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Looping-3Read an integer natural number N. Print the range of numbers(1 to N) followed by the string 'VIT' if the number is a multiple of 5 else print 'VIT'.Sample Input5Sample OutputVITVITVITVIT5 VIT

Question

Looping-3Read an integer natural number N. Print the range of numbers(1 to N) followed by the string 'VIT' if the number is a multiple of 5 else print 'VIT'.Sample Input5Sample OutputVITVITVITVIT5 VIT

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Solution

The problem is asking to print a range of numbers from 1 to N, where N is a natural number input by the user. If the number is a multiple of 5, it should print the number followed by the string 'VIT'. If the number is not a multiple of 5, it should just print 'VIT'.

Here is a step-by-step solution in Python:

  1. First, we need to read an integer from the user. We can use the input() function for this, which reads a line of text from the user. Since this function returns a string, we need to convert this string to an integer using the int() function.
N = int(input())
  1. Next, we need to loop over the range of numbers from 1 to N. In Python, we can use the range() function to generate this range of numbers. Note that the range() function generates numbers up to, but not including, the end value, so we need to add 1 to N.
for i in range(1, N+1):
  1. Inside the loop, we need to check if the current number i is a multiple of 5. In Python, we can use the modulus operator % to find the remainder of the division of i by 5. If the remainder is 0, then i is a multiple of 5.
if i % 5 == 0:
  1. If i is a multiple of 5, we print i followed by the string 'VIT'. In Python, we can use the print() function to print to the console. We can use the str() function to convert i to a string, and then use the + operator to concatenate this string with ' VIT'.
print(str(i) + ' VIT')
  1. If i is not a multiple of 5, we just print 'VIT'.
else:
    print('VIT')

Here is the complete code:

N = int(input())
for i in range(1, N+1):
    if i % 5 == 0:
        print(str(i) + ' VIT')
    else:
        print('VIT')

This problem has been solved

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