CHAPTER 7
1. We write, in order, the Chinese form of 56, 554, 63, and 3282:
2. The largest digit a so that (100a)2 < 142, 884 is a = 3. If we subtract 3002 = 90, 000 from
142,884, the remainder is 52,884. We then need to find b so that 2(100a)(10b) < 52, 884,
or 6000b < 52, 884. We take b = 7 and check that 6000b + (10b)2 < 52, 884. But this
inequality reduces to 42, 000 + 4900 < 52, 884. Since the left side equals 46,900, the
inequality is in fact true. Note that if we had taken b = 8, this second inequality would
not have been true. We now subtract 46,900 from 52,884 to get 5,984. We now need to
find c so that 2(370)c < 5984. We try c = 8 and check that 740c + c2 ≤ 5984. In fact,
740 · 8 + 82 = 5984, so the desired square root is 378.
3. To calculate the cube root x of 12,812,904, we note first that it is a three digit number
beginning with 2: x = 200 + 10b + c. We then temporarily ignore the c and calculate
(200 + 10b)3 = 8, 000, 000 + 3 · 40000 · 10b + 3 · 200 · 100b2 + 1000b3 .
This result must be less than or equal to 12,812,904, or b(1, 200, 000+ 60000b + 1000b2 ) ≤
4, 812, 904. We check that the largest b that satisfies this inequality is b = 3. Thus the
answer begins with 23, and we then repeat the steps to find c:
(230 + c)3 = 12, 167, 000 + 3 · 2302 c + 3 · 230c2 + c3 ≤ 12, 812, 904.
4.
5.
6.
7.
This reduces to c(158, 700 + 690c + c2 ) ≤ 645, 904, and a calculation shows that this is
an equality for c = 4. Thus x = 234.
We add 2 + 4 + 8 + 16 + 32 = 62. So 62 is the divisor. Then on the first day, the weaver
weaves (5 × 2) ÷ 62 = 10/62 chi = 100/62 cun = 1 38/62 cun. On each successive day,
the weaver doubles her output, so we get 3 14/62 cun on the second day, 6 28/62 cun
on the third day, 12 56/62 cun on the fourth day, and 25 50/62 cun on the fifth day.
We would probably set x to be the amount woven on the first day and then solve the
equation x+2x+4x+8x+16x = 50, but this essentially amounts to the same procedure.
560
41
350
12
× 100 = 51 109
; Ty = 1090
× 100 = 32 109
; Tz =
560 + 350 + 180 = 1090. Tx = 1090
180
56
1090 × 100 = 16 109 .
In 15 days, the first channel fills the reservoir 45 times, the second channel 15 times, the
third, 6 times, the fourth, 5 times, and the fifth 3 times. It follows that in 15 days the
reservoir is filled 74 times. To fill it once then requires 15
74 of a day.
If x is the unknown amount, the conditions show that after the first tax the man had
2
4 2
6 4 2
3 x; after the second, he had 5 · 3 x; and after the third, he had 7 · 5 · 3 x. This amount
175
15
must equal 5. It follows that x = 16 = 10 16 pounds.
35
36
Chapter 7
8. We begin with c6 = r = 10 and a6 =
300. Then c12 =
q
c48 =
q
102 − 52 = 8.6603. Therefore, S12 = 21 ·6·10·10 =
(10/2)2 + (10 − 8.6603)2 = 5.1764, and S24 = 21 ·12·10·c12 = 310.5859.
We then get a12 =
2.6105 and S48 =
√
1
2
q
100 − (c12 /2)2 = 9.6593. Next, c24 =
· 24 · 10 · c24 = 313.2629. Next. a24 =
(c24 /2)2 + (10 − a24 )2 = 1.3081. So S96 =
q
we get a48 = 100 − (c48 /2)2 = 9.9786, so c96 =
then get S192 = 12 · 96 · 10 · c96 = 314.1032.
1
q2
q
q
(c12 /2)2 + (10 − a12 )2 =
100 − (c24 /2)2 = 9.9144 and
· 48 · 10 · c48 = 313.9350. Finally,
(c48 /2)2 + (10 − a48 )2 = 0.6544. We
9. From Figure 7.6, we see that one-eighth of the volume of the double box-lid is 0r (r2 −
3
3
3
x2 ) dx = (r2 x − x3 )|r0 = r3 − r3 = 23 r3 . It follows that the entire volume is 16
3r .
