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May 13, 2020

Proper divisors | AIME I, 1986 | Question 8

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




Check the Answer

Answer: is 141.

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



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)


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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