Unknowm

Question

If n = 15 × 30 × 45 × 60............1500, what will be the number of zeros at the end of n?

a) 24                   c) 67

b) 97                   d) 24

Comments

Please check your options, you have given option a & d same.

Answer 1

n = 15 x 30 x 45 x 60 x …….. x 1500.

Let’s calculate the number of terms in the given series.

1500 = 15 + (n-1)15 => n = 100

Total 100 terms.

 

n can also be written as : 

= (15x1) x (15x2) x (15x3) x (15x4) x ………. x (15x100)

=  $15^{100}$ x (1 x 2 x 3 x 4 x 5 x …….. x 100)

= $15^{100}$ x 100!

= $3^{100}$ x $5^{100}$ x 100!

 

As, 10 = 2 x 5

 

Let’s calculate the number of 2’s and 5’s in 100!.

Counting number of 2’s (considering only those terms which can give 2 & rest i will consider as a

constant, K) :

100! = 1 x 2 x 3 x 4 x ……. 99 x 100

= 2 x 4 x 6 x 8 x 100 x K 

= $2^{50}$ x (1 x 2 x 3 x 4 x ……... 50) x K

= $2^{50}$ x (2 x 4 x 6 x 8 x ………. 50) x K 

= $2^{50}$ x $2^{25}$ x (1 x 2 x 3 x 4 x ……. 25) x K

= $2^{75}$ x (2 x 4 x 6 x 8 x 24) x K

= $2^{75 }$ x $2^{12}$ x (1 x 2 x 3 x 4 ….…. x 12) x K

= $2^{87}$ x (2 x 4 x 6 x 8 x 10 x 12) x K

= $2^{87}$ x $2^{6}$ x (1 x 2 x 3 x 4 x 5 x 6) x K 

= $2^{93}$ x (2 x 4 x 6) x K

= $2^{93}$ x $2^{3}$ x (1 x 2 x 3) x K

= $2^{96}$ x 2 x K

= $2^{97}$ x K

So, 2 will appear 97 times.

 

Counting number of 5’s (considering only those terms which can give 5) :

Terms 5, 10, 15, 20, 30, 35, 40, 45, 55, 60, 65, 70, 80, 85, 90, 95 will give 5, only 1 times.

From these terms we will get 5, sixteen times.

Terms 25, 50, 75, 100 will give 5, only 2 times

From these terms we will get 5, eight times.

So, total number of 5’s is 16 + 8 = 24.

 

Now, n = $3^{100}$ x $5^{100}$ x ( $2^{97}$ x $5^{24}$ x K )

= $2^{97}$ x $3^{100}$ x $5^{124}$ x K

 

Clearly, the number of times 5 is appearing in the expression is greater than the number of times 2 will

appear in the expression. So, 2 is the limiting factor.

The number of time 2 is appearing in the expression, will give the number of times 0 will trail in the 

expression.

 

So, the number of times 0 at the end of n is 97.

  

Comments

your expression of n is not correct. it should be $15^{100}*100 !$.

In n we have total 100 terms. 50 which ends with 0, and rest with 5. now for a moment forget about integers which ends with 5 and only consider the integers which ends with 0. so from 30,60,90,...,1500 take 10 common so you'll be left with $10^{50}*(3*6*9...*150)$,

So from this you can conclude that resultant integer will have atleast 50 zeroes at trail.

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@parth023 Thanks!! I have done the necessary corrections.

Answer 2

take 15 common from each term we get

$$n = (15\cdot1) \times (15\cdot2) \times \dots \times (15 \cdot100) \rightarrow 100 terms$$

$$n = 15^{100}(1\cdot2\cdot3 \dots 100)$$

$$n = 3^{100}5^{100} 100!$$

the highest power of a prime $p$ dividing $N!$ is given by

$$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$

$$\text{power of 5 in 100!} = s_p(100!) =\left \lfloor \frac{100}{5} \right \rfloor + \left \lfloor \frac{100}{5^2} \right \rfloor = 24$$

similarly,

$$\text{power of 2 in 100!} = s_p(100!) = 97$$

so we get

$$n = 3^{100}5^{100} 100!$$
$$n = 3^{100}5^{100}(2^{97}5^{24} \cdot \text{something})$$
$$n = \text{something} \cdot5^{100}\cdot5^{24}\cdot2^{97}$$

$$\text{so power of 10 is} = min(100+24,97) = 97$$

there are $97$ zeros at the end

Comments

What is the meaning of sp( 100! ) ? What does it resembles?

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https://math.stackexchange.com/q/141196/534029

see the above link

$S_p(x!) \rightarrow \text{returns the power of prime p in x!}$