### Part I. Section IV. Chapter 3. “Of the Solution of Questions relating to the preceding Chapter.”

585 Question 1. To divide 7 into two such parts that the greater may exceed the less by 3.

Let the greater part be $$x$$, then the less will be $$7-x$$; so that $$x=7-x+3$$, or $$x=10-x$$. Adding $$x$$, we have $$2x=10$$; and dividing by 2, $$x=5$$.

The two parts therefore are 5 and 2.

Question 2. It is required to divide $$a$$ into two parts, so that the greater may exceed the less by $$b$$.

Let the greater part be $$x$$, then the other will be $$a-x$$; so that $$x=a-x+b$$. Adding $$x$$, we have $$2x=a+b$$; and dividing by 2, $$x=\dfrac{a+b}{2}$$.

Another method of solution. Let the greater part $$= x$$; which as it exceeds the less by $$b$$, it is evident that this is less than the other by $$b$$, and therefore must be $$= x - b$$. Now, these two parts, taken together, ought to make $$a$$; so that $$2x-b=a$$; adding $$b$$, we have $$2x=a+b$$, wherefore $$x=\dfrac{a+b}{2}$$, which is the value of the greater part; and that of the less will be $$\dfrac{a+b}{2}-b$$, or $$\dfrac{a+b}{2}-\dfrac{2b}{2}$$, or $$\dfrac{a-b}{2}$$.

586 Question 2. A father leaves 1600 pounds to be divided among his three sons in the following manner; namely, the eldest is to have 200 pounds more than the second, and the second 100 pounds more than the youngest. Required the share of each.

 Let the share of the third son be $$x$$ Then the second's will be $$x+100$$ And the first son's share $$x+300$$

Now, these three sums make up together 1600 pounds; we have, therefore,

$\begin{gather} 3x+400=1600\\ 3x=1200\\ x=400 \end{gather}$
 The share of the youngest is 400 pounds That of the second is 500 pounds That of the eldest is 700 pounds

587 Question 4. A father leaves to his four sons 8600 pounds and, according to the will, the share of the eldest is to be double that of the second, minus 100 pounds; the second is to receive three times as much as the third, minus 200 pounds; and the third is to receive four times as much as the fourth, minus 300 pounds. What are the respective portions of these four sons?

 Call the youngest son's share $$x$$ Then the third son's is $$4x-300$$ The second son's is $$12x-1100$$ And the eldest's $$24x-2300$$

Now, the sum of these four shares must make 8600 pounds We have, therefore, $$41x - 3700 = 8600$$, or $$41x=12300$$, and $$x=300$$.

 Therefore the youngest's share is 300 pounds The third son's 900 pounds The second's 2500 pounds The eldest's 4900 pounds

588 Question 5. A man leaves 11000 crowns to be divided between his widow, two sons, and three daughters. He intends that the mother should receive twice the share of a son, and that each son should receive twice as much as a daughter. Required how much each of them is to receive.

 Suppose the share of each daughter to be $$x$$ Then each son's is consequently $$2x$$ And the widow's $$4x$$

The whole inheritance, therefore, is $$3x+4x+4x$$; or $$1x=11000$$, and, consequently, $$x=1000$$.

 Each daughter, therefore, is to receive 1000 crowns So that the three receive in all 3000 crowns Each son receives 2000 crowns So that the two sons receive 4000 crowns And the mother receives 4000 crowns Sum 11000 crowns

589 Question 6. A father intends by his will, that his three sons should share his property in the following manner: the eldest is to receive 1000 crowns less than half the whole fortune; the second is to receive 800 crowns less than the third of the whole; and the third is to have 600 crowns less than the fourth of the whole. Required the sum of the whole fortune, and the portion of each son.

 Let the fortune be expressed by $$x$$ The share of the first son is $$\frac{1}{2}x-1000$$ That of the second $$\frac{1}{3}x-800$$ That of the third $$\frac{1}{4}x-600$$

So that the three sons receive in all $$\frac{1}{2}x+\frac{1}{3}x+\frac{1}{4}x-2400$$, and this sum must be equal to $$x$$. We have, therefore, the equation $$\frac{13}{12}x-2400=0$$; then adding 2400, we have $$\frac{1}{12}x=2400$$; and, lastly, multiplying by 12, we obtain $$x=28800$$.

