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Solution Manual For History of Mathematics An Introduction, A, 2025 4th Edition by Victor J. Katz
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Solution Manual For History of Mathematics An Introduction, A, 2025 4th Edition by Victor J. Katz
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Voorbeeld 4 van de 172 pagina's
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Solutions 1CHAPTER 1
1. The answers are given in the answer section of the text. For the Egyptian hiero-
glyphics, 375 is three hundreds, seven tens, and five ones, while 4856 is four
thousands, eight hundreds, five tens, and six ones. For Babylonian cuneiform, note that
375 = 6 × 60 + 15, while 4856 = 1 × 3600 + 20 × 60 + 56.
2.
1 34
′
2 68
4 136
8 272
′
16 544
18 612
1 5
10 50′ (multiply by 10)
2 10 (double first line)
4 20 (double third line)
8 40′ (double fourth line)
′
2 22 (halve first line)
′
10 2 (invert third line)
18 2 10 93
3.
1 2 14
2 4 28
4 8 56
1
,2 Solutions
4.
1 28
2 56
4 112
16
5. We multiply 10 by 3 30:
1 3 30
′
2 1 3 15
4 2 3 10 30
′
8 5 2 10
The total of the two marked lines is then 7, as desired.
6.
1 7 2 4 8
2 15 2 4
4 31 2
8 63
3 5 4
Note that the sum of the three last terms in the second column is 99 2 4. We therefore
need to figure out by what to multiply 7 2 4 8 to give 4 so that we get a total of 100.
But since we know from the fourth line that multiplying that value by 8 gives 63, we
also know that multiplying it by 63 gives 8. Thus the required number is double 63,
which is 42 126. Thus the final result of our division is 12 3 42 126.
, Solutions 3
7.
1 7248
2 15 2 4
′
4 31 2
′
8 63
′
3 4 3 3 6 12
12 3 98 2 3 3 6 12
99 2 4
8.
2 ÷ 11 1 11 2 ÷ 23 1 23
3 73 3 15 3
3 33 3 73
′
6 136 6 323
′ ′
66 6 12 1246
′
276 12
6 66 2 12 276 2
9. x + 17 x = 19. Choose x = 7; then 7 + 17 · 7 = 8. Since 19 ÷ 8 = 2 38 , the correct answer
is 2 83 × 7 = 16 58 .
10. (x + 23 x) − 13 (x + 23 x) = 10. In this case, the “obvious” choice for x is x = 9. Then 9
added to 2/3 of itself is 15, while 1/3 of 15 is 5. When you subtract 5 from 15, you get
10. So in this case our “guess” is correct.
, 4 Solutions
11. The equation here is (1 + 13 + 14 )x = 2. Therefore. we can find the solution by dividing
2 by 1 + 13 + 14 . We set up that problem:
1 124
3 1 18
3 2 36
6 4 72
12 8 144
The sum of the numbers in the right-hand column beneath the initial line is 1 141
144 . So
3
we need to find multipliers giving us 144 = 144 72. But 1 3 4 times 144 is 228. It
follows that multiplying 1 3 4 by 228 gives 144 and multiplying by 114 gives 72.
Thus, the answer is 1 6 12 114 228.
12. Since there are 10 hekats to be divided among 10 men, the average for each is 1 hekat.
So to get the largest share, we should add to the average 1/8 times half the number
of differences. Since there are 9 differences, we add 1/8 4 21 times, or 9/16. Therefore,
9
the largest share is 1 16 hekats, which the scribe writes as 1 2 16. We then subtract 1/8
9 7 5
from this value 9 times to get the share of each man. The answers are 1 16 , 1 16 , 1 16 ,
3 1 15 13 11 9 7
1 16 , 1 16 , 16 , 16 , 16 , 16 , and 16 .
45×100
13. Since x must satisfy 100 : 10 = x : 45, we would get that x = 10 ; the scribe breaks
this up into a sum of two parts, 35×
10
100
and 10×100
10 .
14. The ratio of the cross section area of a log of 5 handbreadths in diameter to one of 4
handbreadths diameter is 52 : 42 = 25 : 16 = 1 16 9
. Thus, 100 logs of 5 handbreadths
9
diameter are equivalent to 1 16 × 100 = 156 41 logs of 4 handbreadths diameter.
16. The modern formula for the surface area of a half-cylinder of diameter d and height
h is A = 12 πdh. Similarly, the modern formula for the surface area of a hemisphere of
diameter d is A = 12 πd2 . These formulas are identical if h = d.
17. 7/5 = 1;24 13/15 = 0;52 11/24 = 0;27,30 33/50 = 0;39,36
18. 0;22,30 = 3/8 0;08,06 = 27/200 0;04,10 = 5/72 0;05,33,20 = 5/54
19. Since 60 is (2 21 ) × 24, the reciprocal of 24 is 2;30. Since 60 is (1 78 ) × 32, and 78 can
be expressed as 0;52,30 the reciprocal of 32 is 1;52,30. Since 60 is (1 13 ) × 45, the
reciprocal of 45 is 1:20. Since 60 = 1 91 × 54, and 19 can be expressed as 10 1 1
+ 90 =
6 40
60 + 3600 = 0;06,40, the reciprocal of 54 is 1;06,40. If the only prime divisors of n are
2, 3, 5, then n is a regular sexagesimal.
20. 25 × 1,04 = 1,40 + 25,00 = 26,40. 18 × 1,21 = 6,18 + 18,00 = 24,18. 50 ÷ 18 =
50 × 0;3,20 = 2;30 + 0;16,40 = 2;46,40. 1,21 ÷ 32 = 1,21 × 0;01,52,30 =
1;21 + 1;10,12 + 0;00,40,30 = 2;31,52,30.
21. 1,08,16 × 3,45 = 4,16,0,0. 4,16 × 3,45 = 16,0,0. 16 × 3,45 = 1,0,0.
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