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

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Discussion Overview

The discussion revolves around finding the least possible value of the sum \( a + b \) where \( a \) and \( b \) are positive integers. The conditions specified are that 11 divides \( a + 13b \) and 13 divides \( a + 11b \). The scope includes mathematical reasoning and problem-solving related to divisibility.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Participants seek to determine the least possible value of \( a + b \) under the given divisibility conditions.
  • Some participants clarify that \( a \) and \( b \) should be positive integers, correcting earlier mentions of them being positive.
  • One participant claims that another has provided a solution, although the details of that solution are not included in the posts.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the least possible value of \( a + b \), as the solution has not been detailed in the posts.

Contextual Notes

There is a lack of clarity on the specific solution provided by Albert, and the posts do not include any mathematical steps or reasoning leading to a conclusion.

kaliprasad
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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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