 The fortune, therefore, consists of 28800 crowns The eldest son receives 13400 crowns The second 8800 crowns The youngest 6600 crowns Sum 28800 crowns

590 Question 7. A father leaves four sons, who share his property in the following manner: the first takes the half of the fortune, minus 3000 pounds; the second takes the third, minus 1000 pounds; the third takes exactly the fourth of the property; and the fourth takes 600 pounds and the fifth part of the property. What was the whole fortune, and how much did each son receive?

 Let the whole fortune be represented by $$x$$ Then the eldest son will have $$\frac{1}{2}x-3000$$ The second $$\frac{1}{3}x-1000$$ The third $$\frac{1}{4}x$$ The youngest $$\frac{1}{5}x+600$$

And the four will have received in all

$\frac{1}{2}x+\frac{1}{3}x+\frac{1}{4}x+\frac{1}{5}x-3400,$

which must be equal to $$x$$.

Whence results the equation $$\frac{77}{60}x-3400=x$$; then subtracting $$x$$, we have $$\frac{17}{60}x-3400=0$$; adding 3400, we obtain $$\frac{17}{60}x=3400$$; then dividing by 17, we have $$\frac{1}{60}x=200$$; and multiplying by 60, gives $$x=12000$$.

 The fortune therefore consisted of 12000 pounds The first son received 3000 pounds The second 3000 pounds The third 3000 pounds The fourth 3000 pounds

591 Question 8. To find a number such, that if we add to it its half, the sum exceeds 60 by as much as the number itself is less than 65.

Let the number be represented by $$x$$:

Then $$x+\frac{1}{2}x-60=65-x$$, or $$\frac{3}{2}x-60=65-x$$. Now, by adding $$x$$, we have $$\frac{5}{2}x-60=65$$; adding 60, we have $$\frac{5}{2}x=125$$; dividing by 5, gives $$\frac{1}{2}x=25$$; and multiplying by 2, we have $$x=50$$.

Consequently, the number sought is 50.

592 Question 9. To divide 32 into two such parts, that if the less be divided by 6, and the greater by 5, the two quotients taken together may make 6.

Let the less of the two parts sought be $$x$$; then the greater will be $$32-x$$. The first, divided by 6, gives $$\frac{x}{6}$$; and the second, divided by 5, gives $$\frac{32-x}{5}$$. Now $$\frac{x}{6}+\frac{32-x}{5}=6$$: so that multiplying by 5, we have $$\frac{5}{6}x+32-x=30$$, or $$-\frac{1}{6}x+32=30$$; adding $$\frac{1}{6}x$$, we have $$32=30+\frac{1}{6}x$$; subtracting 30, there remains $$2=\frac{1}{6}x$$; and lastly, multiplying by 6, we have $$x=12$$.

So that the less part is 12, and the greater part is 20.

593 Question 10. To find such a number, that if multiplied by 5, the product shall be as much less than 40 as the number itself is less than 12.

Let the number be $$x$$; which is less than 12 by $$12-x$$; then taking the number $$x$$ five times, we have $$5x$$, which is less than 40 by $$40-5x$$, and this quantity must be equal to $$12-x$$.

We have, therefore, $$40-5x=12-x$$. Adding $$5x$$, we have $$40=12+4x$$; and subtracting 12, we obtain $$28=4x$$; lastly, dividing by 4, we have $$x=7$$, the number sought.

594 Question 11. To divide 25 into two such parts, that the greater may be equal to 49 times the less.

Let the less part be $$x$$, than the greater will be $$25-x$$; and the latter divided by the former ought to give the quotient 49: we have therefore $$\frac{25-x}{x}=49$$. Multiplying by $$x$$, we have $$25-x=49x$$; adding $$x$$, we have $$25=50x$$; and dividing by 50, gives $$x=\frac{1}{2}$$.

