Let N be the largest integer such that 12^N divides n!
If 12^N also divides (n+1)! what is the smallest possible value of n?
12 = 3 * 2^2
We need n! such that 12 can divide it. The smallest value n could take is 4. Any factorial below doesn’t have 2^2
4! = 4 * 3 * 2 * 1 = 24
12^N has to divide 4!, so N has to be 1.
12^1 divides n!. Let’s check if it divides (n+1)!.
12^1 does divide 5!=60.
n=4
This would still be under the assumption that N=1. If N=2,3,4… value of 12^N also increases, and then the numbers would change as well
From my understanding of the question, the goal is to minimise n as much as possible and if that means taking N=1 I can do that right?
This condition is satisfied. So what is wrong with taking N=1?
You’re right if we can assume to minimize n at the smallest possible value.
But if we can leave it to interpretation,
Suppose N=3, we will get 12^3=1728, and then smallest possible of n would be 13 in this case.
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