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This Question being thus stated, it appears by Ruler, Page 176, that it is capable of innumerable Answers; because for any one of these three Letters, a, e, y, there may be taken any Number at Pleasure, provided it be less than 56. But although that may be truly done, yet there are several Ways of arguing about these Sort of Questions, which will limit or bound them to all their proper or poffible Answers in whole Numbers. Thus,

Let ilate-ty=56
And 2 32 a +- 200 t16y=1232

3 lety=56-a
2 32 a 420 +16y=1232 -32
3 x 16 =

16+16 y = 896 - 16а
4 - 5

6

41 = 336 — 16a

7 e=84 4 a; hence as 21 3

5=30 - 28; hence a 7 or 9 From the two laft Steps it appears,' that the Quantity signified by a, ought to be less than 21, and greater than 9 * ; that is, any Number betwixt 9 and 21, may be taken for the Value of a: Consequently there may be eleven Answers to this Question in whole Numbers.

Suppose a = 10, then e = 84 - 40 = 44, per 7th Step; and y= 30 — 28 = 2, per 8th Step. Again, if a=it, thene

- 44 = 40, per 7th Step, and y = 33 — 28 = 5, per 8th Step: and so on for the rest, which will be as in the following Table.

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Thus it will be easy to find out and collect all the limited Anfwers to any Question (of this Kind) wherein there are only three Quantities proposed to be mixed: but when there are more than three, then the Work requires a little more Trouble ; because the fingle Liinits of all the Quantities above two must be found; that is, if there are four Quantities concerned in the Question, the Limits of two of them must be found ; if five Quantities are concerned, then thc Limits of three of them must be found, &c. As in the iullowing Question.

Question Question 35. Suppose it were required to mix four Sorts of Wines together; viz. one Sort warthe 7 s. 4 d. the Gallon, another Sort worth 4 s. 7 d. the Gallon, a third Sort worth 3 s. 8 d. the Gallon, and a fourth Sort worth 2 s. 9 d. the Gallon: How much of each Sort may be taken to make a Mixture of 63 Gallons, so as that the wbale Quantity may be sold for 5 s. 6 d. the Gallon, without Loss, &c.

First, let all these several Rates, and the mean Rate, be reduced to one Denomination, viz. into Pence.

Viz. {3%: 84.344d.

$75. 4 d. = 88 di 45.70. = $50; }and 5 s. 6 d. = 66.

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2 s. 9 d33 d. Put a = the Quantity of that worth 88 d, the Gallon; e= that of 55 d. the Gallon, y = that of 44 d. the Gallon, and u = that of 33 d. the Gallon.

Then latetytu=-63 by the Question.
And 2 88@ +550 + 44 y +33 4 = 4158 = 63 x 66

3 stytu=63-a
288 a 4155€ +44 y + 33 u = 4158 -88 a

3 x 33 533 +33y + 33 u = 2079 — 33 a
4 5

6 22 étuy=2079 - 55a

71 26+y=189-5a; hence a 'f' or 375 3 x 55 8550 + 55y + 554= 3465 -- 55 a 8 - 4 9 y + 22 x=33 4-693

y to 2 u= 32 — 63; hence a 7 1 or 21 From the 7th and 10th Steps it appears, that the Quantity of that Sort of Wine denoted by a, must be less than 37 Gallons, and greater than 21 Gallons: that is, it may be a = any Number of Gallons betwixt 21 and 37$. Whence it follows, that there may be collected 16 Answers to this Question from the Limits of a only. Next to find the Limits of e, y, and u. Suppose a la = 22, then will 5 a=110, and 3 a = 66

But 12120-+y=189—50= 79, per 7th Step.
12 – 2013 y = 79-26; hence est or 39

Again 14 stytu=63-a=41, per 3d Step.
14 - -e15\ytu=41
151311614 =- 38; hence e 7 38

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From the 13th and 16th Steps it appears, that if a = 22, then € = 39, y = 79-21=1, and u =- 38 =I.

Again, Again,

Suppore 1714 = 23, then 5 a = 115, and 3a = 69

But 182e+y=189-50= 74, per 7th Step. 18 20 19 y = 74 - 2 e; hence e === 37

Again 2011 tytu=,63 – 0 = 40, per 3d Step. 20-21 y tu = 40

-19122\u=e— 34, hence e 7 34. From the 19th and 22d Steps it appears, that if a = 13, then e may be either 35 or 36.

21

Once more for a further Illustration.

