Prime factorization

[DiwakarV169]

If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

[VijayS816]

Least Value of N is 11.

990 - 3*3*2*5*11

[DiwakarV169]

Why is it 11? why is it not 3?

[RonPurewal]

take a look at the problem statement again -- you need the smallest n such that the product of all the integers from 1 to n is divisible by 990.

if n is 3, then this is clearly not true (the product of all the integers from 1 to 3 is just 6, which is definitely not divisible by 990).

[RonPurewal]

in the plainest possible terms, the point is that...
...1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 is NOT divisible by 990,
...1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 · 11 IS divisible by 990.

thus the smallest such "n" is 11.