Suppose x is the smallest integer greater than 3 such that when it is divided by 6 or 8, the remainder is 3. y is the smallest integer greater than 2 such that when it is divided by 6 or 8, the remainder is 2. What is x – y ?
x is the smallest integer greater than 3.
When x is divided by 6, the remainder is 3. This can be written as $x \equiv 3 \pmod 6$, or $x – 3$ is divisible by 6.
When x is divided by 8, the remainder is 3. This can be written as $x \equiv 3 \pmod 8$, or $x – 3$ is divisible by 8.
So, $x – 3$ is a common multiple of 6 and 8.
To find the smallest such integer x, $x-3$ must be the smallest positive common multiple of 6 and 8. This is the Least Common Multiple (LCM) of 6 and 8.
Prime factorization of 6 = $2 \times 3$.
Prime factorization of 8 = $2^3$.
LCM(6, 8) = $2^3 \times 3 = 8 \times 3 = 24$.
So, $x – 3$ must be a multiple of 24. $x – 3 = 24k$ for some integer k.
$x = 24k + 3$.
We need the smallest integer x greater than 3.
If k=0, $x = 24(0) + 3 = 3$. This is not greater than 3.
If k=1, $x = 24(1) + 3 = 27$. This is greater than 3 and is the smallest value of x satisfying the condition. So, $x=27$.
For y:
y is the smallest integer greater than 2.
When y is divided by 6, the remainder is 2. This can be written as $y \equiv 2 \pmod 6$, or $y – 2$ is divisible by 6.
When y is divided by 8, the remainder is 2. This can be written as $y \equiv 2 \pmod 8$, or $y – 2$ is divisible by 8.
So, $y – 2$ is a common multiple of 6 and 8.
To find the smallest such integer y, $y-2$ must be the smallest positive common multiple of 6 and 8. This is the LCM(6, 8).
LCM(6, 8) = 24.
So, $y – 2$ must be a multiple of 24. $y – 2 = 24j$ for some integer j.
$y = 24j + 2$.
We need the smallest integer y greater than 2.
If j=0, $y = 24(0) + 2 = 2$. This is not greater than 2.
If j=1, $y = 24(1) + 2 = 26$. This is greater than 2 and is the smallest value of y satisfying the condition. So, $y=26$.
The question asks for the value of $x – y$.
$x – y = 27 – 26 = 1$.
– If a number gives the same remainder when divided by several numbers, then the number minus the remainder is divisible by the LCM of those numbers.
– Finding the smallest integer greater than a given value that satisfies the property.