how to generate a random 10 digit integer with 1 at last place?

savecancel

There are many ways.

Here is a simple one using randint:

sage: 10 * randint(10^8, 10^9 - 1) + 1

We can simply generate a random nine-digits integer, then insert a one at the "last" place. Here is a possibility using the random package:

sage: import random                                                                                                                 
sage: a = random.choice(range(10^8, 10^9)) * 10 + 1                                                                                 
sage: a                                                                                                                             
9768350121

Or we offer directly the corresponding range to the random.choice method:

sage: random.choice(range(10^9 + 1, 10^10, 10))                                                          
8455325531

Here, range(10^9 + 1, 10^10, 10)is a range-object, which "consumed" points to the integers taken from $10^9+1$ with step $10$ going up to maximally (and excluding) $10^{10}$. Then random.choice called with this range instance picks one random element of the corresponding list.

There is also randrange from the random library which is good for that task:

sage: import random
sage: random.randrange(10^9+1, 10^10, 10)
6071633551