The less of the two numbers is ½, and the greater is 24½; dividing therefore the latter by ½, or multiplying by 2, we obtain 49.

595 Question 12. To divide 48 into nine parts, so that every part may be always ½ greater than the part which precedes it.

Let the first, or least part, be $$x$$, then the second will be $$x+\frac{1}{2}$$, the third $$x+1$$, etc.

Now, these parts form an arithmetical progression, whose first term is $$x$$; therefore the ninth and last term will be $$x + 4$$. Adding those two terms together, we have $$2x + 4$$; multiplying this quantity by the number of terms, or by 9, we have $$18x + 36$$; and dividing this product by 2, Ave obtain the sum of all the nine parts $$= 9x + 18$$; which ought to be equal to 48. We have, therefore, $$9x + 18 = 48$$; subtracting 18, there remains $$9x=30$$; and dividing by 9, we have $$x=3\frac{1}{3}$$.

The first part, therefore, is 3⅓, and the nine parts will succeed in the following order:

 1 2 3 4 5 6 7 8 9 3⅓ + 3⅚ + 4⅓ + 4⅚ + 5⅓ + 5⅚ + 6⅓ + 6⅚ + 7⅓

which together make 48.

596 Question 13. To find an arithmetical progression, whose first term is 5, the last term 10, and the entire sum 60.

Here we know neither the difference nor the number of terms; but we know that the first and the last term would enable us to express the sum of the progression, provided only the number of terms were given. We shall therefore suppose this number to be $$x$$, and express the sum of the progression $$\dfrac{15x}{2}$$. We know also, that this sum is 60; so that $$\dfrac{15x}{2}=60$$; or $$\frac{1}{2}x=4$$, and $$x=8$$.

Now, since the number of terms is 8, if we suppose the difference to be $$z$$, we have only to seek for the eighth term upon this supposition, and to make it equal to 10. The second term is $$5 + z$$, the third is $$5 + 2z$$, and the eighth is $$5 + 7z$$; so that

$\begin{gather} 5+7z=10\\ 7z=5\\ z=\frac{5}{7} \end{gather}$

The difference of the progression, therefore, is ⁵⁄₇, and the number of terms is 8; consequently, the progression is

 1 2 3 4 5 6 7 8 5 + 5⁵⁄₇ + 6³⁄₇ + 7⅐ + 7⁶⁄₇ + 8⁴⁄₇ + 9²⁄₇ + 10

the sum of which is 60.

597 Question 14. To find such a number, that if 1 be subtracted from its double, and the remainder be doubled, from which if 2 be subtracted, and the remainder divided by 4, the number resulting from these operations shall be 1 less than the number sought.

Suppose this number to be $$x$$; the double is $$2x$$; subtracting 1, there remains $$2x-1$$; doubling this, we have $$4x-2$$; subtracting 2, there remains $$4x-4$$; dividing by 4, we have $$x-1$$; and this must be 1 less than $$x$$; so that

$x-1=x-1.$

But this is what is called an identical equation; and shows that $$x$$ is indeterminate; or that any number whatever may be substituted for it.

598 Question 15. I bought some ells of cloth at the rate of 7 crowns for 5 ells, which I sold again at the rate of 11 crowns for 7 ells, and I gained 100 crowns by the transaction. How much cloth was there?

Supposing the number of ells to be $$x$$:, we must first see how much the cloth cost; which is found by the following proportion:

$5:x::7:\dfrac{7x}{5}$

This being the expenditure; let us now see the receipt: in order to which, we must make the following proportion

$7 \; \textrm{ells} : 11 \; \textrm{crowns} :: x \; \textrm{ells} : \frac{11}{7}x \; \textrm{crowns};$

and this receipt ought to exceed the expenditure by 100 crowns. We have, therefore, this equation:

$\frac{11}{7}x=\frac{7}{5}x+100.$

Subtracting $$\frac{7}{5}x$$, there remains $$\frac{6}{35}x=100$$; therefore $$6x=3500$$, and $$x=583\frac{1}{3}$$.

