why we will n"t increase the min value of d.
find the least positive value of a+b,where a,b are positive integers and such that 11divides a+13b and 13 divides a+11b
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5 Answers
let a+13b = 11k (k belongs to N)
or a+2b= 11(k-b) = 11c
let a+11b=13l (l belongs to n)
or a-2b = 13(l-b) = 13 d
now a-2b < a+2b (since a,b are +ve)
hence 13d < 11c
hence min. (d)=0
min (c)=1
therefore a-2b=0
and a +2b =11 or a=5.5 (not possible)
hence let us take the next case ,
d=0
c=2
there we get b=5.5(again not possible)
similarly going on we find that for c=4 and d=0 ,
a=22 and b=11
hence min (a+b) =33
I will give you some steps , I think you can complete it .
1 . 13 divides " a - 2 b " .
2 . 13 divides " 6 a + b " .
3 . 11 divides " a + 2 b " .
4 . 11 divides " 6 a + b " .
5 . gcd ( 11 , 13 ) = 1 .
6 . 6 a + b = 143 n . . . . . . . . ( n E N ) .
7 . n + b ≥ 6 .
8 . " a = 23 , b = 5 " works .
By the way , this is an old RMO problem from what I can gather from my memory .