How many integers greater than I0000 have no repeated digits

Solution

We can write 10 = 2*5
So 10¹⁰ becomes

(2*5)¹⁰=2¹⁰ * 5¹⁰

There\'s 21 divisors right there (all 10 powers of 2, all 10 powers of 5, and 1).
The rest come from combinations of the two: 2*5, 2*5², 2*5³, 2⁵*5¹⁰, etc...
That\'s another 100 (10 choices * 10 choices).
2¹⁰ * 5¹⁰
= (10 + 1)(10 + 1 )
= 11 * 11
= 121

So there are 121 distinct positive divisors 10¹⁰ have.