There were, therefore, 583⅓ ells bought for 816⅔ crowns, and sold again for 916⅔ crowns; by which means the profit was 100 crowns.

599 Question 16. A person buys 12 pieces of cloth for 140 pounds; of which two are white, three are black, and seven are blue: also, a piece of the black cloth costs two pounds more than a piece of the white, and a piece of the blue cloth costs three pounds more than a piece of the black. Required the price of each kind.

Let the price of a white piece be $$x$$ pounds; then the two pieces of this kind will cost $$2x$$; also, a black piece costing $$x + 2$$, the three pieces of this color will cost $$3x + 6$$; and lastly, as a blue piece costs $$x + 5$$, the seven blue pieces will cost $$7x+35$$: so that the twelve pieces amount in all to $$12x + 41$$.

Now, the known price of these twelve pieces is 140 pounds; we have, therefore, $$12x+41=140$$, and $$12x=99$$; wherefore $$x=8\frac{1}{4}$$. So that

 A piece of white cloth costs 8¼ pounds A piece of black cloth costs 10¼ pounds A piece of blue cloth costs 13¼ pounds

600 Question 17. A man having bought some nutmegs, says that three of them cost as much more than one penny, as four cost him more than two pence halfpenny. Required the price of the nutmegs?

Let $$x$$ be the excess of the price of three nuts above one penny, or four farthings. Now, if three nutmegs cost $$x + 4$$ farthings, four will cost, by the condition of the question, $$x + 10$$ farthings; but the price of three nutmegs gives that of four in another way, namely, by the Rule of Three, Thus,

$3:x+4 :: 4:\dfrac{4x+16}{3}.$

So that $$\dfrac{4x+16}{3}=x+10$$; or, $$4x+16=3x+30$$; therefore $$x+16=30$$, and $$x=14$$.

Three nutmegs, therefore, cost 4½ pence, and four cost 6 pence: wherefore each cost 1½ pence.

601 Question 18. A certain person has two silver cups, and only one cover for both. The first cup weighs 12 ounces; and if the cover be put on it, it weighs twice as much as the other cup: but when the other cup has the cover, it weighs three times as much as the first. Required the weight of the second cup, and that of the cover.

Suppose the weight of the cover to be $$x$$ ounces; then the first cup being covered it will weigh $$x + 12$$; and this weight being double that of the second, the second cup must weigh $$\frac{1}{2}x + 6$$; and, with the cover, it will weigh $$x + \frac{1}{2}x + 6=\frac{3}{2}x+6$$; which weight ought to be the triple of 12; that is, three times the weight of the first cup. We shall therefore have the equation $$\frac{3}{2}x + 6 = 36$$, or $$\frac{3}{2}x= 30$$; so that $$\frac{1}{x}x=10$$ and $$x=20$$.

The cover, therefore, weighs 20 ounces, and the second cup weighs 16 ounces.

602 Question 19. A banker has two kinds of change: there must be a pieces of the first to make a crown; and $$b$$ pieces of the second to make the same. Now, a person wishes to have $$c$$ pieces for a crown. How many pieces of each kind must the banker give him?

Suppose the banker gives $$x$$ pieces of the first kind; it is evident that he will give $$c - x$$ pieces of the other kind; but the $$x$$ pieces of the first are worth $$\dfrac{x}{a}$$ crown, by the proportion $$a:x :: 1 : \dfrac{x}{a}$$; and the $$c-x$$ pieces of the second kind are worth $$\dfrac{c-x}{b}$$ crown, because we have $$b:-x :: 1: \dfrac{c-x}{b}$$. So that

$\begin{gather} \dfrac{x}{a}+\dfrac{c-x}{b}=1;\\ \dfrac{bx}{a}+c-x=b;\\ bx+ac-ax=ab;\\ bx-ax=ab-ac;\\ x=\dfrac{ab-ac}{b-a};\\ x=\dfrac{a(b-c)}{b-a}; \end{gather}$

consequently, $$c-x$$, the pieces of the second kind, must be equal

$\dfrac{bc-ab}{b-a} = \dfrac{b(c-a)}{b-a}.$

The banker must therefore give $$\dfrac{a(b-c)}{b-a}$$ pieces of the first kind, and $$\dfrac{b(c-a)}{b-a}$$ pieces of the second kind.

