n2, and 13 + 23 + 33 + 43 . . . . 1⁄23, &c. where n denotes the number of terms to which each progreffion is to be continued. In order to investigate the fum of any fuch progreffion, which is the defign of this fection, it will be requifite, first of all, to premise the following

LEMMA.

If any expreffion, or feries, as

--

An + Bn2 + Cn3 + Dn+ &c., involving the pow. — an — bn2 — cn3 dn&c. ers of an indeterminate quantity n, be univerfally equal to nothing, whatsoever be the value of n; then, I say, the fum of the coefficients A-a, Bb, C—e, &c. of each rank of homologous terms, or of the fame powers of n, will also be equal to nothing.

For, in the first place, let the whole equation

An + Bn2 + Cm2, &c. }

-an

bn2

-

cn3

A+

= 0, be divided by n, and

we shall have { _A + B + Cn2 &c. }

a

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bn

-

cn2

= 0; and

this being univerfally fo, be the value of n what it will, let, therefore, n be taken o, and it will

become {A} =0; which being rejected, as

fuch, out of the laft equation, we fhall next have + Bn + Cn2 + Dn3 &c. o; whence, dividing

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