Question

A number which when divided by $10$ leaves a remainder of $9$, when divided by $9$ leaves remainder of $8$, and when divided by $8$ leaves a remainder of $7$, is

• A. $1539$
• B. $5139$
• C. $2519$
• D. $9413$

Answer 1

The correct option is C $2519$

Since the remainders and divisors differ by $1$ in each case:

So, we find the LCM of $(10-1),(9-1),(8-1),...(2-1)$

It just means to find the LCM of $9,8,7,6,5,4,3,2,1$

$9=3^2....8=2^3...7=\mathrm{7...6}=2\times \mathrm{3.....5}=\mathrm{5....4}=2^2....3=\mathrm{3...2}=\mathrm{2...1}=1$

Therefore, LCM $=3^2\times 2^3\times 7\times 5=2520$

Thus, the required number $=2520-1=2519$

Hence, $2519$ is the number when divided by $10$ leaves a remainder $9$, when divided by $9$ leaves remainder $8$, when divided by $8$ leaves remainder $7$ and so on.