# 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

Integers

Divisors

Algebra

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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