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# Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

## Finding smallest positive Integer – AIME I, 1996

Find the smallest positive integer solution to $tan19x=\frac{cos96+sin96}{cos96-sin96}$.

• is 107
• is 159
• is 840
• cannot be determined from the given information

### Key Concepts

Functions

Trigonometry

Integers

AIME I, 1996, Question 10

Plane Trigonometry by Loney

## Try with Hints

$\frac{cos96+sin96}{cos96-sin96}$

=$\frac{sin(90+96)+sin96}{sin(90+96)-sin96}$

=$\frac{sin186+sin96}{sin186-sin96}$

=$\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}$

=$\frac{2sin141cos45}{2cos141sin45}$

=tan141

here $tan(180+\theta)$=$tan\theta$

$\Rightarrow 19x=141+180n$ for some integer n is first equation

multiplying equation with 19 gives

$x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)$ [since 2679 divided by 180 gives remainder 159]

$\Rightarrow x=159$.