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The sum of two positive integers $a$ and $b$ is 1001. What is the largest possible value of $\gcd(a,b)$?

Mellie  Jul 1, 2015

Best Answer 

 #1
avatar+78557 
+20

Note that 1001 factors as 7 x 11 x 13 = 7 * 143

 

So.....143 is the greatest proper divisor of 1001

 

This means that we have  143( m + n)  = 143m + 143m = a + b = 1001  .....where ( m + n) = 7  ....and since m + n are two positive integers whose sum = 7, they are relatively prime to each other

 

Thefore, the two integers 143m, 143n  = a, b  have only the factor of 143 in common.....and it is the gcd of both.

 

{Could some other mathematician check my logic, here ??? }

 

 

CPhill  Jul 2, 2015
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3+0 Answers

 #1
avatar+78557 
+20
Best Answer

Note that 1001 factors as 7 x 11 x 13 = 7 * 143

 

So.....143 is the greatest proper divisor of 1001

 

This means that we have  143( m + n)  = 143m + 143m = a + b = 1001  .....where ( m + n) = 7  ....and since m + n are two positive integers whose sum = 7, they are relatively prime to each other

 

Thefore, the two integers 143m, 143n  = a, b  have only the factor of 143 in common.....and it is the gcd of both.

 

{Could some other mathematician check my logic, here ??? }

 

 

CPhill  Jul 2, 2015
 #2
avatar+26322 
+15

You are correct Chris.  

In fact there are 3 pairs of integers that sum to 1001 and and have the largest gcd of 143, namely: (143, 858), (286, 715), (429, 752)   (All multiples of 143).

 

.

Alan  Jul 2, 2015
 #3
avatar+90970 
+10

That is great logic Chris.  

 

It is so simple and so logical but I would not have thought of it   

Melody  Jul 2, 2015

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