How many numbers are there which only contain digits $4$ and $7$ in them? [closed]

I wanna know how many numbers $$n$$ are there which only contain digits $$4$$ and $$7$$ in them, where $$1 ≤ n ≤ 10^9$$.

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 is $$1 ≤ n ≤ 10^9$$ in this case.

closed as off-topic by user21820, Jyrki Lahtonen, ArsenBerk, Namaste, stressed outOct 5 '18 at 22:50

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• It's unclear what you intend n to represent. – Daniel R Hicks Oct 5 '18 at 14:31
• I really recommend that you try to solve this entirely on you own. This is a excellent introduction to problem solving and is simple enough that anyone no matter how limitted their math skill should be able to figure it out. As a starter try solving for $1 \le n \le 10^k$ were $k = 1,2,3$ first. You will really feel good about yourself when you figure it out on your own. So I'm not giving any hint. Good luck and enjoy. – fleablood Oct 5 '18 at 17:09

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.

• I got it, thanks for the great and straightforward explanation! – Basma Ashour Oct 5 '18 at 8:25

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: $$(2-1)(2+2^2+2^3+\cdots+2^9)=(2^2+2^3+\cdots+2^{10})-(2+2^2+2^3+\cdots+2^9)=2^{10}-2$$

• Nice explanation, thanks for your help. – Basma Ashour Oct 5 '18 at 8:34

One can also derive a recurrence formula for such problem.

• Notice that the number of these numbers less than $$100$$ is $$S=\{\{4,7\},\{44,47,74,77\}\}$$ Notice also that the set $$S$$ is divided into two subsets, call them $$s_1,s_2$$

• For a bound equal to $$1000$$, take $$s_2$$ and augment $$4,7$$ to each number, add them to the set $$S$$ call it $$S_{new} = \{\{4,7\},\{44,47,74,77\},\{444,447,474,477,744,747,774,777\}\}$$ The number of elements of $$S_{new}$$ is $$2*|s_2|+6=14$$. Or more generally: $$|S_{new}| = 2*(|S|-|S_{old}|)+|S|, S_{old} = s_1.$$

• For numbers below $$10^4$$, take the last subset and augment 7,4 to it and add to the set. The total number would be: $$2*8+14 = 30$$. Or $$|S_{new}| = 2*(|S|-|S_{old}|)+|S|, S_{old} = s_1+s_2$$ and so on.

I hope that the recurrence is clear now:

$$f(n) = 2[f(n-1)-f(n-2)]+f(n-1)$$ Or equivalently:

$$f(n) = 3f(n-1)-2f(n-2)$$

Solving this recurrence by Wolfram Alpha check here, the answer is:

$$f(n) = 2(2^n-1)$$

Which is identical to the other answers.

• It's nice to think about it recursively, thank you. – Basma Ashour Oct 5 '18 at 8:34