if n is an integer and n > 1, then n! is the product of n | studysoup
?problem 42e


if n is an integer and n > 1, then n! is the product of n and every other positive integer that is less than n. for example, 5! =5 • 4 • 3 • 2 • 1.
a. write 6! in standard factored form.
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_


b. write 20! in standard factored form.
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_


c. without computing the value of (20!)2 determine how many zeros are at the end of this number when it is written in decimal form. justify your answer.
technical vocab – phil 1010 lecture 2 key: terms – \_\_\_ topics – \_\_\_ important – \_\_\_  what is an argument? o arguments are statements that represent an inference. o premises are statements that support a conclusion.  what is an inference? o inferences are thoughts that we believe is true.  inferences can take place conscious or unconsciously. o inferences are implicit while arguments are explicit. o implicit thoughts are implied, suggested, or understood but not directly expressed. o explicit thoughts are usually overly expressed.  sentences & statements o sentences are grammatically correct string of words. o declarative sentences can be either true or fa
we know that if a number ends with n zeros then its square will have 2n zeroes conversely, if square of a number have 2n zeros at their ends then the number will have n zeroesthe square of a number ends 10 zeroes, then the number will have `(10)/(2)` = 5 zeroes Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4]. use

summation of series of natural numbers -


 

- wherein

sum of first n natural numbers



 

 and

 

summation of series of natural numbers -


 

- wherein

sum of  squares of first n natural numbers



 

 

3+7+14+24+37 15k

let tn=an2+bn+c            because difference is in a.p: [4,7,10,13]

at n=1     (i)

    n=2      (ii)

    n=3        (iii)

    (ii)-(i)        (iv)

    (iii)-(ii)       (v)

    (v)-(iv)      

                    

 

from (i) 

from (ii) 







 



      

 

put n=15   

               

               

               

               

 




option 1)

126

this option is incorrect

option 2)

122

this option is correct

option 3)

81

this option is incorrect

option 4)

119

this option is incorrect sum of a g.p. series is represented by the formula, , when |r|<1. ‘sn’ represents the sum of the g.p. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.here,a = 1r = (ratio between the n term and n-1 term) n terms∴ ⇒∴ (a) for digit



expression : 5n + 1

where n = no of digits

for 5 digits

no of segments required = 5(5) + 1 = 25 + 1 = 26

for 10 digits

no of segments required = 5(10) + 1 = 50 + 1 = 51

for 100 digits

no of segments required = 5(100) + 1 = 500 + 1 = 501

(b) for digit



expression : 3n + 1

where n = no of digits

for 5 digits

no of segments required = 3(5) + 1 = 15 + 1 = 16

for 10 digits

no of segments required = 3(10) + 1 = 30 + 1 = 31

for 100 digits

no of segments required = 3(100) + 1 = 300 + 1 = 301

(c) for digit



expression : 5n + 2

where n = no of digits

for 5 digits

no of segments required = 5(5) + 2 = 25 + 2 = 27

for 10 digits

no of segments required = 5(10) + 2 = 50 + 2 = 52

for 100 digits

no of segments required = 5(100) + 2 = 500 + 2 = 502 Here, this should do it. I just replaced the

 
```
System.out.println(isPrime(n));

```
 part, which would print true or false, with an if

 
```
if (isPrime(n))
    System.out.println("The number is prime.\n");
else 
    System.out.println("The number is not prime.\n");

```
 Here's the code:

 
```
import java.util.Scanner;
public class PrimeNumber{
    public static void main(String args[]){
        int n;
        Scanner Prime = new Scanner(System.in);
        System.out.println(" Please enter a number: ");
        n = Prime.nextInt();
        if (isPrime(n))
            System.out.println("The number is prime.\n");
        else 
            System.out.println("The number is not prime.\n");
    }

    static boolean isPrime(int n) {
        if (n%2==0)return false;
        for(int i=3;i*i<=n;i+=2) {
            if(n%i==0)
                return false;
        }
        return true;
    }
} 

```
 This is a trivial task. You should probably read your coursebook/tutorials, I bet you'll find the answers there. ```
[b]#include[/b] <stdio.h> 

[b]void[/b] [b]foo[/b]([b]int[/b] n)
{
   [b]if[/b] ( n >= 0 )
   {
	  [b]printf[/b]("%d ", n);
	  [b]foo[/b](n - 1);
   }
}

[b]void bar[/b]([b]int[/b] n)
{
   [b]if[/b] ( n >= 0 )
   {
	  [b]bar[/b](n - 1);
	  [b]printf[/b]("%d ", n);
   }
}

[b]int main[/b]() 
{
   [b]int[/b] value = 5;
   [b]foo[/b](value); [b]putchar[/b]('\n');
   [b]bar[/b](value); [b]putchar[/b]('\n');
   [b]return[/b] 0; 
}

/* my output
5 4 3 2 1 0 
0 1 2 3 4 5 
*/
``` if a number is divisible by 55 than it must alsobe divisible by factors of 55 that is 5 and 11.∵ aabb is divisible by 5∴ by divisibility test of 5 b must be 0 or 5.therefore (c) option is correct. applying and factorization, we get, also changing by reversing n and dnote: - try to factorize that term which help in reducing expression. [image] endermaz41:

didn't know what to put in the error portion

first of all, err(…) is a value, not a type. since err is a variant ("constructor" if you like) of enum result, if you invoke it, you get a value of type result. you won't get a value of the error type itself, much less a type.

…