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avatar+129 

When the numbers $\sqrt{2}, \sqrt[3]{3},$ and $\sqrt[5]{5}$ are listed in order from least to greatest, which number is in the middle?

 

Suppose $a$ and $b$ are positive integers and $\dfrac{a}{4}+\dfrac{b}{3}=\dfrac{13}{12}$. What is the value of $a^2+b^2$?

 

What is the largest perfect square less than 12345654320?

 

How many positive integers less than 2000 are of the form $x^n$ for some positive integer $x$ and $n\ge 2?$

 

I have until next Saturday, so take your time!

 Sep 15, 2019
 #1
avatar+111328 
+1

When the numbers \($\sqrt{2}, \sqrt[3]{3},$ and $\sqrt[5]{5}$ \) are listed in order from least to greatest, which number is in the middle?

 

Compare  

 

√2   3√3 , 5√5       raise each  to the 30th power

 

(2^(1/2))^30        (3^(1/3))^30    (5^(1/5))^30

 

2^15                  3^10                  5^6

 

32768              59049                15625

 

So

 

√2    is the middle number

 

 

cool cool cool

 Sep 15, 2019
 #2
avatar+111328 
+1

Suppose a and b are positive integers and \(\dfrac{a}{4}+\dfrac{b}{3}=\dfrac{13}{12}\). What is the value of a^2+b^2?

 

a/4   + b/3  = 13/12

 

[ 3a + 4b]           13

_______   =      ___

    12                  12

 

This implies that  3a + 4b  =  13

 

If we want positive integer values.....then a = 3  and b  = 1

 

So

 

a^2 + b^2   =

 

3^2 + 1^2   =

 

9 +  1  =

 

10

 

 

cool cool cool

 Sep 15, 2019
 #3
avatar+129 
+1

Thank you so much, c-phil!

 

Ironically, just as you answered one of the questions, I figured out one of them too.

 Sep 15, 2019
 #4
avatar+111328 
0

What is the largest perfect square less than 12345654320?

 

Just take the square root of   12345654320  ≈  111110.9

 

So....the largest perfect square  =  (111110)^2 =   12345432100

 

cool cool cool

 Sep 15, 2019
 #5
avatar
0

For number four, I recommend proceeding with casework with n=2, n=3, n=4, etc.

 Sep 15, 2019

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