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.

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

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

Google