Question

The least number which when increased by $4$ is divisible by each of the numbers $10,15,20and25$ is

• A. $296$
• B. $300$
• C. $304$
• D. $308$

Answer 1

The correct option is A $296$

Prime factorization of numbers
$10=2\times 5$
$15=3\times 5$
$20=2\times 2\times 5$
$25=5\times 5$
Smallest number divisble by above $4$ numbers will be LCM = $2^2\times 3\times 5^2$ = $300$
Now , the number we need is $4$ less than $300$, therefore the number is $300-4$ = $296$