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# Problem on Real Numbers | AIME I, 1990| Question 15

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on real numbers. Use sequential hints if required.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Real Numbers.

## Problem on Real Numbers – AIME I, 1990

Find $ax^{5}+by^{5}$ if real numbers a,b,x,y satisfy the equations

ax+by=3

$ax^{2}+by^{2}=7$

$ax^{3}+by^{3}=16$

$ax^{4}+by^{4}=42$

• is 107
• is 20
• is 634
• cannot be determined from the given information

### Key Concepts

Integers

Equations

Algebra

AIME I, 1990, Question 15

Elementary Algebra by Hall and Knight

## Try with Hints

Let S=x+y, P=xy

$(ax^{n}+by^{n})(x+y)$

$=(ax^{n+1}+by^{n+1})+(xy)(ax^{n-1}+by^{n-1})$

or,$(ax^{2}+by^{2})(x+y)=(ax^{3}+by^{3})+(xy)(ax+by)$ which is first equation

or,$(ax^{3}+by^{3})(x+y)=(ax^{4}+by^{4})+(xy)(ax^{2}+by^{2})$ which is second equation

or, 7S=16+3P

16S=42+7P

or, S=-14, P=-38

or, $(ax^{4}+by^{4})(x+y)=(ax^{5}+by^{5})+(xy)(ax^{3}+by^{3})$

or, $42S=(ax^{5}+by^{5})+P(16)$

or, $42(-14)=(ax^{5}+by^{5})+(-38)(16)$

or, $ax^{5}+by^{5}=20$.