R
10. Create the diagram (Fig. 7.7) by beginning with a right triangle (say, the one at the
lower right). Then extend one leg by a segment equal to the other leg and construct the
square on this extended side. If one lays off the first leg along each of the four sides of the
square and then connects the points produced, one gets the square on the hypotenuse.
This is certainly a square since the red triangles are all congruent to one another. The
central square then has the difference of the two legs as its side. From the diagram, it is
clear that the square on the hypotenuse (c2 ) is equal to the square on the difference of the
two legs ((a − b)2) together with twice the rectangle on the two legs (2ab). Algebraically,
c2 = (a − b)2 + 2ab = a2 + b2 − 2ab + 2ab = a2 + b2 . It is not difficult to show by diagrams
that, in fact, (a − b)2 = a2 + b2 − 2ab.
11. Here, we begin with triangle AF J and the squares on AF and AJ as in Figure 7.8.
Because 6 J AB is a right angle, it follows that 6 BAC = 6 F AJ and therefore that B
is on EC extended and that triangle ABC is congruent to triangle AF J . One then
draws the other line segments in the diagram. We now need to show that three pairs
of triangles are congruent. First, triangle J F A is congruent to triangle IKB, because
they are similar and BI = AJ , both being equal to the hypotenuse of the original
triangle. Second, triangle LKI is congruent to triangle HGB since again all the angles
are the same and KI = GB, both being the shorter leg of the original triangle. Finally,
triangle J LE is congruent to triangle ADH, again because they are similar and because
J E = AD, both being the difference between the two legs of the original triangle. Thus
the pieces making up the square on the hypotenuse AJ is equal to the sum of the pieces
making up the two squares on the legs AF and BC = F J .
12. If n is the number of people and p is the price, and if a1 and a2 are the guesses for the
price per person, that is, 8 and 7 respectively, and if f1 , f2 are the surplus and deficiency
respectively, that is, 3 and 4, then, in general, we have the equation a1 n − f1 = a2 n + f2 .
+f2
= 71 = 7. Since 7 × 8 = 56, is a surplus of 3, the actual price is 53.
This gives n = af11−a
2
13. If x is the hypotenuse of a right triangle and 10 and d the legs, then x = d + 1 or
d = x − 1. Then x2 = (x − 1)2 + 100 and x = 50.5.
14. The center of the desired circle is the intersection of the angle bisectors of the two acute
angles of the triangle. To show this, drop perpendiculars to all three sides of the triangle
from that intersection and show that those three lines are all equal by considering the
two pairs of congruent right triangles formed by the three perpendiculars and the angle
bisectors. Let the length of one of the perpendiculars be r, the radius of the circle. Then,
Ancient and Medieval China
37
since the two perpendiculars to the legs of the triangle also produce a square of side r,
we get that r2 + r(b − r) + r(a − r) = 12 ab. This equation reduces to r(a + b) − r2 = 21 ab
and then to 2r(a + b − r) = ab. Since D = 2r, we can rewrite this as
D=
ab
.
a+b−r
But the diagram also shows that c = a−r+b−r = a+b−2r, so that a+b+c = 2(a+b−r).
Therefore, a + b − r = 12 (a + b + c) and, substituting in the formula for D, we get
D=
15.
16.
17.
18.
2ab
,
a+b+c
as desired.
As part of the previous exercise, we have shown that D = 2r = a + b − c = a − (c − b).
2 −2ac+2ab+c2 −2bc+b2 =
Then square both sides. So D2 = a2 −2a(c−b)+(c−b)2 = aq
2(c2 − bc − ac + ab) = 2(c − a)(c − b). It follows that D = 2(c − a)(c − b).
If we set x to be the length of a side of the city, draw a line through the center of the city
extending 20 pu north and 14 pu south, extend a line 1775 pu west from the bottom of
that line, and connect the end of that new line with the end of the line to the north. We
get a right triangle with legs x + 34 and 1775. Since we also have a similar triangle with
legs 20 and x2 , we get the proportion 20 : x2 = (x + 34) : 1775. The resulting equation is
x2 + 34x = 71000 and x = 250.