Remark. These two numbers are easily found by the Rule of Three, when it is required to apply the results which we have obtained. Thus, to find the first we say,

$b-a:a :: b-c : \dfrac{a(b-c)}{b-a};$

and the second number is found thus;

$b-a:b :: c-a : \dfrac{b(c-a)}{b-a}.$

It ought to be observed also, that $$a$$ is less than $$b$$, and that $$c$$ is less than $$b$$; but at the same time greater than $$a$$, as the nature of the thing requires.

603 Question 20. A banker has two kinds of change; 10 pieces of one make a crown, and 20 pieces of the other make a crown; and a person wishes to change a crown into 17 pieces of money: how many of each sort must he have?

We have here $$a=10$$, $$b=20$$, and $$c=17$$, which furnishes the following proportions:

First, 10 : 10 ∷ 3 : 3, so that the number of pieces of the first kind is 3.

Second, 10 : 20 ∷ 7 : 14, and the number of the second kind is 14.

604 Question 21. A father leaves at his death several children, who share his property in the following manner: namely, the first receives a hundred pounds, and the tenth part of the remainder; the second receives two hundred pounds, and the tenth part of the remainder; the third takes three hundred pounds, and the tenth part of what remains; and the fourth takes four hundred pounds, and the tenth part of what then remains; and so on. And it is found that the property has thus been divided equally among all the children. Required how much it was, how many children there were, and how much each received?

This question is rather of a singular nature, and therefore deserves particular attention. In order to resolve it more easily, we shall suppose the whole fortune to be $$z$$ pounds; and since all the children receive the same sum, let the share of each be $$x$$, by which means the number of children will be expressed by $$\dfrac{z}{x}$$. Now, this being laid down, we may proceed to the solution of the question, as follows:

Sum, or property to be divided. Order of the children. Portion of each. Differences.
$$z$$ 1st $$x=100+\dfrac{z-100}{10}$$
$$z-x$$ 2nd $$x=200+\dfrac{z-x-200}{10}$$ $$100-\dfrac{x-100}{10}=0$$
$$z-2x$$ 3rd $$x=300+\dfrac{z-2x-300}{10}$$ $$100-\dfrac{x-100}{10}=0$$
$$z-3x$$ 4th $$x=400+\dfrac{z-3x-400}{10}$$ $$100-\dfrac{x-100}{10}=0$$
$$z-4x$$ 5th $$x=500+\dfrac{z-4x-500}{10}$$ $$100-\dfrac{x-100}{10}=0$$
$$z-5x$$ 6th $$x=600+\dfrac{z-5x-600}{10}$$ $$100-\dfrac{x-100}{10}=0$$

We have inserted, in the last column, the differences which we obtain by subtracting each portion from that which follows; but all the portions being equal, each of the differences must be = 0. As it happens also, that all these differences are expressed exactly alike, it will be sufficient to make one of them equal to nothing, and we shall have the equation $$100-\dfrac{x-100}{10}=0$$. Here, multiplying by 10, we have $$1000-x-100=0$$, or $$900-x=0$$; and, consequently, $$x=900$$.

We know now, therefore, that the share of each child was 900; so that taking any one of the equations of the third column, the first, for example, it becomes, by substituting the value of $$x$$, $$900=100+\dfrac{z-100}{10}$$, whence we immediately obtain the value of $$z$$; for we have $$9000=1000+z-100$$, or $$9000=900+z$$; therefore $$z=8100$$; and consequently $$\dfrac{z}{x}=9$$.

So that the number of children was 9; the fortune left by the father was 8100 pounds; and the share of each child was 900 pounds.

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.