Problem [ ] The ancient Babylonians adopted a sexagesimal (base 60) place-value system for calculation. For example, the number
(
2
,
3
,
17
)
60
{\displaystyle {\left(2,3,17 \right)}_{60} }
(
2
,
3
,
17
)
60
=
2
×
60
2
+
3
×
60
1
+
17
×
60
0
(
2
,
3
,
17
)
60
=
2
×
3600
+
3
×
60
+
17
×
1
(
2
,
3
,
17
)
60
=
7397
{\displaystyle \begin{align}
{(2,3,17)}_{60} &= 2 \times 60^2 + 3 \times 60^1 + 17 \times 60^0 \\[5 pt]
{(2,3,17)}_{60} &= 2 \times 3600 + 3 \times 60 + 17 \times 1 \\[5 pt]
{(2,3,17)}_{60} &= 7397
\end{align}
}
Here, the commas separate the digits of each place value
60
0
,
60
1
,
60
2
,
60
3
,
.
.
.
{\displaystyle 60^0, 60^1, 60^2, 60^3, ... }
(
1
,
2
;
7
,
3
)
60
{\displaystyle {\left(1,2;7,3 \right)}_{60} }
(
1
,
2
;
7
,
3
)
60
=
1
×
60
1
+
2
×
60
0
+
7
60
1
+
3
60
2
(
1
,
2
;
7
,
3
)
60
=
1
×
60
+
2
×
1
+
7
60
+
3
3600
(
1
,
2
;
7
,
3
)
60
=
1
×
60
+
2
×
1
+
7
60
+
1
1200
(
1
,
2
;
7
,
3
)
60
=
62
47
400
{\displaystyle \begin{align}
{(1,2;7,3)}_{60} &= 1 \times 60^1 + 2 \times 60^0 + \frac{7}{60^1} + \frac{3}{60^2} \\[5 pt]
{(1,2;7,3)}_{60} &= 1 \times 60 + 2 \times 1+ \frac{7}{60} + \frac{3}{3600} \\[5 pt]
{(1,2;7,3)}_{60} &= 1 \times 60 + 2 \times 1+ \frac{7}{60} + \frac{1}{1200} \\[5 pt]
{(1,2;7,3)}_{60} &= 62 \frac{47}{400}
\end{align}
}
Here, the semicolon acts like a decimal point and the following commas separate the digits of each place value
60
−
1
,
60
−
2
,
60
−
3
,
60
−
4
,
.
.
.
{\displaystyle {60}^{-1}, {60}^{-2}, {60}^{-3}, {60}^{-4}, ... }
The ancient Babylonians also invented the concept of zero long before the ancient Indians; however, the Babylonian zero served as a placeholder for writing numbers where certain place-values are not a number between 1 and 59.
Problem 1: Convert the number
(
13
,
0
,
21
)
60
{\displaystyle {(13,0,21)}_{60} }
Problem 2: Convert the number
(
3
;
45
,
30
)
60
{\displaystyle {(3;45,30)}_{60} }
Problem 3: Convert the number
(
2
,
3
;
18
,
14
,
24
)
60
{\displaystyle {(2,3;18,14,24)}_{60} }
Solution [ ] Problem 1
(
13
,
0
,
21
)
60
=
13
×
60
2
+
0
×
60
1
+
21
×
60
0
(
13
,
0
,
21
)
60
=
13
×
3600
+
0
×
60
+
21
×
1
(
13
,
0
,
21
)
60
=
46800
+
0
+
21
(
13
,
0
,
21
)
60
=
46821
{\displaystyle \begin{align}
{(13,0,21)}_{60} &= 13 \times 60^2 + 0 \times 60^1 + 21 \times 60^0 \\[5 pt]
{(13,0,21)}_{60} &= 13 \times 3600 + 0 \times 60 + 21 \times 1 \\[5 pt]
{(13,0,21)}_{60} &= 46800 + 0 + 21 \\[5 pt]
{(13,0,21)}_{60} &= 46821
\end{align}
}
Problem 2
(
3
;
45
,
30
)
60
=
3
×
60
0
+
45
60
1
+
30
60
2
(
13
,
0
,
21
)
60
=
3
×
1
+
45
60
+
30
3600
(
13
,
0
,
21
)
60
=
3
+
3
4
+
1
120
(
13
,
0
,
21
)
60
=
3
91
120
{\displaystyle \begin{align}
{(3;45,30)}_{60} &= 3 \times 60^0 + \frac{45}{60^1} + \frac{30}{60^2} \\[5 pt]
