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Serhud [2]
3 years ago
15

Complete the method, print Multiples(), that takes in a positive integer n, and another positive integer, max. Print out all the

multiples of n that are less than or equal to max. Print each multiple on a separate line. For example, if the method is called with n
Computers and Technology
1 answer:
musickatia [10]3 years ago
3 0

Answer:

public static void printMultiples(int n, int max){

    for (int i=1; i<=max; i++){

        if(i%n == 0)

            System.out.println(i);

    }

}

Explanation:

Create a method called printMultiples that takes two parameters n and max

Inside the method, create a for loop that iterates from 1 to max. If i, a number between 1 and max, % n is equal to 0, that means the number is a multiple of n, print the number.

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You have heard that the Linux distribution your are going to install to run a new application runs best on a RISC architecture.
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Answer:

SPARC .

Explanation:

When the user has understood they're required to install the linux distribution to set up a specific program works better on such a RISC architecture. SPARC must therefore find the device from which it would work that following distribution for the Linux. It is a RISC and ISA that was previously designed through both Fujitsu and Sun.

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There is always this thing that comes out and I can't answer one's question. How do I undo this?
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You should enter your email address
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Read 2 more answers
Edhesive assignment 7 calendar
ikadub [295]

Answer:

def leap_year(y):

 if y % 4 == 0:

     return 1

 else:

     return 0

def number_of_days(m,y):

 if m == 2:

     return 28 + leap_year(y)

 elif m == 1 or m == 3 or m == 5 or m == 7 or m == 8 or m ==10 or m == 12:

     return 31

 elif m == 4 or m == 6 or m == 9 or m == 11:

     return 30

def days(m,d):

 if m == 1:

     return 0 + d

 if m == 2:

     return 31 + d

 if m == 3:

     return 59 + d

 if m == 4:

     return 90 + d

 if m == 5:

     return 120 + d

 if m == 6:

     return 151 + d

 if m == 7:

     return 181 + d

 if m == 8:

     return 212 + d

 if m == 9:

     return 243 + d

 if m == 10:

     return 273 + d

 if m == 11:

     return 304 + d

 if m == 12:

     return 334 + d

def days_left(d,m,y):

 if days(m,d) <= 60:

     return 365 - days(m,d) + leap_year(y)

 else:

     return 365 - days(m,d)

print("Please enter a date")

day=int(input("Day: "))

month=int(input("Month: "))

year=int(input("Year: "))

choice=int(input("Menu:\n1) Calculate the number of days in the given month.\n2) Calculate the number of days left in the given year.\n"))

if choice == 1:

 print(number_of_days(month, year))

if choice == 2:

 print(days_left(day,month,year))

Explanation:

Hoped this helped

5 0
3 years ago
There is a simple method for constructing a magic square for any odd value of n:
DaniilM [7]

Answer:

See Explaination

Explanation:

class MagicSquare():

def __init__(self,side):

self.side=side

self.two_dimension=[]

for row in range(1,side+1):

row_line=[]

for col in range(1,side+1):

row_line.append(0)

self.two_dimension.append(row_line)

def display(self):

row=0

col=int((self.side-1)/2)

for i in range(1,self.side**2+1):

self.two_dimension[row][col]=i

row-=1

col+=1

if row==-1:

row=self.side-1

if col==self.side:

col=0

if self.two_dimension[row][col]==0:

continue

else:

row+=1

if row==self.side:

row==0

for line in self.two_dimension:

for num in line:

print("{0:>3}".format(num),end=" ")

print()

def main():

for i in range(1,14,2):

square=MagicSquare(i)

square.display()

print("----------------------------------------------------")

if __name__ == '__main__':

main()

3 0
3 years ago
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