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Post Posted: Dec 04, 2006, 2:25 am • # 1 


If a and b are natural numbers such that a+13b is divisible by 11 and a+11b is divisible by 13, then find the least possible value of a+b.
 
 
Post Posted: Dec 04, 2006, 3:12 am • # 2 


First of all we are speaking about positive natural numbers a and b right?
if so then I think the minimum is achieved by a=23 and b=5.
Actually we have a+2b is divisible by 11 and a-2b is divisible by 13.
if a-2b=0 then a=2b and 3b is divisible by 11 so b\geq 11and a+b is at least 33.if a-2b=13 then a+2b=13+4b is divisible by 11 so from here b\geq 5 and a+b is at least 28. And the last case if a - 2b\geq 26 is easy now a\geq 28
So the minimal sum is 28.
 
 
Post Posted: Dec 04, 2006, 4:02 am • # 3 


11|(a+13b) \iff 11|(a+24b)
13|(a+11b) \iff 13|(a+24b)
Thus 143|(a+24b)
a,b>0 \Rightarrow a+24b\geq 143
Therefore a+b=(a+24b)-23b\geq 143-23b\geq 143-23(5) with equality at b=5, a=23

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