Let $m$ and $n$ be positive integers such that 11 divides $m+13n$ and 13 divides $m+11n$ .
What is the minimum value of $m+n$ ?
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Since $13$ divides $6\left( m+11n \right)=\left( 6m+n \right)+65n$ , $13$ divides $6m+n$ .
Since $11$ divides $6\left( m+13n \right)=6m+n+77n$ , $11$ also divides $6m+n$ .
Hence $11\times 13=143$ divides $6m+n$ , so that $6m+n=143k$ for some positive integer $k$ .
Since $6\left( m+n \right)=143k+5n=6\left( 24k+n \right)\left( k+n \right)$ , $6$ divides $k+n$ so that $k+n\ge 6$ .
Now $6\left( m+n \right)=143k+5n=138k+5\left( k+n \right)\ge 168$ .
Consequently $m+n\ge 28$ , and this is attained if $m=23$ and $n=5$ .
Will u please explain me hoe the instinct of multiplying 6 came to your mind...............was this by practice or its was a kind of hidden hint in the question
And what if i multiply 13n+m by 11n+m and get 0 mod 143 ,and realize that this implies that 24mn+m^2.is equal to 0 mod 143 and so m(24n+m) equal to 0 mod 143 so n equals to 5 and m equal to 23 is the least value?
$m + 13n \equiv 0$ (mod 11)
$m + 11n \equiv 0$ (mod 13)
Starting with the second equation we can deduce that:
$m + 11n \equiv 0$ (mod 13)
$m \equiv 2n$ (mod 13)
$m = 13k + 2n$
This can then substituted into the first equation as follows:
$13k + 2n + 13n \equiv 0$ (mod 11)
$13k + 15n \equiv 0$ (mod 11)
$2k + 4n \equiv 0$ (mod 11)
$k + 2n \equiv 0$ (mod 11)
$9n \equiv k$ (mod 11)
$n \equiv 5k$ (mod 11)
$n = 11j + 5k$
Now we can rewrite the sum $m + n$ as :
$28k + 33j$
Now since both $m$ and $n$ are both positive the sum must be more than 0. Hence setting $k = 1$ and $j = 0$ minimizes the sum to $\boxed{28}$
Why are j and k positive?
11p=m+13n
13q=m+11n
can be written as,
11y=m+2n
13z=m2n
solving,
11y+13z=2m
11y13z=4n
so by substituting for y,z we can know that either y,z both are even or odd. And 11y>13z.
first, possible values for y and z are y=3,z=1 for which m=23 and n=5.
And the next possible values for y and z are y=5,z=3 for which m=47 and n=4 and so on but greater than 28.
so the least possible sum can be 28.
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$m+13n \equiv 0 \pmod {11}$ , $m+11n \equiv 0 \pmod {13}$ . (We add $11n$ and $13n$ to the first and second equations respectively to get)
$m+13n+11n = m+24n \equiv 11n \equiv 0 \pmod{11}$ , $m+13n+11n = m+24n \equiv 13n \equiv 0 \pmod{13}$
Since $m+24n \equiv 0 \pmod {11,13}$ , we can deduce that $m+24n \equiv 0 \pmod {143}$ .
Therefore $m+24n = 143k$ for some integer $k$ .
But since $m,n$ are positive, then to minimize $m+n$ , $k$ must be positive but as small as possible, and the smallest positive integer is $k=1$
Therefore, $m+24n = 143(1) = 143$ . We then maximize $n$ such that $m = 14324n$ is still positive.
The largest satisfying $n$ is $5$ , and the corresponding $m$ is $23$ , so our answer is $23+5 = \boxed{28}$ .