2. The number 23 can be
represented as two rows of ten
and three more.
The number 23 can also be
represented as two rows of
eight and seven more.
We can write this as 27eight
The number 23 can also be
represented as two rows of
nine and five more.
We can write this as 25nine
3. The number 23 can be
represented as two rows of ten
and three more.
The number 23 can also be
represented as three rows of
six and five more.
We can write this as 35six
The number 23 can also be
represented as three rows of
seven and two more.
We can write this as 32seven
4. The number 23 can be
represented as two rows of ten
and three more.
The number 23 can also be
represented as four rows of
five and three more.
We can write this as 43five
5. The number 23 can be
represented as two rows of ten
and three more.
The number 23 can also be
represented as one square of 16
and one row of four and three
more.
We can write this as 113four
The number 23 can also be
represented as two squares of
9 and one row of three and
two more.
We can write this as 212three
6. The base number tells us the size of the squares that we are
counting and the length of the lines that we are counting.
The number of digits that are available is always the same as the base.
The highest digit avilable is always one less than the base. For example in
base four the digits available are 0, 1, 2, and 3.
In base nine the digits available are 0, 1, 2, 3, 4, 5, 6, 7, and 8.
How can we express the number 235six as a base ten number?
The place values for base six are:
64 = 1,296 63 = 216 62 = 36 61 = 6 60 = 1
Writing the number 235six in expanded notation, we will have:
( 2 x 36 ) + ( 3 x 6 ) + ( 5 x 1 ) = 95
235six = 95ten
7. How can we express the number 152six as a base ten number?
The place values for base six are:
64 = 1,296 63 = 216 62 = 36 61 = 6 60 = 1
Writing the number 152six in expanded notation, we will have:
( 1 x 36 ) + ( 5 x 6 ) + ( 2 x 1 ) = 68
152six = 68ten
How can we express the number 2405six as a base ten number?
Writing the number 2405six in expanded notation, we will have:
( 2 x 216 ) + ( 4 x 36 ) + ( 0 x 6 ) + ( 5 x 1 ) = 581
2405six = 581ten
8. How can we express the number 10011two as a base ten number?
The place values for base two are:
24 = 16 23 = 8 22 = 4 21 = 2 20 = 1
Writing the number 10011two in expanded notation, we will have:
( 1 x 16 ) + ( 0 x 8 ) + ( 0 x 4 ) + ( 1 x 2 ) + ( 1 x 1 ) = 19
10011two = 19ten
How can we express the number 365seven as a base ten number?
Writing the number 365seven in expanded notation, we will have:
( 3 x 49 ) + ( 6 x 7 ) + ( 5 x 1 ) = 194
365seven = 194ten
The place values for base seven are:
74 = 2401 73 = 343 72 = 49 71 = 7 70 = 1
9. How can we express the hexadecimal number 3b0f as a base ten number?
The place values for hexadecimal are:
164 = 65,536 163 = 4,096 162 = 256 161 = 16 160 = 1
Writing the number 3b0fsixteen in expanded notation, we will have:
( 3 x 4,096 ) + ( 11 x 256 ) + ( 0 x 16 ) + ( 15 x 1 ) = 15,119
3b0fsixteen = 15,119ten
How can we express the number facesixteen as a base ten number?
Writing the number facesixteen in expanded notation, we will have:
( 15 x 4,096 ) + ( 10 x 256 ) + ( 12 x 16 ) + ( 14 x 1 ) = 64,206
facesixteen = 64,206ten
We can use bases that are larger than 10. Hexadecimal is base 16. The digits used are:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, f when a =10, b = 11, c = 12, d = 13, e = 14, f = 15
10. Here is a Mickey Mouse problem. How
would Mickey express the base ten number
348?
What base would Mickey use?
The place values for base eight are:
84 = 4,096 83 = 512 82 = 64 81 = 8 80 = 1
The largest place value will be 64. It will be a three digit octal number.
28
320
5
34864
Divide the remainder by 8 to get the next digit.
4
24
3
288
The remainder will be the one’s digit.
348 = 534eight
( 5 x 64 ) + ( 3 x 8 ) + ( 4 x 1 ) = 348
Divide 348 by 64 to get the 64 digit.
11. Write 437 as a binary number.
The place values for base two are:
29 = 512 28 = 256 27 = 128 26 = 64 25 = 32 24 = 16 23 = 8
22 = 4 21 = 2 20 = 1
The largest place value will be 256. It will be a nine digit binary number.
181
256
1
437256
Divide 437 by 256 to get the 256 digit.
53
128
1
181128
The remainder will be the one’s digit.
437 = 110110101two
( 1 x 256 ) + ( 1 x 128 ) + ( 0 x 64 ) + ( 1 x 32 ) + ( 1 x 16 ) + ( 0 x 8 ) + ( 1 x 4 ) + ( 0 x 2 ) + ( 1 x 1 ) = 437
Divide the remainder by 128 to get the next digit.
Divide the remainder by 64 to get the next digit (it will be zero.)
Divide the remainder by 32 to get the next digit.
21
32
1
5332
Divide the remainder by 16 to get the next digit.
5
16
1
2116
Divide the remainder by 8 to get the next digit (it will be zero.)
Divide the remainder by 4 to get the next digit.
1
4
1
54
Divide the remainder by 2 to get the next digit ( it will be zero.)
12. What will 190 be in hexadecimal?
The place values for hexadecimal are:
164 = 65,536 163 = 4,096 162 = 256 161 = 16 160 = 1
14
16
30
160
11
19016
The remainder will be the one’s digit.
190 = besixteen
( 11 x 16 ) + ( 14 x 1 ) = 190
Divide 190 by 16 to get the 16 digit.
The largest place value will be 16. It will be a two digit hexadecimal number.
a =10, b = 11, c = 12, d = 13, e = 14, f = 15
13. There is an easier way to convert from base 10 to another base.
Divide the number by the base that you want and keep track of the
remainders. To convert 382 to base 5 divide 382 by 5 and continue.
382 3012ten five
76
5 382
15
5 76
3
5 15
0
5 3
0
1
3
2
Remainders
14. One more time. To convert 382 to base 6, divide by 6 and keep
track of the remainders.
382 1434ten six
63
6 382
10
6 63
1
6 10
0
6 1
4
3
1
4
Remainders
