Question

Find the smallest number which when divided by 3 5 7 9 11 leaves a remainder 2 4 6 8 10

Answer 1

Let the number be N.

N leaves remainder 2, 4, 6, 8 and 10 on division by 3, 5, 7, 9 and 11 respectively.

⇒  As N leaves remainder 2 on division by 3, let  N = 3a + 2.

⇒  As N leaves remainder 4 on division by 5, let  N = 5b + 4.

⇒  As N leaves remainder 6 on division by 7, let  N = 7c + 6.

⇒  As N leaves remainder 8 on division by 9, let  N = 9d + 8.

⇒  As N leaves remainder 10 on division by 11, let  N = 11e + 10.

So,

N  =  3a + 2  =  5b + 4  =  7c + 6  =  9d + 8  =  11e + 10

Adding 1 to each,

    N + 1  =  3a + 2 + 1  =  5b + 4 + 1  =  7c + 6 + 1  =  9d + 8 + 1  =  11e + 10 + 1

⇒  N + 1  =  3a + 3  =  5b + 5  =  7c + 7  =  9d + 9  =  11e + 11

⇒  N + 1  =  3(a + 1)  =  5(b + 1)  =  7(c + 1)  =  9(d + 1)  =  11(e + 1)

Here it seems that  N + 1  is divisible by 3, 5, 7, 9 and 11. Hence the smallest possible value for  N + 1  will be the LCM of 3, 5, 7, 9 and 11.

$\begin{tabular}{c|l}3&3,\ 5,\ 7,\ 9,\ 11\\ \cline{2-}&1,\ 5,\ 7,\ 3,\ 11\\ \cline{2-}\end{tabular}$

∴   N + 1 = 3 × 5 × 7 × 3 × 11

⇒  N + 1 = 3465

⇒  N = 3465 - 1

⇒  N = 3464

Hence 3464 is the answer.