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Breaking Down the Info:

As our factorials get higher, the number of trailing zeros would get higher and the number would start jumping up when we reach certain thresholds.

For example, from 99! to 100! we created 2 more trailing zeros, hence a 2-number jump. We would like to know the lowest number of trailing zeros needed for a 4-number jump.

To create a trailing zero, we need a factor of 2 and a factor of 5. The factor of 5 is much less common than the factor of 2, so we only need to gather factors of 5's to find the number of trailing zeros.

Therefore using the logic above, we will need \(5^4\) as the next factorial so that we instantly jump 4 trailing zeros. \(5^4 = 625\). Hence 625! is our threshold.

Finally, we need to count the number of trailing zeros in 625 or the number of multiples of 5s. We can count the layers to make this easier:

Multiple of 5's: 625/5 = 125.

Multiple of 25's: 625/25 = 25

Multiple of 125's: 625/125 = 5.

Multiple of 625's: 625/625 = 1.

Adding everything up gives us 156. Our count of trailing zeros jumped from 152 to 156 with 624! to 625!, then n here is 153.

Answer: B
­JerryAtDreamScore can you please explain the thought process behind identifying that we need to find the lowest number of trailing zeroes to get a 4 number jumper?
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