Let|23|a= 24, then 5 a=-120, and 3a = 72

But 24/26+y=189 – 5a = 69, per 7th Step. 24 24 25 y =69 -- 2e; hence e or 34 *

Again | 26 ety tu=63-a=39, per 3d Step26127 ytu 3927 - 25128 4 - 30, hence € 7 30.

From hence it appears, that if a = 24, then e may be either 31, 32, 33, or 34, viz. it may be any Number betwixt 30 and 34 } by the 25th and 28th Steps; from whence the Values of y and u may be easily found.

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Proceeding on in this manner with all the other fingle Values of 48. there may be found above 1 20 Answers to this Question in whole Numbers : and if you please to put a Fractions, there may be found an innumerable Set of Answers; whereas the Rule of Alligation in l'ulgar Arithmetick affords but only one Answer in Fractions; to wit, that of a =31i, e = 10, y=10, u= 103; as may be casily tried per Rule Page 115, &c.

These two Examples being well understood (especially if the last be thoroughly pursued) may fuffice to shew the Method of limiting the Answers to all sorts of Questions of this Kind. I shall therefore conclude this Chapter of Questions with giving a Solution to the Enigma (or Riddle) proposed (but not answered) by Mr fhn Kersey, in the Close of the Appendix to his Arithmetick,

which affords several pretty Questions, the Solution whereof will discover a certain Sentence consisting of three Words, which must be found by the Help of Figures placed (or supposed to be placed) over the twenty-four Letters of the Alphabet.

3 4 5:

6
7

&c. called Indices.
Thus
b d

f. 8

&c. to the last Letter. So that if the Index to that Letter be once found, the Letter to which it belongs is consequently known.

The Enigma.

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1. If the Difference between the Indices of the second Letter of the second Word, and the third Letter of the first Word, be multiplied into the Difference of their Squares, the Product will be 576; and if their Sum be multiplied into the Sum of their Squares, that Product will be 2336, the Index of the said third Letter being the greatest.

Let Ila = the greater Index, or that of the 3d Letter.

And 2e = the lesser, or that of the ad Letter.
Then 3

e xa a -e = 576.

{

4a texa a Fée=2336} by the Question.

6 +7

3

3 x saaa-aac-aeetee

eee =576
4 x 6 aa a tanetaretece = 2336
6
5 7

2 a aetzee = 1760
sla a a + 3a a e + 3a ee teee = 4096
8
9 a te= V 4096 = 16

2336 2336 4 • atelola a teos

146

ate 16 90uaa + 2a e tee = 256

2 ae ITO 10_12113 a 2 2 0 0 toee = 36

13 ww2 14|--= V 36 = 6 9 +14 15/20 = 22 From henie it appears, that the 3d 15 • 21 16

Letter of the ist Word is l, and 9-16117

5 the ad Letter of the zu Iord is e.

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II

Note, In order to set down the Letters (as they become found) in their proper Places, it may be convenient to supply the vacant Places with Stars.

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* * * * *

2. The

2. The Indices laft found, are the two Extreams of four Numbers in Arithmetical Progression, the lesser Mean being the Index of the first Letter of the third Word; and the greater Mean is the Index of the fourth and laft Letter of the first Word. Viz. $: 7.9. Il are the four Terms in Arithmetical Progression. Whence it appears, that G (whose Index is 7) is the firš Letter of the third Word; and that i (whose Index is 9) is the fourth or laft Letter of the first Word; which being placed down, will stand thus,

* * * *

G **** 3. The second Letter of the third Word is the same with the third Letter of the first Word; and the fifth Letter of the third Word is the same with the last Letter of the first Word: whence the Letters will stand thus, * * li.

*e * * *

** li.

GI * *

4. The Sum of the Squares of the Indices of the first and second Letters of the first Word is 520, and the Product of the same Indices is seven Ninths of the Square of the greater Index, which is the Index of the said first Letter.

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5 x 81

49 aa

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Let a = the greater, and e = the lesser Index.:

Then aa+%= 520} according to the Data.

And 2
2 :a 31 izja
3 ou 4e2=

4 ce=&a a
- 4 5 a 2 = 520 - *?a a

6181aa = 421 20 -
6 + 49 a al 71130 a a = 421 20
7130 8 a 2 = 41% 324

8 9a=V 324 = 18, whose Letter is s. 3 and glrole=a=14, whose Letter is o. Hence the Letters will stand thus,

Soli. * * * * Gl. * i ++ 5. The Difference between the two laft Indices, is the Index of the first Letter of the second Word, viz. 18–14 34 being the Index of the Letter D. Then the Letters will stand thus, Soli. De * * *. GI. * * i *.

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