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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Proper divisors | AIME I, 1986 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Proper divisors.

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

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