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

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The discussion focuses on finding the least possible value of a + b, where a and b are positive integers that satisfy the conditions of divisibility by 11 and 13. The key equations derived from the problem are that 11 divides a + 13b and 13 divides a + 11b. Participants explore various integer combinations to meet these criteria. Albert provides a solution that identifies the minimum value of a + b. The final answer is confirmed through calculations and logical reasoning.
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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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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