Mixing
Trying to compute the greatest value of $x$.
$begingroup$
$a,b,c,x,k$ are positive natural numbers.
$$86 = ax+k$$
$$142 = bx + k$$
$$252 = cx+k$$
I'm trying to compute the greatest value of $x$.
Let's assume $ k = 1$ (we want x to take its greatest value), we have that
$$85 = ax$$
$$141 = bx$$
$$251 = cx$$
However, this makes no literal sense. Could you assist me?
Regards
greatest-common-divisor
asked Jan 31 at 18:53
EnzoEnzo
30617
$endgroup$
add a comment |
$begingroup$
$a,b,c,x,k$ are positive natural numbers.
$$86 = ax+k$$
$$142 = bx + k$$
$$252 = cx+k$$
I'm trying to compute the greatest value of $x$.
Let's assume $ k = 1$ (we want x to take its greatest value), we have that
$$85 = ax$$
$$141 = bx$$
$$251 = cx$$
However, this makes no literal sense. Could you assist me?
Regards
greatest-common-divisor
asked Jan 31 at 18:53
EnzoEnzo
30617
$endgroup$
add a comment |
$begingroup$
$a,b,c,x,k$ are positive natural numbers.
$$86 = ax+k$$
$$142 = bx + k$$
$$252 = cx+k$$
I'm trying to compute the greatest value of $x$.
Let's assume $ k = 1$ (we want x to take its greatest value), we have that
$$85 = ax$$
$$141 = bx$$
$$251 = cx$$
However, this makes no literal sense. Could you assist me?
Regards
greatest-common-divisor
asked Jan 31 at 18:53
EnzoEnzo
30617
$endgroup$
$a,b,c,x,k$ are positive natural numbers.
$$86 = ax+k$$
$$142 = bx + k$$
$$252 = cx+k$$
I'm trying to compute the greatest value of $x$.
Let's assume $ k = 1$ (we want x to take its greatest value), we have that
$$85 = ax$$
$$141 = bx$$
$$251 = cx$$
However, this makes no literal sense. Could you assist me?
Regards
greatest-common-divisor
greatest-common-divisor
asked Jan 31 at 18:53
EnzoEnzo
30617
asked Jan 31 at 18:53
EnzoEnzo
30617
asked Jan 31 at 18:53
EnzoEnzo
30617
asked Jan 31 at 18:53
EnzoEnzo
30617
asked Jan 31 at 18:53
EnzoEnzo
30617
30617
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First note that $x$ cannot be any larger than gcd$(142-86, 252-142)$ $=$ gcd$(56,110)$ $=2$.
Yet 2 works: $a=42,b=70,c=125, k=2$.
answered Jan 31 at 18:58
MikeMike
4,621512
$endgroup$
$begingroup$
Can you be more clear about how you got these?
$endgroup$
– Enzo
Jan 31 at 19:36
1
$begingroup$
@Enzo If $x$ divides $86-k$ and also divides $142-k,$ it must divide $(142-k)-(86-k)$ too.
$endgroup$
– David K
Jan 31 at 19:42
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First note that $x$ cannot be any larger than gcd$(142-86, 252-142)$ $=$ gcd$(56,110)$ $=2$.
Yet 2 works: $a=42,b=70,c=125, k=2$.
answered Jan 31 at 18:58
MikeMike
4,621512
$endgroup$
$begingroup$
Can you be more clear about how you got these?
$endgroup$
– Enzo
Jan 31 at 19:36
1
$begingroup$
@Enzo If $x$ divides $86-k$ and also divides $142-k,$ it must divide $(142-k)-(86-k)$ too.
$endgroup$
– David K
Jan 31 at 19:42
add a comment |
$begingroup$
First note that $x$ cannot be any larger than gcd$(142-86, 252-142)$ $=$ gcd$(56,110)$ $=2$.
Yet 2 works: $a=42,b=70,c=125, k=2$.
answered Jan 31 at 18:58
MikeMike
4,621512
$endgroup$
$begingroup$
Can you be more clear about how you got these?
$endgroup$
– Enzo
Jan 31 at 19:36
1
$begingroup$
@Enzo If $x$ divides $86-k$ and also divides $142-k,$ it must divide $(142-k)-(86-k)$ too.
$endgroup$
– David K
Jan 31 at 19:42
add a comment |
$begingroup$
First note that $x$ cannot be any larger than gcd$(142-86, 252-142)$ $=$ gcd$(56,110)$ $=2$.
Yet 2 works: $a=42,b=70,c=125, k=2$.
answered Jan 31 at 18:58
MikeMike
4,621512
$endgroup$
First note that $x$ cannot be any larger than gcd$(142-86, 252-142)$ $=$ gcd$(56,110)$ $=2$.
Yet 2 works: $a=42,b=70,c=125, k=2$.
answered Jan 31 at 18:58
MikeMike
4,621512
edited Jan 31 at 18:59
answered Jan 31 at 18:58
MikeMike
4,621512
answered Jan 31 at 18:58
MikeMike
4,621512
answered Jan 31 at 18:58
MikeMike
4,621512
4,621512
$begingroup$
Can you be more clear about how you got these?
$endgroup$
– Enzo
Jan 31 at 19:36
1
$begingroup$
@Enzo If $x$ divides $86-k$ and also divides $142-k,$ it must divide $(142-k)-(86-k)$ too.
$endgroup$
– David K
Jan 31 at 19:42
add a comment |
$begingroup$
Can you be more clear about how you got these?
$endgroup$
– Enzo
Jan 31 at 19:36
1
$begingroup$
@Enzo If $x$ divides $86-k$ and also divides $142-k,$ it must divide $(142-k)-(86-k)$ too.
$endgroup$
– David K
Jan 31 at 19:42
$begingroup$
Can you be more clear about how you got these?
$endgroup$
– Enzo
Jan 31 at 19:36
$begingroup$
Can you be more clear about how you got these?
$endgroup$
– Enzo
Jan 31 at 19:36
1
1
$begingroup$
@Enzo If $x$ divides $86-k$ and also divides $142-k,$ it must divide $(142-k)-(86-k)$ too.
$endgroup$
– David K
Jan 31 at 19:42
$begingroup$
@Enzo If $x$ divides $86-k$ and also divides $142-k,$ it must divide $(142-k)-(86-k)$ too.
$endgroup$
– David K
Jan 31 at 19:42
add a comment |
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