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How many numbers are there which only contain digits 4 and 7 in them?
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I wanna know how many numbers n are there which only contain digits 4 and 7 in them, where n (1  ≤  n  ≤ 10 9).
Ex: 4, 7, 44, 47, 74, 77, ....
I am trying to find a general equation to compute the numbers, given how many digits which is 2 in this case and the range which (1  ≤  n  ≤ 10 9).
combinatorics discrete-mathematics
asked 53 mins ago
Basma Ashour
134
add a comment |Â
up vote
2
down vote
favorite
I wanna know how many numbers n are there which only contain digits 4 and 7 in them, where n (1  ≤  n  ≤ 10 9).
Ex: 4, 7, 44, 47, 74, 77, ....
I am trying to find a general equation to compute the numbers, given how many digits which is 2 in this case and the range which (1  ≤  n  ≤ 10 9).
combinatorics discrete-mathematics
asked 53 mins ago
Basma Ashour
134
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I wanna know how many numbers n are there which only contain digits 4 and 7 in them, where n (1  ≤  n  ≤ 10 9).
Ex: 4, 7, 44, 47, 74, 77, ....
I am trying to find a general equation to compute the numbers, given how many digits which is 2 in this case and the range which (1  ≤  n  ≤ 10 9).
combinatorics discrete-mathematics
asked 53 mins ago
Basma Ashour
134
I wanna know how many numbers n are there which only contain digits 4 and 7 in them, where n (1  ≤  n  ≤ 10 9).
Ex: 4, 7, 44, 47, 74, 77, ....
I am trying to find a general equation to compute the numbers, given how many digits which is 2 in this case and the range which (1  ≤  n  ≤ 10 9).
combinatorics discrete-mathematics
combinatorics discrete-mathematics
asked 53 mins ago
Basma Ashour
134
asked 53 mins ago
Basma Ashour
134
asked 53 mins ago
Basma Ashour
134
asked 53 mins ago
Basma Ashour
134
asked 53 mins ago
Basma Ashour
134
134
add a comment |Â
add a comment |Â
3 Answers
3
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oldest
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up vote
4
down vote
Well, how do you create a number of $i$ digits where all the digits are $4$ and $7$? For each of the $i$ digits you have to choose if it will be $4$ or $7$, so for each digit you have $2$ options. Hence the number of such numbers is $2^i$. Now, how is that related to your problem? You are interested in finding how many numbers are there that contain only digits $4$ and $7$ when the number of digits is between $1$ and $9$. Well, there are $2^1$ numbers with $1$ digit, $2^2$ with $2$ digits and so on. So the final answer is:
$sum_k=1^9 2^k=2(2^9-1)$
I used the formula for geometric sum.
answered 46 mins ago
Mark
3,995114
I got it, thanks for the great and straightforward explanation!
– Basma Ashour
41 mins ago
add a comment |Â
up vote
1
down vote
There are $2$ with length $1$, $2^2$ with length $2$, $2^3$ with length $3$, et cetera.
That observation leads to a total of:$$2+2^2+2^3+cdots+2^9=2^10-2$$
The last equation on base of: $$(2-1)(2+2^2+2^3+cdots+2^9)=(2^2+2^3+cdots+2^10)-(2+2^2+2^3+cdots+2^9)=2^10-2$$
answered 39 mins ago
drhab
90.1k542123
Nice explanation, thanks for your help.
– Basma Ashour
32 mins ago
add a comment |Â
up vote
0
down vote
You may consider the following recursive formula:
$$f(n) = 2f(n-1)+f(n-1), f(1) = 2$$
answered 38 mins ago
Maged Saeed
1597
1
It's nice to think about it recursively, thank you.
– Basma Ashour
33 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Well, how do you create a number of $i$ digits where all the digits are $4$ and $7$? For each of the $i$ digits you have to choose if it will be $4$ or $7$, so for each digit you have $2$ options. Hence the number of such numbers is $2^i$. Now, how is that related to your problem? You are interested in finding how many numbers are there that contain only digits $4$ and $7$ when the number of digits is between $1$ and $9$. Well, there are $2^1$ numbers with $1$ digit, $2^2$ with $2$ digits and so on. So the final answer is:
$sum_k=1^9 2^k=2(2^9-1)$
I used the formula for geometric sum.
answered 46 mins ago
Mark
3,995114
I got it, thanks for the great and straightforward explanation!
– Basma Ashour
41 mins ago
add a comment |Â
up vote
4
down vote
Well, how do you create a number of $i$ digits where all the digits are $4$ and $7$? For each of the $i$ digits you have to choose if it will be $4$ or $7$, so for each digit you have $2$ options. Hence the number of such numbers is $2^i$. Now, how is that related to your problem? You are interested in finding how many numbers are there that contain only digits $4$ and $7$ when the number of digits is between $1$ and $9$. Well, there are $2^1$ numbers with $1$ digit, $2^2$ with $2$ digits and so on. So the final answer is:
$sum_k=1^9 2^k=2(2^9-1)$
I used the formula for geometric sum.
answered 46 mins ago
Mark
3,995114
I got it, thanks for the great and straightforward explanation!
