Perfect numbers from 1 to n

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Python Practice › Loop Programs › Program 38

Perfect numbers from 1 to n

MediumLoop Programs · 38 of 90

Write a Python program to print all perfect numbers from 1 to n.

Input
n = 10000
Output
[6, 28, 496, 8128]

Approach

  1. Find divisors in pairs up to √n to keep the loop fast.

Solution

Python
def divisor_sum(n):    total = 1    d = 2    while d * d <= n:        if n % d == 0:            total += d            if d != n // d:                total += n // d        d += 1    return total print([n for n in range(2, 10001) if divisor_sum(n) == n])
Output
[6, 28, 496, 8128]

Line by line

Line 4

Only divisors up to √n are tried.

Lines 5-8

Each divisor d comes with a partner n // d; both are added, except when they are the same number.

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Perfect number check

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Python Practice › Loop Programs › Program 37

Perfect number check

EasyLoop Programs · 37 of 90

Write a Python program to check whether a number is perfect: the sum of its factors (excluding itself) equals the number, like 28 = 1 + 2 + 4 + 7 + 14.

Input
28, 12
Output
28 is perfect
12 is not perfect (factor sum 16)

Approach

  1. Add every divisor smaller than n and compare.

Solution

Python
for n in [28, 12]:    total = 0    for d in range(1, n):        if n % d == 0:            total += d    print(n, "is perfect" if total == n else f"is not perfect (factor sum {total})")
Output
28 is perfect
12 is not perfect (factor sum 16)

Line by line

Line 3

Divisors are tested up to n – 1, so the number itself is excluded.

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Armstrong numbers from 1 to n

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Python Practice › Loop Programs › Program 36

Armstrong numbers from 1 to n

MediumLoop Programs · 36 of 90

Write a Python program to print all Armstrong numbers from 1 to n.

Input
n = 10000
Output
[1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474]

Approach

  1. Apply the Armstrong test to every number.

Solution

Python
def is_armstrong(n):    digits = str(n)    return n == sum(int(d) ** len(digits) for d in digits) print([n for n in range(1, 10001) if is_armstrong(n)])
Output
[1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474]

Line by line

Line 3

The sum of the powered digits is compared with the number itself.

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Prime factors

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Python Practice › Loop Programs › Program 34

Prime factors

MediumLoop Programs · 34 of 90

Write a Python program to find all prime factors of a number.

Input
360
Output
[2, 2, 2, 3, 3, 5]
[2, 3, 5]

Approach

  1. Divide by 2, 3, 4, … as many times as possible; every successful divisor is a prime factor.

Solution

Python
n = 360factors = []d = 2while n > 1:    while n % d == 0:        factors.append(d)        n //= d    d += 1print(factors)print(sorted(set(factors)))
Output
[2, 2, 2, 3, 3, 5]
[2, 3, 5]

Line by line

Lines 5-7

The inner loop removes a factor completely (360 → 180 → 90 → 45 for d = 2).

Line 8

Composite divisors like 4 never match, because their prime parts were already removed.

Line 10

set gives the distinct prime factors.

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Sum of primes from 1 to n

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Python Practice › Loop Programs › Program 33

Sum of primes from 1 to n

EasyLoop Programs · 33 of 90

Write a Python program to find the sum of all prime numbers from 1 to n.

Input
n = 100
Output
Sum of primes up to 100: 1060

Approach

  1. Reuse a prime test and add the primes.

Solution

Python
def is_prime(n):    return n > 1 and all(n % d for d in range(2, int(n ** 0.5) + 1)) total = 0for n in range(1, 101):    if is_prime(n):        total += nprint("Sum of primes up to 100:", total)
Output
Sum of primes up to 100: 1060

Line by line

Line 2

all(n % d ...) is True when every remainder is non-zero, meaning no divisor exists.

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Primes from 1 to n

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Python Practice › Loop Programs › Program 32

Primes from 1 to n

EasyLoop Programs · 32 of 90

Write a Python program to print all prime numbers from 1 to n.

Input
n = 50
Output
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

Approach

  1. Test each number with a helper function.

Solution

Python
def is_prime(n):    if n < 2:        return False    for d in range(2, int(n ** 0.5) + 1):        if n % d == 0:            return False    return True print([n for n in range(1, 51) if is_prime(n)])
Output
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

Line by line

Line 9

The comprehension keeps the numbers that pass the test.

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Prime check with a loop

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Python Practice › Loop Programs › Program 31

Prime check with a loop

EasyLoop Programs · 31 of 90

Write a Python program to check whether a number is prime using a loop.

Input
97, 91
Output
97 is prime
91 is not prime, divisible by 7

Approach

  1. Use for ... else: the else runs only if no divisor was found.

Solution

Python
for n in [97, 91]:    if n < 2:        print(n, "is not prime")        continue    for d in range(2, int(n ** 0.5) + 1):        if n % d == 0:            print(n, "is not prime, divisible by", d)            break    else:        print(n, "is prime")
Output
97 is prime
91 is not prime, divisible by 7

Line by line

Lines 5-8

break stops the search at the first divisor.

Lines 9-10

The else belongs to the for; it runs only when the loop finished without break.

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Factorial with a loop

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Python Practice › Loop Programs › Program 30

Factorial with a loop

EasyLoop Programs · 30 of 90

Write a Python program to calculate the factorial of a number using a loop.

Input
7
Output
7! = 5040

Approach

  1. Multiply 1 × 2 × … × n.

Solution

Python
n = 7fact = 1for i in range(1, n + 1):    fact *= iprint(f"{n}! = {fact}")
Output
7! = 5040

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Factors of a number

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Python Practice › Loop Programs › Program 29

Factors of a number

EasyLoop Programs · 29 of 90

Write a Python program to find all the factors of a number.

Input
36
Output
[1, 2, 3, 4, 6, 9, 12, 18, 36]

Approach

  1. Test every number from 1 to n; keep those that divide n exactly.

Solution

Python
n = 36factors = []for d in range(1, n + 1):    if n % d == 0:        factors.append(d)print(factors)
Output
[1, 2, 3, 4, 6, 9, 12, 18, 36]

Line by line

Line 4

n % d == 0 means d is a factor.

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Power using a loop

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Python Practice › Loop Programs › Program 28

Power using a loop

EasyLoop Programs · 28 of 90

Write a Python program to calculate base raised to a power using a for loop.

Input
base = 3, power = 4
Output
3^4 = 81 | check: 81

Approach

  1. Multiply a result by the base power times.

Solution

Python
base, power = 3, 4result = 1for _ in range(power):    result *= baseprint(f"{base}^{power} = {result}", "| check:", base ** power)
Output
3^4 = 81 | check: 81

Line by line

Line 3

The loop body runs exactly power times.

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