Question

Find the least number which when divided by $2,3,4,5$ and $6$ leaves $1,2,3,4$ and $5$ as remainders respectively but when divided by $7$ leaves no remainder.

• A. $210$
• B. $119$
• C. $126$
• D. $154$

Answer 1

The correct option is B $119$

Common difference between divisors and respective remainders $=(2-1)=(3-2)=(4-3)=(5-4)=(6-5)=1$
LCM of $(2,3,4,5,6)=60$

$\therefore$ Required number $=60k-1$
Now we have to find the least value of $k$ for which $60k-1$ is divisible by $7$

By inspection, we find that for $k=2,4\times 2-1=7$
$\therefore$ Required number $=60\times 2-1=120-1=119$