Question

how many 2s, 3s and 5s are there in 100 factorial...
wrong answers to be reported and right ones will be marked as brainliest...​

Answer 1

Step-by-step explanation:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. It can be solved in two ways -

Let’s look at how trailing zeros are formed in the first place. A trailing zero is formed when a multiple of 5 is multiplied with a multiple of 2. Now all we have to do is count the number of 5’s and 2’s in the multiplication.

Each pair of 2 and 5 will cause a trailing zero. Since we have only 24 5’s, we can only make 24 pairs of 2’s and 5’s thus the number of trailing zeros in 100 factorial is 24.

2. The question can also be answered using the simple formula given below:

The above formula gives us the exact number of 5s in n! because it will take care of all multiples of 5 which are less than n. Not only that it will take care of all multiples of 25, 125, etc. (higher powers of 5).

Tip: Instead of dividing by 25, 125, etc. (higher powers of 5); it would be much faster if you divided by 5 recursively.

Let us use this to solve a few examples:

Q) What is the number of trailing zeroes in 100! ?

[100/5] = 20

Now we can either divided 100 by 25 or the result in the above step i.e. 20 by 5.

[20/5 ]= 4. It is less than 5, so we stop here.

The answer is - 20+ 4 = 24 (direct answer in just few secs)

Q) What is the number of trailing zeroes in 200! ?

[200/5] = 40

Now we can either divided 200 by 25 or the result in the above step i.e. 40 by 5.

[40/5 ]= 8

[8/5]=1. It is less than 5, so we stop here.

The answer is - 40 + 8 + 1= 49

Hope it will help you

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Answer 2

Given:

100!

To Find:

how many 2s, 3s and 5s are there

Solution:

For the above sum, we should know how to calculate for any factorial, how many a specific natural number is in it.

Say we have 50! and we need to find the number of 5s for that we will divide 50 consecutively with the increasing power of 5 and adding them, like

$5s=\frac{50}{5} +\frac{50}{25}\\=10+2\\=12$

so the number of 5s in 50! is 12, we stop taking the power after the denominator will increase the numerator and we also take the approximate values when dividing,

Similarly for 100!

Number of 2s,

$2s=\frac{100}{2} +\frac{100}{4} +\frac{100}{8} +\frac{100}{16} +\frac{100}{32} +\frac{100}{64}\\=50+25+12+6+3+1\\=97$

Number of 3s,

$3s=\frac{100}{3} +\frac{100}{9} +\frac{100}{27}+ \frac{100}{81}\\=33+11+3+1\\=48$

Number of 5s,

$5s=\frac{100}{5} +\frac{100}{25} \\=20+4\\=24$

Hence, there are 97 2s, 48 3s and 24 5s in 100!