Question

Find the least number which when divided by $6,7,8,9$ and $12$ leaves the remainder $1$ in each case.

Answer 1

Writing the prime factorization:
$6=2\times 3$
$7=1\times 7$

$8=2\times 2\times 2$
$9=3\times 3$
$12=2\times 2\times 3$
the least common multiple therefore is:
$2\times 7\times 2\times 2\times 3\times 3=504$
So numbers which on being divided by $6,7,8,9and12$ leave a remainder $1$ are : $504+1=505$