INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More

# 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.

## 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

Integers

Equations

Algebra

## Check the Answer

AIME I, 1990, Question 15

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

Let S=x+y, P=xy

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

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

Second Hint

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

Final Step

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

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.

### Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
HALL OF FAMESUPER STARSBOSE OLYMPIADBLOG
CAREERTEAM
support@cheenta.com