# Thread: Least possible value of a+b

1. ## Least possible value of a+b

Find the least pssible value of $a + b$, where $a, b$ are positive integers such that $11$ divides $a + 13b$ and $13$ divides $a + 11b$.

2. Originally Posted by alexmahone
Find the least possible value of $a + b$, where $a, b$ are positive integers such that $11$ divides $a + 13b$ and $13$ divides $a + 11b$.
I make the answer 28 (a=23, b=5), but I don't have a neat argument to prove that it's minimal.

3. Originally Posted by Opalg
I make the answer 28 (a=23, b=5), but I don't have a neat argument to prove that it's minimal.
here's a proof:

$a+13b \equiv 0 \mod 11$ and $a+11b \equiv 0 \mod 13$ are equivalent to say that $a+2b=11k$ and $a-2b=13 \ell,$ for some integers $k, \ell.$ clearly we must have: $k > 0.$ we also get:

$2a=11k + 13 \ell, \ \ 4b=11k - 13 \ell \equiv -k - \ell \mod 4 .$ so: $k + \ell = 4m.$ hence: $\ell=4m - k,$ which gives us: $a=26m - k, \ b=6k - 13m.$ clearly $a>0, \ k > 0,$ implies $m > 0.$

if $m \geq 2,$ then since $b > 0,$ we will have: $k > \frac{13m}{6} > 4.$ thus: $a+b=13m + 5k > 13 \times 2 + 5 \times 4=46 > 28.$

if $m=1,$ then $a+b=5k+13.$ also from $b > 0,$ we get: $k > \frac{13}{6} > 2.$ thus the minimum possible value of $k$ is $3.$ therefore: $a=26 - k = 23, \ b = 6k - 13 = 5, \ a+b=28. \ \ \ \Box$