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# Finding Greatest Integer | AMC 10A, 2018 | Problem No 14

Try this beautiful Problem on Algebra based on finding greatest integer from AMC 10 A, 2018. You may use sequential hints to solve the problem.

## Finding Greatest Integer - AMC-10A, 2018- Problem 14

What is the greatest integer less than or equal to $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

• $80$
• $81$
• $96$
• $97$
• $625$

Algebra

greatest integer

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-14

#### Check the answer here, but try the problem first

$80$

## Try with Hints

#### First Hint

The given expression is $\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$

We have to find out the greatest integer which is less than or equal to the given expression .

Let us assaume that $x=3^{96}$ and $y=2^{96}$

Therefore the given expression becoms $\frac{81 x+16 y}{x+y}$

Now can you finish the problem?

#### Second Hint

Now $\frac{81 x+16 y}{x+y}$

=$\frac{16 x+16 y}{x+y}+\frac{65 x}{x+y}$

$=16+\frac{65 x}{x+y}$

Now if we look very carefully we see that $\frac{65 x}{x+y}<\frac{65 x}{x}=65$

Therefore $16+\frac{65 x}{x+y}<16+65=81$

Now Can you finish the Problem?

#### Third Hint

Therefore less than $$81$$ , the answer will be $$80$$

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### One comment on “Finding Greatest Integer | AMC 10A, 2018 | Problem No 14”

1. Can you explain to me :
$\frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?$
These symbols are All Greek to me!