# Find the smallest integer that produces remainder of 2,4,6&1 when it is divided by 3,5,7&11 respectively?

##### 1 Answer
May 2, 2017

$419$

#### Explanation:

Call the number $n$.

• In order to give a remainder of $2$ when divided by $3$, $n$ must be one less than a multiple of $3$.

• In order to give a remainder of $4$ when divided by $5$, $n$ must be one less than a multiple of $5$.

• In order to give a remainder of $6$ when divided by $7$, $n$ must be one less than a multiple of $7$.

So $n$ is one less than a multiple of the LCM of $3$, $5$ and $7$, namely:

$3 \cdot 5 \cdot 7 = 105$

So:

$n = 105 k - 1 \text{ }$ for some integer $k$.

In order to give a remainder of $1$ when divided by $11$, $n$ must be of the form $11 m + 1$ for some integer $m$.

So we have:

$11 m + 1 = 105 k - 1$

That is:

$105 k = 11 m + 2$

Now:

$\frac{105}{11} = 9 \text{ }$ with remainder $6$

So what are the multiples of $6$ modulo $11$?

$1 \cdot 6 = 6 = 0 \cdot 11 + 6$

$2 \cdot 6 = 12 = 1 \cdot 11 + 1$

$3 \cdot 6 = 18 = 1 \cdot 11 + 7$

$4 \cdot 6 = 24 = 2 \cdot 11 + 2$

So the smallest possible positive value of $k$ which gives us the required remainder is $k = 4$

So:

$n = 105 k - 1 = 105 \cdot 4 - 1 = 419$