Question

Find the smallest and the largest four-digit numbers which when lessened by $12$ are exactly divisible by $16,24and40$.

• A. $1208,9848$
• B. $1200,9840$
• C. $1212,9852$
• D. $1188,9828$

Answer 1

The correct option is C $1212,9852$

Prime factorisation of $12=2^2\times 3$
Prime factorisation of $16=2^4$
Prime factorisation of $24=2^3\times 3$
Prime factorisation of $40=2^3\times 5$
So, LCM of $12,16,24,40=2^4\times 3\times 5=240$
Multiples of $240$ are also multiples of all of these given numbers
So, $240,480,720,960,\mathrm{1200...}$ are all multiples of $12,16,24,40$
Of these, as $1200$ is the smallest $4-$ digit number divisible by $12,16,24,40$, the number $1200+12=1212$ will be exactly divisible by these numbers when lessened by $12$
The highest $4-$ digit number is $9999$.
$9999$ when divided by $240$ gives a remainder $159$, so $9999-159=9840$ is the largest $4-$ digit number divisible by $12,16,24,40$,
And the number $9840+12=9852$ will be exactly divisible by these numbers when lessened by $12$