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# Algebra and Positive Integer | AIME I, 1987 | Question 8 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

## Algebra and Positive Integer - AIME I, 1987

What is the largest positive integer n for which there is a unique integer k such that $\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}$?

• is 107
• is 112
• is 840
• cannot be determined from the given information

### Key Concepts

Digits

Algebra

Numbers

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

## Try with Hints

First hint

Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

Second Hint

for 91n-104k<n+k, K>$\frac{6n}{7}$

and 0<91n-104k gives k<$\frac{7n}{8}$

Final Step

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

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Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

## Algebra and Positive Integer - AIME I, 1987

What is the largest positive integer n for which there is a unique integer k such that $\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}$?

• is 107
• is 112
• is 840
• cannot be determined from the given information

### Key Concepts

Digits

Algebra

Numbers

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

## Try with Hints

First hint

Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

Second Hint

for 91n-104k<n+k, K>$\frac{6n}{7}$

and 0<91n-104k gives k<$\frac{7n}{8}$

Final Step

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

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