Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1986 based on Proper divisors.
Let S be the sum of the base 10 logarithms of all the proper divisors (all divisors of a number excluding itself) of 1000000. What is the integer nearest to S?
Integers
Divisors
Algebra
Answer: is 141.
AIME I, 1986, Question 8
Elementary Number Theory by David Burton
1000000=\(2^{6}5^{6}\) or, (6+1)(6+1)=49 divisors of which 48 are proper
\(log1+log2+log4+....+log1000000\)
\(=log(2^{0}5^{0})(2^{1}5^{0})(2^{2}5^{0})....(2^{6}5^{6})\)
power of 2 shows 7 times, power of 5 shows 7 times
total power of 2 and 5 shows=7(1+2+3+4+5+6)
=(7)(21)=147
for proper divisor taking out \(2^{6}5^{6}\)=147-6=141
or, \(S=log2^{141}5^{141}=log10^{141}=141\).
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