'Trailing zeros' concept

GRE Quant Problem Solving
1

I was going through the quant mountain and got stuck on the ‘trailing zeros’ topic.

I get it that whatever the power of 10 is, gives the

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number of trailing zeros, and to find the ‘number of numbers’ or ‘number of power’ a prime number is needed in the denominator. However, when 10 is factorized it gives us 2*5. How can we omit 2 and write 5^x.
Could you elaborate on this or recommend a video to understand this?

2

Think of it this way: would you have more 2s or 5s in 270!?

3

I still don’t get it. Does 2s or 5s mean having 2s or 5s in the units place or does it mean digits 2s or 5s in the value of ‘270!’ ? What does it mean to have 2s or 5s?

4

If we spread out 270!, it looks like the below right?

270 \times 269 \times 268 \times 267 ... \times 3 \times 2 \times 1

In this, do we have more multiples of 2 or 5?

To learn this concept properly, watch the following:

# of Numbers in Factorials

Shortcut for Large Factorials

# of Numbers in Factorials II

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