Question

Find the smallest and the largest three-digit numbers which when divided by $22,33$ and $55$ leave a remainder of $5$ in each case.

• A. $340,980$
• B. $335,995$
• C. $330,990$
• D. $325,985$

Answer 1

Answer: (B)

$335,995$

Prime factorisation of $22=2\times 11$
Prime factorisation of $33=3\times 11$
Prime factorisation of $55=5\times 11$
So, LCM $22,33,55=2\times 3\times 5\times 11=330$
As $330$ is the smallest $3-$ digit number divisible by $22,33,55$, the number $330+5=335$ will give a remainder $5$ when divided by these numbers.
The highest $3-$ digit number is $999$.
$999$ when divived by $330$ gives a remainder $9$, so $999-9=990$ is the largest $3-$ digit number divisible by $22,33,55$, the number $990+5=995$ will give a remainder $5$ when divided by these numbers.