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Suppose that a and b are positive integers such that $a-b=6$ and $\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9$. Find the smallest possible value of b.

 Oct 2, 2019
 #1
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Can you write your question in "plaintext"? I can't read it !.

 Oct 2, 2019
 #2
avatar+2361 
+1

\(a-b=6\)

 

\(\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\)

 

\(\frac{(a+b)(a+b)^2}{(a+b)}\)

 

\((a+b)^2\)

 

 

So now it is \(\text{gcd}\left((a+b)^2, ab\right) = 9\)

 

So now we can plug in a - b = 6

 

a = 6 + b

 

\(\text{gcd}((2b+6)^2,b^2+6b)\)

 

Ok notice how b^2 + 6b must be divisible by 9, so we just keep guess and checking until we hit our first integer that is divisible by nine for b.

 

1 = 7

 

2 = 16

 

3 = 27

 

So b=3 works for b^2 + 6b,

 

now we test for (2b+6)^2

 

We get 144, which is divisible by 9.

 

So the answer should be 3.

 

 

 

 

My solution is very bogus, it may be wrong, or it may be right! But most likely wrong. Can someone check?

.
 Oct 2, 2019
edited by CalculatorUser  Oct 3, 2019
 #3
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a=1; b=1;c= gcd((a^3 + b^3)/(a + b), a * b); if(c==9, goto4, goto5);printc, a, b; a++;if(a<100, goto2, 0);a=1;b++;if(b<100, goto2, discard=0;

 

a = 9    and   b = 3

 Oct 3, 2019
 #4
avatar+2361 
0

Nice job guest! Cool how you can easily program something that can solve problems like this.

CalculatorUser  Oct 3, 2019

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