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 \u2013 y ?<\/p>\n
[amp_mcq option1=”1″ option2=”2″ option3=”3″ option4=”4″ correct=”option1″]<\/p>\n
For y:
\ny is the smallest integer greater than 2.
\nWhen 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.
\nWhen 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.
\nSo, $y – 2$ is a common multiple of 6 and 8.
\nTo 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).
\nLCM(6, 8) = 24.
\nSo, $y – 2$ must be a multiple of 24. $y – 2 = 24j$ for some integer j.
\n$y = 24j + 2$.
\nWe need the smallest integer y greater than 2.
\nIf j=0, $y = 24(0) + 2 = 2$. This is not greater than 2.
\nIf j=1, $y = 24(1) + 2 = 26$. This is greater than 2 and is the smallest value of y satisfying the condition. So, $y=26$.<\/p>\n