If x is the depth of the well, the similarity relationship gives 5−0.4
= 0.4
x
5 . Thus x =
5·4.6
0.4 = 57.5.
The simplest way is to set y = DC, x = CE. Then
10
y
1 =
x + 13 3
13 13
93
3 120
y
=
.
x+5
5
and
Simplifying these equations gives 40y = 30x + 400; 200y = 151x + 755. Solving these
simultaneously gives the solution x = 1245, y = 943 43 .
19. If x is the yield of good grain, y the yield of ordinary grain, and z the yield of worst
grain, then the system of equations is
2x +
y
3y
x
+ z
+ 4z
= 1
= 1
= 1
In matrix form we get, in turn
1
0
4
1
0
3
1
1
It follows that 25z = 4 or z =
9
or x = 25
.
2
1
0
1
4
25 ;
0
−1
8
1
0
3
1
1
2
1
0
1
3y + z = 1, 3y =
0
0
25
4
21
25 ,
0
3
1
1
or y =
2
1
0
1
7
25 ;
2x + y = 1, 2x =
18
25 ,
38
Chapter 7
20. By the Pythagorean theorem, the altitude h of the lower triangle is given by
q
q
b2 − ( 2c )2 .
c
c 2
2
2 b − ( 2 ) as stated. Similarly, the area
q
c
c 2
4
4
2
2 a − ( 2 ) . If x = A + B, then x = (A + B) =
B 4 = 2(A4 + 2A3 B + 2A2 B 2 + 2AB 3 + B 4 ) − (A4 −
The area B of that triangle is then B =
A of the upper triangle is A =
A4 + 4A3 B + 6A2 B 2 + 4AB 3 +
2A2 B 2 + B 4 ) = 2(A2 + B 2 )(A2 + 2AB + B 2 ) − (A2 − B 2 )2 = 2(A2 + B 2 )x2 − (A2 − B 2 )2 .
The equation
for x follows√immediately. If a = 39, b = 25, and c =√30, we have
√
A = 15 1521 − 225 = 15 1296 = 15 · 36 = 540. Similarly, B = 15 400 = 300.
Then 2(A2 + B 2 ) = 2(291, 600 + 90, 000) = 763, 200 and (A2 − B 2 )2 = (201, 600)2 =
40, 642, 560, 000 as desired.
21. In this case, we see by trial that the solution is between 6 and 7. So we use 6 in the
synthetic division procedure:
6|
16
6|
16
6|
16
|16
192 −1863.2
96
1728
288 |−135.2
96
|384
For the next step, we will use decimals. The (positive) solution to 16x2 + 384x − 135.2
is between 0 and 1. Again, we find by trial that the value is between 0.3 and 0.4. So our
next chart is as follows:
.3 |
16
.3 |
16
.3 |
16
|16
384
−135.2
4.8
116.64
388.8 |−18.56
4.8
|393.6
For a third step, we will try values between 0 and 0.1. Again, the closest value seems to
be 0.5, as in the following chart:
.05 | 16 393.6 −18.56
.8
19.72
16 394.4 |0.96
Since the last value is relatively close to 0, we will leave the solution as 6.35. If we
wanted to go further, we could have used 0.4 in this last step and continued to find the
next decimal place.
22. Qin’s method gives the following diagrams to produce 234 as the answer. We begin by
Ancient and Medieval China
39
noting that the answer is a three-digit number beginning with a 2.
200 |
1
0
200
0
40, 000
−12, 812, 904
8, 000, 000
200 |
1
200
200
40, 000
80, 000
|−4, 812, 904
200 |
1
400
200
|120, 000
1 |600
|1
The second digit is a 3.
30 |
1
600
30
120, 000
18, 900
−4, 812, 904
4, 167, 000
30 |
1
630
30
138, 900
19, 800
|−645, 904
30 |
1
660
30
|158, 700
1 |690
|1
The final digit is a 4.
4 | 1 690 158, 700
4
2, 776
−645, 904
645, 904
1 694 161, 476
|0
The third order coefficients occur, for example, in 600 = 3 × 200, 120, 000 = 3 × 2002 ,
and in 690 = 3 × 230 and 158, 700 = 3 × 2302 . Note that these numbers occur in the
solution to exercise 3 above.