{(13,0,21)}_{60} &= 3 \times 1 + \frac{45}{60} + \frac{30}{3600} \\[5 pt]
{(13,0,21)}_{60} &= 3 + \frac{3}{4} + \frac{1}{120} \\[5 pt]
{(13,0,21)}_{60} &= 3 \frac{91}{120}
\end{align}
}
Problem 3
(
2
,
3
;
18
,
14
,
24
)
60
=
2
×
60
1
+
3
×
60
0
+
18
60
1
+
14
60
2
+
24
60
3
(
2
,
3
;
18
,
14
,
24
)
60
=
2
×
60
+
3
×
1
+
18
60
+
14
3600
+
24
216000
(
2
,
3
;
18
,
14
,
24
)
60
=
120
+
3
+
3
10
+
7
1800
+
1
9000
(
2
,
3
;
18
,
14
,
24
)
60
=
123
38
125
{\displaystyle \begin{align}
{(2,3;18,14,24)}_{60} &= 2 \times 60^1 + 3 \times 60^0 + \frac{18}{60^1} + \frac{14}{60^2} + \frac{24}{60^3} \\[5 pt]
{(2,3;18,14,24)}_{60} &= 2 \times 60 + 3 \times 1+ \frac{18}{60} + \frac{14}{3600} + \frac{24}{216000} \\[5 pt]
{(2,3;18,14,24)}_{60} &= 120 + 3 + \frac{3}{10} + \frac{7}{1800} + \frac{1}{9000} \\[5 pt]
{(2,3;18,14,24)}_{60} &= 123 \frac{38}{125}
\end{align}
}
![{\displaystyle {\begin{aligned}{(2,3,17)}_{60}&=2\times 60^{2}+3\times 60^{1}+17\times 60^{0}\\[5pt]{(2,3,17)}_{60}&=2\times 3600+3\times 60+17\times 1\\[5pt]{(2,3,17)}_{60}&=7397\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/21af588a60ec14e7fb569e9b122954409fd1604f)
![{\displaystyle {\begin{aligned}{(1,2;7,3)}_{60}&=1\times 60^{1}+2\times 60^{0}+{\frac {7}{60^{1}}}+{\frac {3}{60^{2}}}\\[5pt]{(1,2;7,3)}_{60}&=1\times 60+2\times 1+{\frac {7}{60}}+{\frac {3}{3600}}\\[5pt]{(1,2;7,3)}_{60}&=1\times 60+2\times 1+{\frac {7}{60}}+{\frac {1}{1200}}\\[5pt]{(1,2;7,3)}_{60}&=62{\frac {47}{400}}\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/8d18464813af353b7c51f65552e682f9fc912caf)
![{\displaystyle {\begin{aligned}{(13,0,21)}_{60}&=13\times 60^{2}+0\times 60^{1}+21\times 60^{0}\\[5pt]{(13,0,21)}_{60}&=13\times 3600+0\times 60+21\times 1\\[5pt]{(13,0,21)}_{60}&=46800+0+21\\[5pt]{(13,0,21)}_{60}&=46821\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/555abf39b4db2b9cd30aa1f930d8cf18977a99f4)
![{\displaystyle {\begin{aligned}{(3;45,30)}_{60}&=3\times 60^{0}+{\frac {45}{60^{1}}}+{\frac {30}{60^{2}}}\\[5pt]{(13,0,21)}_{60}&=3\times 1+{\frac {45}{60}}+{\frac {30}{3600}}\\[5pt]{(13,0,21)}_{60}&=3+{\frac {3}{4}}+{\frac {1}{120}}\\[5pt]{(13,0,21)}_{60}&=3{\frac {91}{120}}\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/97dbad5bd93cbedceab3ddf6a88ab6fb24250a7a)
![{\displaystyle {\begin{aligned}{(2,3;18,14,24)}_{60}&=2\times 60^{1}+3\times 60^{0}+{\frac {18}{60^{1}}}+{\frac {14}{60^{2}}}+{\frac {24}{60^{3}}}\\[5pt]{(2,3;18,14,24)}_{60}&=2\times 60+3\times 1+{\frac {18}{60}}+{\frac {14}{3600}}+{\frac {24}{216000}}\\[5pt]{(2,3;18,14,24)}_{60}&=120+3+{\frac {3}{10}}+{\frac {7}{1800}}+{\frac {1}{9000}}\\[5pt]{(2,3;18,14,24)}_{60}&=123{\frac {38}{125}}\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/ed78de9869225129f84beb5e92df04be23ef63f2)