– Basma Ashour
41 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Well, how do you create a number of $i$ digits where all the digits are $4$ and $7$? For each of the $i$ digits you have to choose if it will be $4$ or $7$, so for each digit you have $2$ options. Hence the number of such numbers is $2^i$. Now, how is that related to your problem? You are interested in finding how many numbers are there that contain only digits $4$ and $7$ when the number of digits is between $1$ and $9$. Well, there are $2^1$ numbers with $1$ digit, $2^2$ with $2$ digits and so on. So the final answer is:
$sum_k=1^9 2^k=2(2^9-1)$
I used the formula for geometric sum.
answered 46 mins ago
Mark
3,995114
Well, how do you create a number of $i$ digits where all the digits are $4$ and $7$? For each of the $i$ digits you have to choose if it will be $4$ or $7$, so for each digit you have $2$ options. Hence the number of such numbers is $2^i$. Now, how is that related to your problem? You are interested in finding how many numbers are there that contain only digits $4$ and $7$ when the number of digits is between $1$ and $9$. Well, there are $2^1$ numbers with $1$ digit, $2^2$ with $2$ digits and so on. So the final answer is:
$sum_k=1^9 2^k=2(2^9-1)$
I used the formula for geometric sum.
answered 46 mins ago
Mark
3,995114
answered 46 mins ago
Mark
3,995114
answered 46 mins ago
Mark
3,995114
answered 46 mins ago
Mark
3,995114
3,995114
I got it, thanks for the great and straightforward explanation!
– Basma Ashour
41 mins ago
add a comment |Â
I got it, thanks for the great and straightforward explanation!
– Basma Ashour
41 mins ago
I got it, thanks for the great and straightforward explanation!
– Basma Ashour
41 mins ago
I got it, thanks for the great and straightforward explanation!
– Basma Ashour
41 mins ago
add a comment |Â
up vote
1
down vote
There are $2$ with length $1$, $2^2$ with length $2$, $2^3$ with length $3$, et cetera.
That observation leads to a total of:$$2+2^2+2^3+cdots+2^9=2^10-2$$
The last equation on base of: $$(2-1)(2+2^2+2^3+cdots+2^9)=(2^2+2^3+cdots+2^10)-(2+2^2+2^3+cdots+2^9)=2^10-2$$
answered 39 mins ago
drhab
90.1k542123
Nice explanation, thanks for your help.
– Basma Ashour
32 mins ago
add a comment |Â
up vote
1
down vote
There are $2$ with length $1$, $2^2$ with length $2$, $2^3$ with length $3$, et cetera.
That observation leads to a total of:$$2+2^2+2^3+cdots+2^9=2^10-2$$
The last equation on base of: $$(2-1)(2+2^2+2^3+cdots+2^9)=(2^2+2^3+cdots+2^10)-(2+2^2+2^3+cdots+2^9)=2^10-2$$
answered 39 mins ago
drhab
90.1k542123
Nice explanation, thanks for your help.
– Basma Ashour
32 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There are $2$ with length $1$, $2^2$ with length $2$, $2^3$ with length $3$, et cetera.
That observation leads to a total of:$$2+2^2+2^3+cdots+2^9=2^10-2$$
The last equation on base of: $$(2-1)(2+2^2+2^3+cdots+2^9)=(2^2+2^3+cdots+2^10)-(2+2^2+2^3+cdots+2^9)=2^10-2$$
answered 39 mins ago
drhab
90.1k542123
There are $2$ with length $1$, $2^2$ with length $2$, $2^3$ with length $3$, et cetera.
That observation leads to a total of:$$2+2^2+2^3+cdots+2^9=2^10-2$$
The last equation on base of: $$(2-1)(2+2^2+2^3+cdots+2^9)=(2^2+2^3+cdots+2^10)-(2+2^2+2^3+cdots+2^9)=2^10-2$$
answered 39 mins ago
drhab
90.1k542123
answered 39 mins ago
drhab
90.1k542123
answered 39 mins ago
drhab
90.1k542123
answered 39 mins ago
drhab
90.1k542123
90.1k542123
Nice explanation, thanks for your help.
– Basma Ashour
32 mins ago
add a comment |Â
Nice explanation, thanks for your help.
– Basma Ashour
32 mins ago
Nice explanation, thanks for your help.
– Basma Ashour
32 mins ago
Nice explanation, thanks for your help.
– Basma Ashour
32 mins ago
add a comment |Â
up vote
0
down vote
You may consider the following recursive formula:
$$f(n) = 2f(n-1)+f(n-1), f(1) = 2$$
answered 38 mins ago
Maged Saeed
1597
1
It's nice to think about it recursively, thank you.
– Basma Ashour
33 mins ago
add a comment |Â
up vote
0
down vote
You may consider the following recursive formula:
$$f(n) = 2f(n-1)+f(n-1), f(1) = 2$$
answered 38 mins ago
Maged Saeed
1597
1
It's nice to think about it recursively, thank you.
– Basma Ashour
33 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You may consider the following recursive formula:
$$f(n) = 2f(n-1)+f(n-1), f(1) = 2$$
answered 38 mins ago
Maged Saeed
1597
You may consider the following recursive formula:
$$f(n) = 2f(n-1)+f(n-1), f(1) = 2$$
answered 38 mins ago
Maged Saeed
1597
answered 38 mins ago
Maged Saeed
1597
answered 38 mins ago
Maged Saeed
1597
answered 38 mins ago
Maged Saeed
1597
1597
1
It's nice to think about it recursively, thank you.
– Basma Ashour
33 mins ago
add a comment |Â
1
It's nice to think about it recursively, thank you.
– Basma Ashour
33 mins ago
1
1
It's nice to think about it recursively, thank you.
– Basma Ashour
33 mins ago
It's nice to think about it recursively, thank you.
– Basma Ashour
33 mins ago
add a comment |Â
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