Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1986 based on Proper divisors.

Proper Divisor – AIME I, 1986


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?

  • is 107
  • is 141
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisors

Algebra

Check the Answer


But try the problem first…

Answer: is 141.

Source
Suggested Reading

AIME I, 1986, Question 8

Elementary Number Theory by David Burton

Try with Hints


First hint

1000000=\(2^{6}5^{6}\) or, (6+1)(6+1)=49 divisors of which 48 are proper

\(log1+log2+log4+….+log1000000\)

\(=log(2^{0}5^{0})(2^{1}5^{0})(2^{2}5^{0})….(2^{6}5^{6})\)

Second Hint

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

Final Step

for proper divisor taking out \(2^{6}5^{6}\)=147-6=141

or, \(S=log2^{141}5^{141}=log10^{141}=141\).

Subscribe to Cheenta at Youtube