23. The diagram is as follows, where the solution is x = 23. We begin by noting that the
first digit is a 2.
0
0
0
−279, 841
20 | 1
20
400
8, 000
160, 000
20 |
1
20
20
400
800
8, 000
24, 000
20 |
1
40
20
1, 200
1, 200
|32, 000
20 |
1
60
20
|2, 400
20 |
1 |80
|1
|−119, 841
40
Chapter 7
The second digit is a 3.
3 | 1 80 2, 400 32, 000 −119, 841
3
249
7, 947
119, 841
1 83 2, 649 39, 947
|0
The fourth order coefficients show up in 80 = 4 × 20, 2400 = 6 × 202 , 32, 000 = 4 × 203 ,
and 160, 000 = 1 × 204 .
24. We are given that b2 − [c − (b − a)] = ba, a2 + c + b − a = ac, and a2 + b2 = c2 . Also,
x = b and y = a + c. Then y − x = a + c − b = c − (b − a). From a2 + b2 = c2 , we get b2 =
2
c2 − a2 = (c + a)(c − a), so c − a = b2 /(a + c) = x2 /y. Then 2a = (c + a) − (c − a) = y − xy ,
2
2
so a = 12 (y − xy ). Thus the first equation becomes x2 − [y − x] = x[ 21 (y − xy )]. If we
multiply by 2y, we get 2x2 y − 2y 2 + 2xy = xy 2 − x3, or x3 + 2yx2 + 2xy − xy 2 − 2y 2 = 0,
2
2
the first of our desired equations. Next, we have c = y − a = y − 12 (y − xy ) = 12 (y + xy ).
2
2
4
So ac = 12 (y − xy ) 12 (y + xy ) = 14 (y 2 − xy2 ). If we now substitute this in the second equation
in a, b, and c, we get
"
1
x2
y−
2
y
!#2
x2
1 2 x4
+ +x =
y − 2
y
4
y
!
1 2
x4
x2
1 2 x4
or
y − 2x2 + 2 + +x =
y − 2 .
4
y
y
4
y
!
!
Now multiply by 4y 2 . We get y 4 − 2x2 y 2 + x4 + 4x2 y + 4xy 2 = y 4 − x4 , or 2x4 −
2x2 y 2 + 4x2 y + 4xy 2 = 0. If we divide by 2x, we get the second desired equation:
x4 + 2yx − xy 2 + 2y 2 = 0.
25. Using the notation from the description of Qin Jiushao’s method, we first note that
M = 12. Then M1 = 12 ÷ 3 = 4 and M2 = 12 ÷ 4 = 3. Also P3 = 1, and P4 = 3.
Therefore, we need to solve two congruences: x1 ≡ 1 (mod 3), and 3x2 ≡ 1 (mod 4).
The solutions are x1 = 1 and x2 = 3. Therefore N = 0 · 4 · 1 + 1 · 3 · 3 = 9 ≡ 9 (mod 12).
26. Using the notation from the description of Qin Jiushao’s method, we first calculate
M = 11 · 5 · 9 · 8 · 7 = 27720. Then M1 = M ÷ 11 = 2520; M2 = M ÷ 5 = 5544;
M3 = M ÷ 9 = 3080; M4 = M ÷ 8 = 3465; and M5 = M ÷ 7 = 3960. We then calculate
that M1 ≡ 1 (mod 11); M2 ≡ 4 (mod 5); M3 ≡ 2 (mod 9); M4 ≡ 1 (mod 8);
and M5 ≡ 5 (mod 7). We next need to solve congruences: the solution to 1x1 ≡ 1
(mod 11) is x1 = 1; to 4x2 ≡ 1 (mod 5) is x2 = 4; to 2x3 ≡ 1 (mod 9) is x3 = 5;
to 1x4 ≡ 1 (mod 8) is x4 = 1; and to 5x5 ≡ 1 (mod 7) is x5 = 3. Given that 3 of
the ri are equal to 0, we calculate N simply as n = 4 · 3080 · 5 + 6 · 3465 · 1 = 82, 390.
Subtracting off twice M, we get the solution as N = 82, 390 − 2 · 27, 720 = 26, 950, where
the answer is taken modulo 27,720.