Babylonian Numerals

## **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} }

 in base 60 is equal to the number 7397 in base 10.

(
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, ... }

 etc. Another example, the number

(

1
,
2
;
7
,
3

)

60

{\displaystyle {\left(1,2;7,3 \right)}\_{60} }

 in base 60 is equal to the number 62.1175 in base 10.

(
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}, ... }

 etc.

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} }

 to base 10.

**Problem 2:** Convert the number

(
3
;
45
,
30
)

60

{\displaystyle {(3;45,30)}\_{60} }

 to base 10.

**Problem 3:** Convert the number

(
2
,
3
;
18
,
14
,
24
)

60

{\displaystyle {(2,3;18,14,24)}\_{60} }

 to base 10.

## **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}
}