Question

What is the least number that must be added to 2,345 to make it exactly divisible by (i) 5, (ii) 6 and (iii) 7?​

Answer 1

Answer:

${\huge {\green {\boxed {\mathfrak {Answer}}}}}$

$\mathbb { A number/ to /be /exactly /divisible /by /5, /6/ and /7, /must/ be /divisible/ by/ 210 /(i.e. /5 /× /6 /× /7)/, i.e/}$ $\large{\frac{2345}{210}}$.

Thus the remainder, 35 is the least number which must be added to 2345 to be divided by 5, 6 and 7.