L.C.M. of 3,4,5,6 = 60

Number is of the form = $60 k_1 + 2$ -------------(i)

When divided by 11, it leaves 0 remainder so number will also be of the form = $11 k_2$ ---------(ii)

Hence equating (i) and (ii), we get,

$60k_1 + 2 = 11k_2$

$60k_1 - 11k_2 = -2$ or $11k_2 - 60k_1 = 2$ -----------(iii)

It means $60k_1$ will leave remainder 9 when divide by 11.

Lets consider values for 60k1, if k1=1, 60k1=60, reminder is 60mod11=5

120mod11 will be 5+5=10, 180mod11 will be 5+5+5=15, since 15>11, reminder will be 15-11=4

240mod11 reminder will be 4+5=9

$\therefore$ By remainder root $\frac{4 k_1}{11}$ should leave remainder as 9 or -2

=> Possible values of $K_1 = 4, 15, 26, 37, 48, 59$ (As 11 and 60 are co-prime)

$\therefore$ Required value = $60 \times 59 + 2 = 3540 + 2 = 3542$

Alternatively,
L.C.M. of 3,4,5,6 = 60
As the number 60k+2 is divisible by 11, 60k leaves a reminder of 9
60mod11=5, 120mod11=10, 180mod11=4, 240mod11=9
Hence the first number where both conditions are satisfied as 242.
As 60 and 11 are co-prime, the next number where this is true is 242+60*11
Hence, the numbers are in the form 242+660k
For 6th number, k=5 => 3300+242=3542

An ascending series of numbers satisfies the following conditions: i. When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2.Ii. When divided by 11, the numbers leave no remainder. The 6th number in this series will be:

An ascending series of numbers satisfies the following conditions:

i. When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2.
Ii. When divided by 11, the numbers leave no remainder.

The 6th number in this series will be: