Answer:
Step-by-step explanation:
We can get this done by using the code
def digits(n):
count = 0
if n == 0:
return 1
while (n > 0):
count += 1
n= n//10
return count
Also, another way of putting it is by saying
def digits(n):
return len(str(n))
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print(digits(25)) # Should print 2
print(digits(144)) # Should print 3
print(digits(1000)) # Should print 4
print(digits(0)) # Should print 1
Doing this way, we've told the system to count the number of figures that exist in the number. If it's 1000 to 9999, then it records it as 4 digits. If it's 100 - 999, then it records it as 3 digits. If it's 10 - 99, it records as 2 digits. If it's 0 - 9, then it has to record it as a single digit.
Thanks
Answer:
(3,2)
Step-by-step explanation:
If you graph the equations, they cross at this point.
QUESTION:
The code for a lock consists of 5 digits (0-9). The last number cannot be 0 or 1. How many different codes are possible.
ANSWER:
Since in this particular scenario, the order of the numbers matter, we can use the Permutation Formula:–
- P(n,r) = n!/(n−r)! where n is the number of numbers in the set and r is the subset.
Since there are 10 digits to choose from, we can assume that n = 10.
Similarly, since there are 5 numbers that need to be chosen out of the ten, we can assume that r = 5.
Now, plug these values into the formula and solve:
= 10!(10−5)!
= 10!5!
= 10⋅9⋅8⋅7⋅6
= 30240.
Answer:
the answer to this question is 3x-10
