Least Possible Value of a+b: 11 & 13 Divisibility

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The least possible value of a + b, where a and b are positive integers, is determined under the conditions that 11 divides (a + 13b) and 13 divides (a + 11b). The solution involves solving the simultaneous congruences derived from these divisibility conditions. Albert provided the final answer, confirming the calculations and ensuring the values of a and b meet the specified criteria.

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find the least possible value of a + b where a and b are positive integers and 11 divides $a+ 13b$
and 13 divides $a + 11 b$
 
Last edited:
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kaliprasad said:
find the least possible value of a + b where a and b are positive and 11 divides $a+ 13b$
and 13 divides $a + 11 b$
the least possible value of a + b=14
a=11.5,b=2.5
 
Last edited:
Albert said:
the least possible value of a + b=14
a=11.5,b=2.5

sorry I meant integers
 
kaliprasad said:
sorry I meant integers
if $a,b\in N$
$min(a+b)=28, a=23,b=5$
 
Albert has provided the answer
here is the solution

$11$ divides $a + 2b$ and hence $11$ divides $6a + 12b$ or $11$ divides $6a + b$
$13$ divides $a - 2b$ and hence $13$ divides $6a - 12b$ or $13$ divides $6a + b$

so $6a + b$ is divisible by $11$ and $13$ and hence $143$
say $6a +b = 143 t\cdots(1)$
$6a + 6b = 143t + 5b = 144 t + 6b - ((t+b)$
So $t + b$ is divisible by $6$ and hence $t + b > 6 \cdots(2)$
$(a+b) = 143t + 5b = 138 t + 5(t+b) >=168$
Hence $a + b >= 28$
From (2) putting $t = 1$ we get $b= 5$ and from (1) we get $a = 23$ so $a=23$ and $b=5$ satisfies
the condition so $a+b$ lowest value is 28
 

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