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## Positive divisor | AIME I, 1988 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Positive divisor.

## Positive Divisors- AIME I, 1988

Let $\frac{m}{n}$ in lowest term, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$. Find m+n.

• is 107
• is 634
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

DIvisors

Algebra

AIME I, 1988, Question 5

Elementary Number Theory by David Burton

## Try with Hints

$10^{99}=2^{99}5^{99}$

or, (99+1)(99+1)=10000 factors

those factors divisible by $10^{88}$

are bijection to number of factors $10{11}=2^{11}5^{11}$ has, which is (11+1)(11+1)=144

one required probability =$\frac{m}{n}=\frac{144}{10000}=\frac{9}{625}$

m+n=634.

Categories

## 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

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

$log1+log2+log4+….+log1000000$

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

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

for proper divisor taking out $2^{6}5^{6}$=147-6=141

or, $S=log2^{141}5^{141}=log10^{141}=141$.