Question

list the first twenty counting numbers in base seven? (only digits 0 thru 6 are used in base 7)??????????????????????????????????????

Answer 1

0, 1, 2, 3, 4, 5, 6
10, 11, 12, 13, 14, 15, 16
20, 21, 22, 23, 24, 25

is this right?

Answer 2

1
2
3
4
5
6
7=7^1=10
8=7^1+7^0=11
9=7^1+2*7^0=12
10=7^1+3*7^0=13
11=7^1+4*7^0=14
12=7^1+5*7^0=15
13=7^1+6*7^0=16
14=2*7^1+0*7^0=20
15=2*7^1+1*7^0=21
16=2*7^1+2*7^0=22
17=2*7^1+3*7^0=23
18=2*7^1+4*7^0=24
19=2*7^1+5*7^0=25
20=2*7^1+6*7^0=26

Answer 3

21=30
22=31
23=32
24=33
25=34
26=35
27=36
28=40
29=41
do you see a pattern?

Answer 4

THE NUMBERS KEEP INCREASING
THAT PATTERN?

Answer 5

MYININAYA CAN U EXPLAIN FURTHER PLEASEE

Answer 6

the 7^0 digit only stays between 0 and 6

Answer 7

7/\1 MEANS 7 AND 1 CORRECT?

Answer 8

7^0=1
7^1=7
7^2=49

Answer 9

those are powers

Answer 10

\[7^x\]=7^x

Answer 11

YES OK
I UNDERSTAND THAT
NOW WE ARE TALKING OF BASE SEVEN CORRECT

Answer 12

Just like in base 10 you have the

ones column, the sevens column, the 49's column, etc.

So if you take a number (like 56) It has 1 49 (with 17 left over) 2 7's (with 3 left over) and 3 ones.

\(123_7 = 56_{10}\)

Answer 13

say we are given a number c and we want to write it in base 7
first we would figure out the coefficients a_i of
\[c=a_n7^n+a_{n-1}7^{n-1}+a_{n-2}7^{n-2}+\cdot \cdot \cdot + a_37^3+a_27^2+a_17^1+a_07^0\]
then we can write c as
\[c=a_na_{n-1}a_{n-2}....a_3a_2a_1a_0\]
where he have written c in base 7

Answer 14

where we have written c in base 7*

Answer 15

\[c=a_na_{n-1}a_{n-2}....a_3a_2a_1a_0 \]
this is not multiplication
these are the digits of c

Answer 16

in order of course

Answer 17

maybe i should say
\[c_{10}=(a_na_{n-1}a_{n-2}....a_3a_2a_1a_0)_{7}\]

Answer 18

ust like in base 10 you have the

ones column, the sevens column, the 49's column, etc.

So if you take a number (like 56) It has 1 49 (with 17 left over) 2 7's (with 3 left over) and 3 ones.

1237=5610- I THINK I CAN UNDERSTAND THIS POINT

Answer 19

\[56_{10}=1 \cdot 7^2+1 \cdot 7^1+0 \cdot 7^0=110_{7}\]

Answer 20

ok, Myininaya u just lost me. your going to fast

Answer 21

go back to my original question

Answer 22

the answer is irrelevant. just tell me how u got it

Answer 23

what are the counting numbers of a base?

Answer 24

because 49+7=56

Answer 25

say you want to write a number c in base 7 and you are given a number in base 10
you could do this to figure out what you highest power would be
\[\log_7c=\frac{lnc}{\ln7}\]

so for example we want to write 56 in base 7
so we do
\[\log_7(56)=\frac{\ln(56)}{\ln7}=2.068621561\]
take the floor of this number which is 2

and that is the highest power you should begin with
so we have
the 7^2 part

\[56=1 \cdot 7^2+ a_1 \cdot 7^1+a_2 \cdot 7^0\]

we need to figure out these other digits

so we already have 49
and 56-49=7
so we should put a 1 in front of 7^1 to get the 7 and a 0 in front of 7^0

so we have
\[56=1 \cdot 7^2+ 1 \cdot 7^1+0 \cdot 7^0=(110)_7\]

Answer 26

\[\frac{56}{7}=8=7+1\]
\[(\frac{56}{7}=7+1)7\]
\[56=7(7)+1(7)=7^2+7\]

Answer 27

1
2
3
4
5
6
7=7^1=10
8=7^1+7^0=11
9=7^1+2*7^0=12
10=7^1+3*7^0=13
11=7^1+4*7^0=14
12=7^1+5*7^0=15
13=7^1+6*7^0=16
14=2*7^1+0*7^0=20
15=2*7^1+1*7^0=21
16=2*7^1+2*7^0=22
17=2*7^1+3*7^0=23
18=2*7^1+4*7^0=24
19=2*7^1+5*7^0=25
20=2*7^1+6*7^0=26
explain this answer to me. how is 7 to the first power = to 10
how did u get the first six numbers

Answer 28

10=7+3

Answer 29

but 7 to the first power is not 7 + 3

Answer 30

\[10=7^1+3 \cdot 7^0=(10)_7\]

Answer 31

no 7^1 does not = 7+3
but 7^1+3*7^0=7+3

Answer 32

this is what u have......7=7^1=10

Answer 33

write u said to write it in base 7

Answer 34

\[7=1 \cdot 7^1+0 \cdot 7^0=10_7\]

Answer 35

right u said to write in base 7*

Answer 36

\[1_{10}=1 \cdot 7^0=1_7\]
the other first 5 work like this as well

Answer 37

\[2_{10}=2 \cdot 7^0=2_{7}\]

Answer 38

\[6_{10}=6 \cdot 7^0=6_7\]

Answer 39

when i am asked to find the counting numbers in a base.......what is that?

Answer 40

oh yes forgot to say one more thing you want all the a_i's to state between 0 and 7 not including 7

Answer 41

HELP???????\[78000/1+0.08\left(\begin{matrix}17 \\ 12\end{matrix}\right)\]