Proper divisors

Number Theory Level pending

What is the sum of the largest proper divisors of each of the integers from $2$ to $100$ ?

Note: a proper divisor of $n$ is a number diferent of $n$

The answer is 1690.

2 solutions

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Paola Ramírez
Jan 23, 2015

The maximum proper divisor of each number will be the number divided by the least divisor prime.

Case 1:

$\frac{1}{2}(2+4+6...+100)=1275$

Case2:

$\frac{1}{3}(3+9+15...+99)=289$

Case 3:

$\frac{1}{5}(5+25+35...+95)=73$

Case 4: $\frac{1}{7}(7+49+77+91)=32$

Case prime:

There are $21$ primes from $1$ to $100$ this maximum proper divisor is $1$

$\therefore$ the sum of the maximum prime divisors is $1275+289+73+32+21=\boxed{1690}$

Nice problem. I had to think for a moment about how to deal with 1, as it has no proper divisors and hence no maximum proper divisor. It seemed logical to just assign 0 as this value, but you may want to consider going from 2 to 100 rather than 1 to 100 to avoid this minor issue. Just a thought. :)

Brian Charlesworth - 6 years, 4 months ago

i like it :)

Avinash Singh - 6 years, 4 months ago

Brock Brown
Jan 27, 2015

Lazy Python approach:

def big_proper(number):
    test = number / 2
    while number % test != 0:
        test -= 1
    return test
total = 0
for i in xrange(2, 101):
    total += big_proper(i)
print "Answer:", total

Running time: 0.00277 seconds

Even lazier! :)

Bill Bell - 6 years, 4 months ago

Proper divisors

The answer is 1690.

2 solutions

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